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292
Chapter 13
4 Find the following integr
als:
a ∫(4x3 − 3x−4 + r)dx b ∫(x + x − 1 _ 2 + x − 3 _ 2 )dx
c ∫( px4 + 2t + 3x−2)dx
5 Find the following integr
als:
a ∫(3t2 − t−2)dt b ∫(2t2 − 3 t − 3 _ 2 + 1)dt c ∫( pt3 + q2 + px3)dt
6 Find the following integr
als:
a ∫ (2x3 + 3) ________ x2 dx b ∫(2x + 3)2 dx c ∫(2x + 3) √ __
x dx
7 Find ∫f(x)dx when f(x) is given by the following:
a (x + 1 __ x ) 2
b ( √ __
x + 2)2 c ( 1 ___ √ __
x + 2 √ __
x )
8 Find the following integr
als:
a ∫ ( x 2 _ 3 + 4 __ x3 ) dx b ∫ ( 2 + x _____ x3 + 3) dx c ∫(x2 + 3)(x − 1)dx
d ∫ (2x +
1)2 ________ √ __
x dx e ∫ (3 + √ __
x + 6x3 ________ x ) dx f ∫ √ __
x ( √ __
x + 3)2 dx
9 Find the following integr
als:
a ∫ ( A __ x2 − 3) dx b ∫ ( √ ___ Px + 2 __ x3 ) dx c ∫ ( p
__ x2 + q √ __
x + r) dx
10 Given tha
t f(x) = 6 __ x 2 + 4 √ __
x − 3x + 2, x > 0, find ∫f(x)dx. (5 marks)
11 Find ∫ (8x3 + 6x − 3 ___ √ __
x ) dx, giving each term in its simplest form. (4 marks)
12 a Show that (2 + 5 √ __
x ) 2 can be written as 4 + k √ __
x + 25x, where k is a constant to be
found. (2 marks)
b Hence find ∫(2 + 5 √ __
x )2 dx. (3 marks)
13 Given tha
t y = 3 x 5 − 4 ___ √ __
x , x > 0, find ∫y dx in its simplest f orm. (3 marks)
14 ∫ ( p ___ 2x2 + pq) dx = 2 __ x + 10x + c (5 marks)
Find the value of
p and the value of q.E
E
E/P
E
Integrate the expression on the left-hand
side, treating p and q as constants, then compare the result with the right-hand side.Problem-solving E/P In Q4 part c you are
integrating with respect to x , so
treat p and t as constants.Hint | [
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293
Integration
Only one of these curves passes through
this point. Choosing a point on the curve determines the value of c . 13.3 Finding functions
You can find the constant of integration, c, when
you are given (i) any point (x, y) that the curve of the function passes through or (ii) any value that
the function takes. For example, if
dy ___ dx = 3x2 then
y = x3 + c. There are infinitely many curves with
this equation, depending on the value of c .
O(1.5, 5.375)
28
–2y = x3 – 2y = x3 + 2y = x3 + 8y
x
■ To find the constant of integration, c
• Integrate the function
• Substitute the values ( x, y) of a point on the curve, or the value of the function at a given
point f( x) = k, into the integrated function
• Solve the equation to find c15 f(x) = (2 − x)10
Given that x is small, and so terms in x3 and higher powers of x can be ignored:
a find an appro
ximation for f(x) in the
form A + Bx + Cx2 (3 marks)
b find an appro
ximation for ∫f(x)dx. (3 marks)E/P
Find the first three terms
of th
e binomial expansion of
(2 − x )10. ← Section 8.3Hint
Example 6
The curve C with equation y = f(x) passes through the point (4, 5). Given that f ′(x ) = x2 − 2 ______ √ __
x , find
the equation of C.
f ′(x) = x2 − 2 ______ √ __
x = x 3 __ 2 − 2 x − 1 __ 2
So f(x)
= x 5 __ 2 ___
5 __ 2 − 2 x 1 __ 2 ____
1 __ 2 +
c
= 2 __ 5 x 5 __ 2 − 4 x 1 __ 2 + c
But f(
4) = 5First write f ′(x) in a fo rm suitable for integration.
Use the fact that the curve passes through (4, 5).Integrate as normal and don’t forget the + c . | [
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294
Chapter 13
So 5 = 2 __ 5 × 25 − 4 × 2 + c
5 = 64 ___ 5 − 8 + c
5 = 24 ___ 5 + c
So c =
1 __ 5
So y
= 2 __ 5 x 5 __ 2 − 4 x 1 __ 2 + 1 __ 5 Remember 4 5 _ 2 = 25.
Exercise 13C
1 Find the equation of the curv e with the given derivative of y with respect to x that passes
through the given point:
a dy ___ dx = 3x2 + 2x; point (2, 10) b dy ___ dx = 4x3 + 2 __ x3 + 3; point (1, 4)
c dy ___ dx = √ __
x + 1 __ 4 x2; point (4, 11) d dy ___ dx = 3 ___ √ __
x − x; point (4, 0)
e dy ___ dx = (x + 2)2; point (1, 7) f dy ___ dx = x2 + 3 ______ √ __
x ; point (0, 1)
2 The curve C
, with equation y = f(x), passes through the point (1, 2) and f ′(x
) = 2x3 − 1 __ x2
Find the equation of
C in the form y = f(x).
3 The gradient of a particular curv
e is given by dy ___ dx = √ __
x + 3 ______ x2 . Given that the curve passes through
the point (9, 0), find an equation of
the curve.
4 The curve with equation
y = f(x) passes through the point (−1, 0). Given that
f ′(
x) = 9x2 + 4x − 3, find f(x). (5 marks)
5 dy ___ dx = 3 x − 1 _ 2 − 2x √ __
x , x > 0.
Given that y = 10 at x = 4, find y in terms of x, giving each term in its simplest form. (7 marks)
6 Given tha
t 6x + 5 x 3 _ 2 ________ √ __
x can be written in the form 6x p + 5x q,
a write down the va
lue of p and the value of q. (2 marks)
Given tha
t dy ___ dx = 6x +
5 x 3 _ 2 ________ √ __
x and that y = 100 when x = 9,
b find y in ter
ms of x, simplifying the coefficient of each term. (5 marks)E
E/P
E/PSolve for c .
Explore the solution using
GeoGe
bra.OnlineFinally write down the equation of the curve. | [
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295
Integration
7 The displacement of a particle at time
t is given by
the function f(t), where f(0) = 0.
Given that the velocity of the particle is given by f
′(
t) = 10 − 5t,
a find f(t)
b determine the displacement of the particle w
hen t = 3.
8 The height, in metres, of
an arrow fired horizontally from the top of a castle is modelled by the
function f(t), where f(0) = 35. Given that f ′(t
) = −9.8t,
a find f(t).
b determine the height of the arr
ow when t = 1.5.
c write down the height of the castle accor
ding to this model.
d estimate the time it will tak
e the arrow to hit the ground.
e state one assumption used in your ca
lculation.P
You don’t need any specific knowledge
of mechanics to answer this question. You are told that the displacement of the particle at time t is given by f( t).Problem-solving
P
1 A set of curves, where each curve passes through the origin, has eq
uations y = f1(x), y = f2(x), y = f3(x) … where f ′n(x) = fn − 1(x) and
f1(x) = x2.
a Find
f2(x), f3(x).
b Sug
gest an expression for fn(x).
2 A set o
f curves, with equations y = f1(x), y = f2(x), y = f3(x), … all pass
through the point (0, 1) and they are related by the property f
′n(x) = fn − 1(x) and f1(x) = 1. Find f2(x), f3(x), f4(x).Challenge
13.4 Definite integr als
You can calculate an integral between two limits. This is called a definite integral. A definite integral
usually produces a value whereas an indefinite integral always produces a function.
Here are the steps for integrating the function 3x2 between the limits x = 1 and x = 2.
∫ 3x2 dx = [x3]2
1
= (23) − (13)
= 8 − 1
= 72
1
Evaluate the integral at
the upper limit.Write the integral in [ ] brackets.
Write this step in ( ) brackets.
Evaluate the integral at the
lower limit.The limits of the integral are from x = 1 to x = 2. | [
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296
Chapter 13
There are three stages when you work out a definite integral:
Write the definite integral
statement with its limits, a and b.Integrate, and write the
integral in square bracketsEvaluate the definite integral
by working out f( b) − f( a).
∫ … dxb
a[ … ]b
a (…) − (…)
■ If f ′(x) is the derivative of f( x) for all
values of x in the interval [ a, b], then the
definite integral is defined as
∫ b
a f9(x)dx = [f( x)]b
a = f(b) − f( a).
Example 7
Example 8Evaluate
∫
0 1 ( x 1 _ 3 − 1) 2 dx
Given tha
t P is a constant and ∫
1 5 ( 2Px + 7)dx = 4P2, show that there are two possible values for P
and find these values. ∫ 0 1 ( x 1 __ 3 − 1) 2dx
= ∫ 0 1 ( x 2 __ 3 − 2 x 1 __ 3 + 1) dx
=
[ x 5 __ 3 ___
5 __ 3 − 2 x 4 __ 3 ___
4 __ 3 +
x
] 1
0
= [ 3 __ 5 x 5 __ 3 − 3 __ 2 x 4 __ 3 + x] 1
0
= ( 3 __ 5 − 3 __ 2 + 1) − (0 + 0 + 0)
= 1 __ 10
∫ 1 5 ( 2Px + 7)d x = [Px2 + 7x] 1 5
= (25 P
+ 35) − ( P + 7)
= 24 P + 28
24P + 28 = 4 P2
4P2 − 24 P − 28 = 0
P2 − 6 P − 7 = 0
(P + 1)( P − 7) = 0
P = − 1 or 7Divide every term by 4 to simplify.The relationship between the derivative
and the integral is called the fundamental theorem of calculus .Problem-solving
Simplify each term.For definite integrals you don’t need to include +c in your square brackets.First multiply out the bracket to put the expression in a form ready to be integrated.
You are integrating with respect to x so treat P as a constant. Find the definite integral in terms of P then it equal to 4 P
2. The fact that the question
asks for ‘two possible values’ gives you a clue that the resulting equation will be quadratic.Problem-solving | [
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297
Integration
Exercise 13D
1 Evaluate the following definite integrals:
a ∫
2 5 x 3 dx b ∫
1 3 x 4 dx
c ∫
0 4 √ __
x dx d ∫
1 3 3 __ x2 dx
2 Evalua
te the following definite integrals:
a ∫
1 2 ( 2 __ x3 + 3x) dx b ∫
0 2 ( 2x3 − 4x + 5)dx c ∫
4 9 ( √ __
x − 6 __ x2 ) dx d ∫
1 8 ( x − 1 _ 3 + 2x − 1)dx
3 Evalua
te the following definite integrals:
a ∫
1 3 x3 + 2x2 ________ x dx b ∫
3 6 (x − 3 __ x ) 2
dx c ∫
0 1 x 2 ( √ __
x + 1 __ x ) dx d ∫
1 4 2 + √ __
x ______ x2 dx
4 Given tha
t A is a constant and ∫
1 4 (6 √ __
x − A) dx = A2, show that there are two possible values for
A and find these values. (5 marks)
5 Use calculus to find the va
lue of ∫
1 9 (2x − 3 √ __
x ) dx. (5 marks)
6 Eva
luate ∫
4 12 2 ___ √ __
x dx, giving your answer in the form a + b √ __
3 , where a and b are integers. (4 marks)
7 Given tha
t ∫
1 k 1 ___ √ __
x dx = 3, calculate the value of k. (4 marks)
8 The speed, v ms−1, of a train at time t seconds is given
by v = 20 + 5t, 0 < t < 10.
The distance, s metres, travelled by the train in 10 seconds is given by s =
∫
0 10 ( 20 + 5 t)dt. Find the value of s. You m ust not use a calculator to
work out definite integrals in your exam. You
need to use calculus and show clear algebraic working.Watch out
E/P
E
E
E/P
You might encounter a definite integral with an unknown in the limits. Here, you can find an expression for the definite integral in terms of k then set that expression equal to 3.Problem-solving
Given that ∫
k 3k 3x + 2 ______ 8 dx = 7 and k > 0, calculate the value of k .Challenge
13.5 Areas under curves
Definite integration can be used to find the area under a curve. y
y = f(x)
A(x)
x OFor any curve with equation y = f( x), you can define the area
under the curve to the left of x as a function of x called A (x).
As x increases, this area A (x) also increases (since x moves
further to the right). | [
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298
Chapter 13
Example 9
Find the area of the finite region between the curve with equation y = 20 − x − x 2 and the x-axis.
y = 20 − x − x2 = (4 − x )(5 + x )
–5 /four.ss01 O20
xy
Area = ∫ −5 4 ( 20 − x − x2)dx
= [20x − x 2 ___ 2 − x 3 ___ 3 ]
−5 4
= (80
− 8 − 64 ___ 3 ) − (−100 − 25 ___ 2 + 125 ____ 3 )
= 243 _____ 2 Factorise the expression.
Draw a sketch of the graph. x = 4 and x = − 5 are
the points of intersection of the curve and the
x-axis.
You don’t normally need to give units when you are finding areas on graphs.If you look at a small increase in x, say δ x, then the area
increases by an amount δ A = A(x + δx) − A(x).
This increase in the δ A is approximately rectangular
and of magnitude y δx. (As you make δ x smaller any error
between the actual area and this will be negligible.)
So you have δA ≈ yδx
or δA ___ δx ≈ y
an
d if you take the limit lim
δx→0 ( δA ___ δx ) then you will see that dA ___ dx = y.
Now if you know that dA ___ dx = y, then to find A you have to integrate, giving A = ∫y dx.
■ The area between a positive curve, the x-axis
and the lines x = a and x = b is given by
Area = ∫ a b y dx
where y = f(x) is the equation of the curve. y
xy = f(x)
a O by
y = f(x) A(x)δA
x x + δx OThis vertical height
will be y = f( x). | [
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299
Integration
Exercise 13E
1 Find the area between the curv e with equation y = f(x), the x-axis and the lines x = a and
x = b in each of the following cases:
a f(x
) = −3x2 + 17x − 10; a = 1, b = 3
b f(x
) = 2x3 + 7x2 − 4x; a = −3, b = −1
c f(x
) = −x4 + 7x3 − 11x2 + 5x; a = 0, b = 4
d f(x
) = 8 ___ x 2 ; a = −4, b = −1
2 The sketch shows part of
the curve with equation y = x(x2 − 4).
Find the area of the shaded region.
3 The diagram sho
ws a sketch of the curve with equation
y = 3x + 6 __ x2 − 5, x > 0.
The r
egion R is bounded by the curve, the x-axis and the
lines x = 1 and x = 3.
Find the area of R.
4 Find the area of the finite r
egion between the curve with equation y = (3 − x)(1 + x) and the
x-axis.
5 Find the area of the finite r
egion between the curve with equation y = x(x − 4)2 and the x-axis.
6 Find the area of the finite r
egion between the curve with equation y = 2x2 − 3x3 and the x-axis.
7 The shaded area under the gra
ph
of the function f(x) = 3x2 − 2x + 2,
bounded by the curve, the x-axis
and the lines x = 0 and x = k, is 8. Work out the value of k.
8
The finite region R
is bounded by the x-axis and
R
B AO
y = –x2 + 2x + 3y
x
the curve with equation y = −x2 + 2x + 3, x > 0.
The curve meets the x-axis at points A and B.
a Find the coordinates of
point A and point B. (2 marks)
b Find the area of the r
egion R. (4 marks) For part c , f (x) = − x(x − 1)2(x − 5) Hint
y
x–2y = x(x2 – 4)
O
y
xR
O1 3y = 3x +6
x2– 5
P
Oy
x ky = 3x2 – 2x +2P
E ∫
0 k (3 x2 − 2x + 2) dx = 8Problem-solving | [
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300
Chapter 13
9 The graph sho
ws part of the
curve C with equation y = x2(2 − x).
The region R, shown shaded,
is bounded by C and the x-axis.
Use calculus to find the exact
area of R. (5 marks)E
13.6 Areas under the x-axis
You need to be careful when you are finding areas below the x-axis.
■ When the area bounded by a curve and the x-axis is below the x-axis, ∫y dx gives a negative
answer.
Example 10
Find the area of the finite region bounded by the curve y = x(x − 3) and the x -axis.
First sketch the curve.
It is -shaped and crosses the x-axis at 0 and 3.
The limits on the integral will therefore be 0 and 3.
State the area as a positive quantity.When x = 0, y = 0
When y = 0, x = 0 or 3
y
x O 3y = x(x – 3)
Area = ∫ 0 3 x (x − 3)dx
= ∫ 0 3 ( x2 − 3 x)dx
= [ x3 ___ 3 − 3x2 ____ 2 ]
0 3
= ( 27 ___ 3 − 27 ___ 2 ) − (0 − 0)
= − 27 ___ 6 or − 9 __ 2 or −4.5
So the area is 4.5
The following example shows that great care must be taken if you are
trying to find an area which straddles the x-axis such as the shaded region.
For examples of this type you need to draw a sketch, unless one is
given in the question.Multiply out the brackets.
Integrate as usual.
The area is below the x -axis so the definite
integral is negative. If a
qu
estion says “use
calculus” then you need
to use integration or differentiation, and show clear algebraic working.Watch out
RC
O 2y
xy = x2(2 – x)
Check your solution using
yo
ur calculator.Online | [
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301
Integration
Example 11
Sketch the curve with equation y = x(x − 1)(x + 3) and find the area of the finite region bounded by
the curve and the x -axis.
When x = 0, y = 0
When y = 0, x = 0, 1 or − 3
x → ∞ , y → ∞
x → − ∞, y → − ∞
y
x –3 O 1y = x (x – 1)(x + 3)
The area is given by ∫ −3 0 y dx − ∫ 0 1 y dx
Now ∫y dx = ∫(x3 + 2x2 − 3 x)dx
= [ x4 ___ 4 + 2x3 ____ 3 − 3x2 ____ 2 ]
So ∫ −3 0 y dx = (0) − ( 81 ___ 4 − 2 __ 3 × 27 − 3 __ 2 × 9)
= 45 ___ 4
an
d ∫ O 1 y dx = ( 1 __ 4 + 2 __ 3 − 3 __ 2 ) − (0)
= − 7 ___ 12
So th
e area required is 45 ___ 4 + 7 ___ 12 = 71 ___ 6 Find out where the curve cuts the axes.
Multiply out the brackets.Find out what happens to y when x is large and
positive or large and negative.
Exercise 13F
1 Sketch the following and find the total area of the finite region or regions bounded by the curves
and the x
-axis:
a y = x
(x + 2) b y = (
x + 1)(x − 4) c y = (
x + 3)x(x − 3)
d y = x2(x − 2) e y = x (x − 2)(x − 5)
2 The graph sho
ws a sketch of part of the curve C with equation
O
CABy
xy = x(x + 3)(2 – x)
y = x(x + 3)(2 − x).
The curve C crosses the x-axis at the origin O and at points
A and B.
a Write down the
x-coordinates of A and B. (1 mark)
The finite region, shown shaded, is bounded b
y the curve C and the x-axis.
b Use integration to find the tota
l area of the finite shaded region. (7 marks)E If yo u try to calculate the area as a
single definite integral, the positive and negative
areas will partly cancel each other out. Watch outAlways draw a sketch, and use the points of intersection with the x-axis as the limits for your integrals.Problem-solving
Since the area between x = 0 and 1 is below the axis the integral between these points will give a negative answer. | [
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302
Chapter 13
3 f(x
) = − x 3 + 4 x 2 + 11x − 30
Oy
xy = –x3 + 4x2 – 11x – 30The graph shows a sketch of part of the curve with
equation y = − x 3 + 4 x 2 + 11x − 30.
a Use the factor theorem to show tha
t (x + 3) is a
factor of f(x).
b Write f(x
) in the form (x + 3)(Ax2 + Bx + C).
c Hence, factorise f(x
) completely.
d Hence, determine the
x-coordinates where the curve intersects the x-axis.
e Hence, determine the tota
l shaded area shown on the sketch.
1 Given that f( x) = x(3 − x ), find the area of the finite region bounded
by the x -axis and the curve with equation
a y =
f(x) b y =
2f(x) c y =
af(x)
d y =
f(x + a ) e y =
f(ax).
2 The g
raph shows a sketch of
OC
ABy
xy = x (x – 1)(x + 2)
part of the curve C with
equation y = x (x − 1)( x + 2).
The curve C crosses the x-axis at the origin O and
at point B .
The shaded areas above and below the x -axis are equal.
a
Sho
w that the x -coordinate of A satisfies the equation
(x − 1)2(3x2 + 10 x + 5) = 0
b Hen
ce find the exact coordinates of A , and interpret geometrically
the other roots of this equation.Challenge
13.7 Areas between curves and lines
■ You can use definite integration together with areas of trapeziums and triangles to find
more complicated areas on graphs.
Example 12
The diagram shows a sketch of part of the curve with equation
y = x(4 − x) and the line with equation y = x.
Find the area of the region bounded by the curve and the line.y
x O 4y = x(4 – x) y = x | [
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-0.03... |
303
Integration
Example 13
The diagram shows a sketch of the curve with equation
xy
OC
A
(a, 0)B
(b, 0)y = 2xy = x(x – 3)
y = x(x − 3) and the line with equation y = 2x.
Find the area of the shaded region OAC .x(4 − x ) = x
3x − x2 = 0
x(3 − x ) = 0
x = 0 or 3
Area beneath curve = ∫ 0 3 ( 4x − x2)dx
= [2 x 2 − x 3 ___ 3 ]
0 3
= 9
Ar
ea beneath triangle = 1 __ 2 × 3 × 3
= 9 __ 2
Sha
ded area = 9 − 9 __ 2 = 9 __ 2
Look for ways of combining triangles, trapeziums
and direct integrals to find the missing area.
xy
OC
A
(a, 0)B
(b, 0)Problem-solvingFirst, find the x -coordinate of the points of
intersection of the curve y = x(4 − x ) and the
line y = x .
[2 x 2 − x 3 __ 3 ]
0 3
= (18 − 27 ___ 3 ) − (0 − 0) = 18 − 9Shaded area = area beneath curve − area
beneath triangle
y
x O3y
xO 33
The required area is given by:
Area of triangle OBC − ∫ a b x (x − 3)dx
The curve cuts the x -axis at x = 3
(and x = 0) so a = 3.
The curve meets the line y = 2x when2x = x (x − 3).
So
0 =
x2 − 5 x
0 = x(x − 5)
x = 0 or 5, so b = 5
The point C is (5, 10).Area of triangle OBC =
1 __ 2 × 5 × 10 = 25.
Area between curve, x -axis and the line
x = 5 is
∫ 3 5 x( x − 3)dx = ∫ 3 5 ( x2 − 3 x)dx
= [ x3 ___ 3 − 3x2 ____ 2 ]
3 5
Substituting x = 5 into the equation of the line
gives y = 2 × 5 = 10.
Work out the definite integral separately. This will help you avoid making errors in your working. | [
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304
Chapter 13
= ( 125 _____ 3 − 75 ___ 2 ) − ( 27 ___ 3 − 27 ___ 2 )
= ( 25 ___ 6 ) − (− 27 ___ 6 )
= 26 ___ 3
Sha
ded region is therefore = 25 − 26 ___ 3 = 49 ___ 3
Exercise 13G
1 The diagram shows part of the curve with equation
xy
y = 6y = x2 + 2
AB
O26
y = x2 + 2 and the line with equation y = 6.
The line cuts the curve at the points A and B.
a Find the coordinates of
the points A and B.
b Find the area of the finite r
egion bounded
by line AB and the curve.
2 The diagram sho
ws the finite region, R, bounded by the curve
with equation y = 4x − x2 and the line y = 3.
The line cuts the curve at the points A and B.a
Find the coordinates of
the points A and B.
b Find the area of
R.
3 The diagram sho
ws a sketch of part of the curve with equation
y = 9 − 3x − 5x2 − x3 and the line with equation y = 4 − 4x.
The line cuts the curve at the points A (−1, 8) and B(1, 0).Find the area of the shaded region between AB and the curve.
4
Find the area of the finite r
egion bounded by the curve with
equation y = (1 − x)(x + 3) and the line y = x + 3.
5 The diagram sho
ws the finite region, R, bounded by the
curve with equation y = x(4 + x), the line with equation
y = 12 and the y-axis.
a Find the coordinates of
the point A where the line
meets the curve.
b Find the area of
R.xy
AB Ry = 3y = 4x – x2
O 4
A
B
y = 4 – 4xy = 9 – 3x – 5x2 – x3
xy
OP
P
y = 12
xA
Ry
Oy = x(4 + x) | [
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-0.... |
305
Integration
6 The diagram sho
ws a sketch of part of the curve with
equation y = x2 + 1 and the line with equation y = 7 − x.
The finite region, R1 is bounded by the line and the curve.
The finite region, R2 is below the curve and the line and is
bounded by the positive x - and y -axes as shown in the diagram.
a Find the area of
R1.
b Find the area of
R2.
7 The curve C
has equation y = x 2 _ 3 − 2 ___
x 1 _ 3 + 1.
a Verify that
C crosses the x-axis at the point (1, 0).
b Show that the point
A(8, 4) also lies on C.
c The point B is (4, 0). Find the equa
tion of the line through AB.
The finite region R is bounded by C, AB and the positive x-axis.
d Find the area of
R.
8 The diagram sho
ws part of a sketch of the curve with equation
y = 2 __ x2 + x. The points A and B have x-coordinates 1 _ 2 and 2
respectiv
ely.
Find the area of the finite region between AB and the curve.
9 The diagram sho
ws part of the curve with equation
y = 3 √ __
x − √ __
x3 + 4 and the line with equation
y
= 4 − 1 _ 2 x.
a Verify that the line and the curv
e cross at the
point A(4, 2).
b Find the area of the finite r
egion bounded by
the curve and the line.
10 The sketch shows part of
the curve with equation
y = x2(x + 4). The finite region R1 is bounded by the
curve and the negative x-axis. The finite region R2 is
bounded by the curve, the positive x-axis and AB,
where A(2, 24) and B(b, 0).
The area of R1 = the area of R2.
a Find the area of
R1.
b Find the value of
b.xR1
R2y
y = x2 + 1
y = 7 – x
77
OP
P
xy
A
B
O 2 1
2y = + x2
x2P
xy
A
O4√
y = 4 – 1
2y = 3 x – x3 + 4√
xP
xy
A (2, 24)
B (b, 0)R1 R2
Oy = x2(x + 4)P
Split R2 into two areas by drawing
a vertical line at x = 2.Problem-solving | [
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306
Chapter 13
11 The line with equation y
= 10 − x cuts the curve with
A
B
xyy = 2x2 – 5x + 4
y = 10 – xOR
equation y = 2x2 − 5x + 4 at the points A and B,
as shown.
a Find the coordinates of
A and the coordinates
of B. (5 marks)
The shaded region R
is bounded by the line and the
curve as shown.b
Find the exact area of
R. (6 marks)E/P
1 Find:a
∫(x + 1)(2x − 5)dx b ∫( x 1 _ 3 + x − 1 _ 3 )dx
2 The gradient of a curv
e is given by f ′(x
) = x2 − 3x − 2 __ x2 . Given that the curve passes through
the point (1, 1), find the equation of
the curve in the form y = f(x).
3 Find:
a ∫(8x3 − 6x2 + 5)dx b ∫(5x + 2) x 1 _ 2 dx
4 Giv
en y = (x +
1)(2x − 3) _____________ √ __
x , find ∫y dx.
5 Given tha
t dx ___ dt = (t + 1)2 and that x = 0 when t = 2, find the value of x when t = 3.
6 Given tha
t y 1 _ 2 = x 1 _ 3 + 3:
a show that
y = x 2 _ 3 + A x 1 _ 3 + B, where A and B are constants to be found. (2 marks)
b hence find ∫y dx. (3 marks)
7 Given tha
t y 1 _ 2 = 3 x 1 _ 4 − 4 x − 1 _ 4 (x > 0):
a find dy ___ dx (2 marks)
b find ∫y dx. (3 marks)
8 ∫ ( a ___ 3x3 − ab) dx = − 2 ___ 3x2 + 14x + c
Find the value of a and the value of b.
9 A rock is dropped off
a cliff. The height in metres of the rock above the ground after t seconds
is given by the function f(t). Given that f(0) = 70 and f ′(t
) = −9.8t, find the height of the rock
above the ground after 3 seconds.P
P
E/P
E/P
P
PMixed exercise 13 | [
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307
Integration
10 A cyclist is tra
velling along a straight road. The distance in metres of the cyclist from a fixed
point after t seconds is modelled by the function f(t), where f ′(t
) = 5 + 2t and f(0) = 0.
a Find an expression f
or f(t).
b Calculate the time ta
ken for the cyclist to travel 100 m.
11 The diagram sho
ws the curve with equation
y = 5 + 2x − x2 and the line with equation y = 2.
The curve and the line intersect at the points A and B.
a Find the x-coor
dinates of A and B.
b The shaded region R
is bounded by the curve and
the line. Find the area of R.
12 a Find ∫( x 1 _ 2 − 4)( x − 1 _ 2 − 1)dx. (4 marks)
b Use your answ
er to part a to evaluate
∫
1 4 ( x 1 _ 2 − 4)( x − 1 _ 2 − 1)dx
giving your answer as an exact fraction. (2 marks)
13 The diagram sho
ws part of the curve with equation y
AB
R
x Oy = x3 – 6x2 + 9x
y = x3 − 6x2 + 9x. The curve touches the x -axis at A and
has a local maximum a
t B.
a Show that the equa
tion of the curve may be written
as y = x(x − 3)2, and hence write down the
coordinates of A. (2 marks)
b Find the coordinates of
B. (2 marks)
c The shaded region R
is bounded by the
curve and the x-axis. Find the area of R. (6 marks)
14 Consider the function y =
3 x 1 _ 2 − 4 x − 1 _ 2 , x > 0.
a Find dy ___ dx . (2 marks)
b Find ∫y dx. (3 marks)
c Hence show that ∫
1 3 y dx = A + B √ __
3 , where A and B are integers to be found. (2 marks)
15 The diagram sho
ws a sketch of the curve with equation
y = 12 x 1 _ 2 − x 3 _ 2 for 0 < x < 12.
a Show that dy ___ dx = 3 __ 2 x − 1 _ 2 (4 − x ). (2 marks)
b At the point B on the curv
e the tangent to the
curve is parallel to the x-axis. Find the coordinates
of the point B. (2 marks)
c Find, to 3 significant figures, the ar
ea of the finite
region bounded by the curve and the x -axis. (6 marks)P
y
ABR
y = 2
y = 5 + 2x – x2x O
E/P
E
E
E/P y
xB
O 12y = 12x – x1
232 | [
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0.04... |
308
Chapter 13
16 The diagram sho
ws the curve C with equation
y = x(8 − x) and the line with equation y = 12
which meet at the points L and M.
a Determine the coordina
tes of the point M. (2 marks)
b Given tha
t N is the foot of the perpendicular from
M on to the x-axis, calculate the area of the shaded
region which is bounded by NM, the curve C and the x-axis.
(6 marks)
17 The diagram sho
ws the line y = x − 1 meeting the y
BC
A Oy = (x – 1)(x – 5)y = x – 1
x
curve with equation y = (x − 1)(x − 5) at A and C. The curve meets the x-axis at A and B.
a
Write down the coor
dinates of A and B and find
the coordinates of C. (4 marks)
b Find the area of the shaded r
egion bounded by
the line, the curve and the x-axis. (6 marks)
18 The diagram sho
ws part of the curve with equation
y = p + 10x − x2, where p is a constant, and part
of the line l with equation y = qx + 25, where q is a
constant. The line l cuts the curve at the points A and B. The x-coordinates of A and B are 4 and 8 respectively. The line through A parallel to the x-axis intersects the curve again at the point C.
a
Show that
p = −7 and calculate the value of q. (3 marks)
b Calculate the coor
dinates of C. (2 marks)
c The shaded region in the diagr
am is bounded by
the curve and the line segment AC . Using integration
and showing all your working, calculate the area of the
shaded region. (6 marks)
19 Given tha
t f(x) = 9 __ x2 − 8 √ __
x + 4x − 5, x > 0, find ∫f (x)dx. (5 marks)
20 Given tha
t A is constant and ∫
4 9 ( 3 ___ √ __
x − A) dx = A2 show that there are two possible values
for A and find these values. (5 marks)
21 f ′(
x) = (2 − x 2 ) 3 _______ x 2 , x ≠ 0
a Show that f ′(x) = 8x−2 − 12 + Ax2 + Bx4, where A and B are constants to be found. (3 marks)
b Find f ″(
x).
Given that the point (−2, 9) lies on the curve with equation y = f(x),
c find f(x). (5 marks)E/P y
LM
N O12y = 12
y = x(8 – x)
x
E/P
E/P y
DAC
B
O xy = qx + 25
y = p + 10x – x2
E
E/P
E/P | [
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309
Integration
22 The finite region S
, which is shown shaded, is
bounded by the x-axis and the curve with
equation y = 3 − 5x − 2x2.
The curve meets the x-axis at points A and B.
a Find the coordinates of
point A and
point B. (2 marks)
b Find the area of the r
egion S. (4 marks)
23 The graph sho
ws a sketch of part of the curve C with
RAC
B Oy
xy = (x – 4)(2x + 3)
equation y = (x − 4)(2x + 3).The curve C crosses the x-axis at the points A and B.a
Write down the
x-coordinates of A and B. (1 mark)
The finite region R
, shown shaded, is bounded by C
and the x-axis.b
Use integration to find the ar
ea of R. (6 marks)
24 The graph sho
ws a sketch of part of the curve C
C
B A Oy
x
y = x(x – 3)(x + 2)
with equation y = x(x − 3)(x + 2).The curve crosses the x-axis at the origin O and
the points A and B.
a
Write down the
x-coordinates of the points
A and B. (1 mark)
The finite region shown shaded is bounded b
y the curve C and the x-axis.
b Use integration to find the tota
l area of this region. (7 marks)E
ABS
Oy
xy = 3 – 5x – 2x2
E
E
The curve with equation y = x2 − 5x + 7 cuts the curve with
xy
y = x2 – 5x + 7
ORy = x2 – x + 71
252equation y = 1 _ 2 x2 − 5 _ 2 x + 7. The shaded region R is
bounded by the curves as shown.
Find the exact area of R .Challenge | [
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310
Chapter 13
1 If dy ___ dx = xn, then y = 1 _____ n + 1 x n + 1 + c, n ≠ −1.
Using function notation, if f ′(x
) = xn, then f(x) = 1 _____ n + 1 x n + 1 + c, n ≠ −1.
2 If dy ___ dx = kxn, then y = k _____ n + 1 x n + 1 + c, n ≠ −1.
Using function notation, if f ′(x
) = kxn, then f(x) = k _____ n + 1 x n + 1 + c, n ≠ −1.
When integrating polynomials, apply the rule of integration separately to each term.
3 ∫f ′(x)dx = f(x) + c
4 ∫ (f(x) + g(x) ) dx = ∫f(x)dx + ∫g(x)dx
5 To find the constant o
f integration, c
• Integrate the function
• Sub
stitute the values (x
, y) of a point on the curve, or the value of the function at a given
point f(x) = k into the integrated function
• Solve the equation to find c
6 If f ′(
x) is the derivative of f(x) for all values of x in the interval [a, b], then the definite integral
is defined as ∫ a b f ′(x)dx = [f(x)]ba = f(b) − f(a)
7 The area betw
een a positive curve, the x-axis and the lines x = a and x = b is given by
Area = ∫ a b y dx
wher
e y = f(x) is the equation of the curve.
8 When the area bounded by a cur
ve and the x-axis is below the x-axis, ∫y dx gives a negative
answer
.
9 You can use definite int
egration together with areas of trapeziums and triangles to find more
complicated areas on graphs.Summary of key points | [
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311
Exponentials and
logarithms
After completing this unit you should be able to:
● Sketch gr
aphs of the form y = ax, y = ex, and transformations of
these graphs → pages 312–317
● Differentiate ekx and understand why this result is important
→ pages 314–317
● Use and interpret m odels that use exponential functions
→ pages 317–319
● Recognise the relationship between exponents and logarithms
→ pages 319–321
● Recall and apply the laws of logarithms → pages 321–324
● Solve equations of the form ax = b → pages 324–325
● Describe and use the natural l ogarithm function → pages 326–328
● Use logarithms to estimate the values of constants in non-linear
models → pages 328–333Objectives
1 Given that x = 3 and y = −1, evaluate
these expressions without a calculator.
a 5x b 3y c 22x−1 d 71−y e 11x+3y
← GC SE Mathematics
2 Simplify these expressions, writing each
answ
er as a single power.
a 68 ÷ 62 b y3 × (y9)2 c 2 5 × 2 9 ______ 2 8 d √ ___
x 8
← Sec tions 1.1, 1.4
3 Plot the following data on a scatter graph
and dra
w a line of best fit.
x 1.2 2.1 3.5 4 5.8
y 5.8 7.4 9.4 10.3 12.8
Determine the gradient and intercept of your line of best fit, giving your answers to one decimal place.
← GCSE MathematicsPrior knowledge check
Logarithms are used to report and compare earthquakes. Both the Richter scale and the newer moment magnitude scale use base 10 logarithms to express the size of seismic activity.
→ Mixed exercise Q1514 | [
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312
Chapter 14
14.1 Exponential functions
Functions of the form f( x) = a x, where a is a constant, are called exponential functions . You should
become familiar with these functions and the shapes of their graphs.
For an example, look at a table of values of y = 2x.
x −3 −2 −1 0 1 2 3
y 1 _ 8 1 _ 4 1 _ 2 1 2 4 8
The value of 2x tends towards 0 as x decreases, and
grows without limit as x increases.
The graph of y = 2x is a smooth curve that looks like this:
x –1 –2 –3 1 2 31
O2345678y
–1 In the e xpression 2x,
x can be called an index , a power
or an exponent .Notation
Recall that 20 = 1 and that
2 −3 = 1 __ 2 3 = 1 __ 8 ← Section 1.4Links
Example 1
a On the same axes sketch the gr aphs of y = 3x, y = 2x and y = 1.5x.
b On another set of axes sk
etch the graphs of y = ( 1 _ 2 ) x and y = 2x.
Whenever a > 1, f( x) = ax is an increasing
function. In this case, the value of ax grows
without limit as x increases , and tends towards 0
as x decreases .The x -axis is an asymptote
to the curve.
a For all three graphs, y = 1 when x = 0.
Wh
en x > 0, 3x > 2x > 1.5x.
When x < 0, 3x < 2x < 1.5x.
x –1 –2 –31 23 Oy
y = 2x
y = 1.5xy = 3x
b The graph of y = ( 1 __ 2 ) x is a reflection in the
y-axis of the graph of y = 2x.
x–1 –2 –31 23 Oy
y = 2xy = ( )x1
2a0 = 1
Work out the relative positions of the three
graphs.
Since 1 _ 2 = 2−1, y = ( 1 _ 2 ) x
is the same as
y = (2−1)x = 2−x.
Whenever 0 < a < 1, f( x) = ax is a decreasing
function. In this case, the value of ax tends
towards 0 as x increases , and grows without limit
as x decreases . | [
-0.0008238779846578836,
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0.03936227411031723,
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0.06271... |
313
Exponentials and logarithms
Example 2
Sketch the graph of y = ( 1 _ 2 ) x − 3
. Give the coordinates of the point where the graph crosses the y-axis.
If f(x) = ( 1 _ 2 ) x then y = f( x − 3).
The graph is a translation of the graph
y = ( 1 _ 2 ) x by the vector ( 3 0 ) .
The graph crosses the y -axis when x = 0.
y = ( 1 _ 2 ) 0 − 3
y = 8
The graph crosses the y -axis at (0, 8).
22
46810 046810y
xYou can also consider this graph as a stretch of
the graph y = ( 1 _ 2 ) x
y =
( 1 _ 2 ) x − 3
= ( 1 _ 2 ) x
× ( 1 _ 2 ) −3
= ( 1 _ 2 ) x
× 8
= 8 ( 1 _ 2 ) x
= 8f(x)
So the graph of y = ( 1 _ 2 ) x − 3 is a vertical stretch of
the graph of y = ( 1 _ 2 ) x
with scale factor 8.If you have to sketch the graph of an unfamiliar
function, try writing it as a transformation of a familiar function.
← Section 4.5Problem-solving
Exercise 14A
1 a Draw an accurate graph of y = (1.7)x, for −4 < x < 4.
b Use your gra
ph to solve the equation (1.7)x = 4.
2 a Draw an accur
ate graph of y = (0.6)x, for −4 < x < 4.
b Use your gra
ph to solve the equation (0.6)x = 2.
3 Sketch the gra
ph of y = 1x.
4 For each of these sta
tements, decide whether it is true or false, justifying your answer or offering
a counter-example.
a The graph of
y = a x passes through (0, 1) for all positive real numbers a.
b The function f(x)
= ax is always an increasing function for a > 0.
c The graph of
y = a x, where a is a positive real number, never crosses the x-axis.
5 The function f(x) is defined as f(
x) = 3x, x ∈ ℝ. On the same axes, sketch the graphs of:
a y =
f(x) b y =
2f(x) c y =
f(x) − 4 d y =
f ( 1 _ 2 x)
Write down the coor
dinates of the point where each graph crosses the y-axis, and give the
equations of any asymptotes.
6 The graph of
y = kax passes through the
points (1, 6) and (4, 48). Find the values
of the constants k and a.P
P Substitute the coordinates into y = ka x to create
two simultaneous equations. Use division to
eliminate one of the two unknowns.Problem-solving | [
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0.03193579614162445,
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-0.0857396051287651,
-0.00827698688954115,
0.0067... |
314
Chapter 14
7 The graph of
y = pqx passes through the points (−3, 150) and (2, 0.048).
a By drawing a sk
etch or otherwise, explain why 0 < q < 1.
b Find the values of
the constants p and q.P
14.2 y = e x
Exponential functions of the form f(x) = ax have a special mathematical property. The graphs of their
gradient functions are a similar shape to the graphs of the functions themselves.
y
y = 3x = 1.099… × 3x
x Ody
dxy
y = 4x= 1.386… × 4x
x Ody
dxyy = 2x = 0.693… × 2x
x Odydx
In each case f9(x) = kf(x), where k is a constant. As the value of a increases, so does the value of k.
Something unique happens between a = 2 and
a = 3. There is going to be a value of a where the gradient function is exactly the same as the original function. This occurs when a is approximately equal to 2.71828. The exact value is represented by the letter e. Like
π, e is both an
important mathematical constant and an irrational number.
■ For all real values of x:
• If f(x) = e x then f’( x) = e x
• If y = e x then dy ___ dx = e x
A similar result holds for functions such as e5x, e−x and e 1 _ 2 x .
■ For all real values of x and for any constant k:
• If f(x) = e kx then f’( x) = ke kx
• If y = e kx then dy ___ dx = ke kxFunction Gradient function
f(x) = 1xf9(x) = 0 × 1x
f(x) = 2xf9(x) = 0.693… × 2x
f(x) = 3xf9(x) = 1.099… × 3x
f(x) = 4xf9(x) = 1.386… × 4xSketch the graph of y = 2x − 2 + 5. Give the coordinates of the point where
the graph crosses the y -axis.Challenge
Explore the relationship between
ex
ponential functions and their derivatives
using GeoGebra.Online | [
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0... |
315
Exponentials and logarithms
Example 3
Differentiate with respect to x.
a e4x b e − 1 _ 2 x c 3e2x
a y = e4x
dy ___ dx = 4e4x
b y = e − 1 __ 2 x
dy ___ dx = − 1 __ 2 e − 1 __ 2 x
c y =
3e2x
dy ___ dx = 2 × 3e2x = 6e2x
Example 4
Sketch the graphs of the following equations. Give the coordinates of any points where the graphs
cross the axes, and state the equations of any asymptotes.
a y =
e2x b y = 10e−x c y = 3 + 4 e 1 _ 2 x
a y = e2x
When x = 0, y = e2 × 0 = 1 so the graph
crosses the y -axis at (0, 1).
The x-axis ( y
= 0) is an asymptote.
1yy = e2x
y = ex
x O
b y = 10e−x
When x = 0, y = 10e−0. So the graph
crosses the y -axis at (0, 10).
The x-axis ( y
= 0) is an asymptote.
110y
y = 10e–xy = e–x
x OThe graph of y = ex has been shown in purple on
this sketch.Use the rule for differentiating ekx with k = 4.
To differentiate aekx multiply the whole function
by k. The derivative is kaekx.
This is a stretch of the graph of y = ex, parallel to
the x -axis and with scale factor 1 _ 2
← Section 4.6
Negative powers of e x, such as e−x or e−4x, give
rise to decreasing functions.
The graph of y = e x has been reflected in the
y-axis and stretched parallel to the y-axis with
scale factor 10. | [
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0.05994642898440361,
0.0017932092305272818,
-0.06664608418941498,
-0.03586585447192192,
0.029398... |
316
Chapter 14
c y = 3 + 4 e 1 __ 2 x
Whe
n x = 0, y = 3 + 4 e 1 __ 2 × 0 = 7 so
the graph crosses the y -axis at (0, 7).
The line y = 3 is an asymptote.
y
y = 3 + 4e x
x7
3
O1
2
The graph of y = e 1 _ 2 x has been stretched parallel to
the y-axis with scale factor 4 and then translated
by ( 0 3 ) .If you have to sketch a transformed graph with
an asymptote, it is often easier to sketch the asymptote first.Problem-solving
Exercise 14B
1 Use a calculator to find the value of ex to 4 decimal places when
a x =
1 b x =
4 c x =
−10 d x =
0.2
2 a Draw an accur
ate graph of y = ex for −4 < x < 4.
b By drawing a
ppropriate tangent lines, estimate the gradient at x = 1 and x = 3.
c Compare your ans
wers to the actual values of e and e3.
3 Sketch the gra
phs of:
a y =
ex + 1 b y = 4e−2x c y = 2ex − 3
d y =
4 − ex e y = 6 + 10 e 1 _ 2 x f y = 100e−x + 10
4 Each of the sketch gr
aphs below is of the form y = Aebx + C, where A, b and C are constants.
Find the values of A and C for each graph, and state whether b is positive or negative.
a
6
5
Oy
x b
4
Oy
x c
28
Oy
x
5 Rearrange f(x) = e3x + 2 into the form f(x) = Aebx, where A and b
are constants whose values are to be found. Hence, or otherwise,
sketch the graph of y = f(x).
6 Differentiate the f
ollowing with respect to x.
a e6x b e − 1 _ 3 x c 7e2x
d 5e0.4x e e3x + 2ex f e x(e x + 1) You do not have
en
ough information
to work out the value
of b, so simply state
whether it is positive or negative.Hint
e m + n = em × en Hint
For part f , sta rt
by expanding the bracket.Hint Use GeoGebra to draw
tra
nsformations of y = ex.Online | [
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0.00767015665769577,
0.04277471825480461,
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0.041750531643629074,
0.062... |
317
Exponentials and logarithms
7 Find the gradient of the curv
e with equation y = e3x at the point where
a x =
2 b x =
0 c x =
−0.5
8 The function f is defined as f(x
) = e0.2x, x ∈ ℝ. Show that the tangent to the curve at the point
(5, e) goes through the origin.P
14.3 Exponential modelling
You can use ex to model situations such as population growth, where the rate of increase is
proportional to the size of the population at any given moment. Similarly, e−x can be used to model
situations such as radioactive decay, where the rate of decrease is proportional to the number of
atoms remaining.
Example 5
The density of a pesticide in a given section of field, P mg/m2, can be modelled by the equation
P = 160e−0.006t
where t is the time in days since the pesticide was first applied.
a Use this model to estimate the density of
pesticide after 15 days.
b Interpret the meaning of the v
alue 160 in this model.
c Show that dP ___ dt = kP, where k is a constant, and state the value of k.
d Interpret the significance of the sign of
your answer to part c.
e Sketch the gra
ph of P against t.
a After 15 days, t = 15.
P
= 160e−0.006 × 15
P = 146.2 mg /m2
b When t = 0, P = 160e0 = 160, so 160 mg/ m2
is the initial density of pesticide in the field.
c P =
160e−0.006 t
dP ___ dt = −0.96e−0.006 t, so k = − 0.96
d As k
is negative, the density of pesticide
is decreasing (there is exponential decay).
e
160
OP
t Work this out in one go using the
e button on your calculator.OnlineSubstitute t = 15 into the model.
The v alue given by a model when
t = 0 is called the initial value .Notation
Use your answers to parts a and d to help you
draw the graph. To check what happens to P in the long term, substitute in a very large value of t .If y = ekx then dy ___ dx = kekx | [
0.048632360994815826,
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0.07192396372556686,
0.07050491124391556,
-0.07559532672166824,
-0.04693979024887085,
0.05... |
318
Chapter 14
Exercise 14C
1 The value of a car is modelled by the formula
V
= 20 000 e − t __ 12
where
V is the value in £s and t is its age in years from new.
a State its va
lue when new.
b Find its value (to the near
est £) after 4 years.
c Sketch the gra
ph of V against t.
2 The population of a country is mode
lled using the formula
P = 20 + 10 e t __ 50
where
P is the population in thousands and t is the time in years after the year 2000.
a State the population in the y
ear 2000.
b Use the model to predict the popula
tion in the year 2030.
c Sketch the gra
ph of P against t for the years 2000 to 2100.
d Do you think that it w
ould be valid to use this model to predict the population in the year
2500? Explain your answer.
3 The number of people infected with a disease is mode
lled by the formula
N = 300 − 100e−0.5t
where N is the number of people infected with the disease and t is the time in years after
detection.
a How many people w
ere first diagnosed with the disease?
b What is the long ter
m prediction of how this disease will spread?
c Sketch the gra
ph of N against t for t > 0.
4 The number of r
abbits, R, in a population after m months is modelled by the formula
R = 12e 0.2m
a Use this model to estimate the n umber of rabbits after
i 1 month ii 1 year
b Interpret the meaning of the constant 12 in this mode
l.
c Show that after 6 months
, the rabbit population is increasing by almost 8 rabbits per month.
d Suggest one reason wh
y this model will stop giving valid results for large enough values of t.P
P
P
Your answer to part b must refer
to the context of the model.Problem-solving | [
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0.01795... |
319
Exponentials and logarithms
5 On Earth, the atmospheric pressur
e, p, in bars can be modelled approximately by the formula
p = e−0.13h where h is the height above sea level in kilometres.
a Use this model to estimate the pr
essure at the top of Mount Rainier, which has an altitude of
4.394 km. (1 mark)
b Demonstrate tha
t dp ___ dh = kp where k is a constant to be found. (2 marks)
c Interpret the significance of the sign of
k in part b. (1 mark)
d This model predicts tha
t the atmospheric pressure will change by s % for ev
ery
kilometre gained in height. Calculate the value of s. (3 marks)
6 Nigel has bought a tractor f
or £20 000. He wants to model the depr
eciation of the value of his
tractor, £T, in t years. His friend suggests two models:
Model 1: T = 20 000e −0.24t
Model 2: T = 19 000e −0.255t + 1000
a Use both models to predict the v
alue of the tractor after one year.
Compare your results. (2 marks)
b Use both models to predict the v
alue of the tractor after ten years.
Compare your results. (2 marks)
c Sketch a gra
ph of T against t for both models. (2 marks)
d Interpret the meaning of the 1000 in mode
l 2, and suggest why this might make
model 2 more realistic. (1 mark)E/P
E/P
14.4 Logarithms
The inverses of exponential functions are called logarithms. A relationship which is expressed using
an exponent can also be written in terms of logarithms.
■ loga n = x is equivalent to ax = n (a ≠ 1) a is ca lled the
base of the logarithm.Notation
Example 7
Rewrite each statement using a power.
a log3 81 = 4 b log2 ( 1 _ 8 ) = −3
a log3 81 = 4, so 34 = 81
b log2 ( 1 __ 8 ) = −3, so 2−3 = 1 __ 8 Example 6
Write each statement as a logarithm.
a 32 = 9 b 27 = 128 c 6 4 1 _ 2 = 8
a 32 = 9, so log3 9 = 2
b 27 = 128, so log2 128 = 7
c 6 4 1 __ 2 = 8, so log64 8 = 1 __ 2 Logarithms can take fractional or negative values.In words, you would say ‘the logarithm of 9 to the
base 3 is 2’. | [
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0.00000... |
320
Chapter 14
Example 8
Without using a calculator, find the value of:
a log3 81 b log4 0.25 c log0.5 4 d loga (a5)
a log3 81 = 4
b log4 0.25 = −1
c log0.5 4 = −2
d loga (a5) = 5Because 34 = 81.
Because 4−1 = 1 _ 4 = 0 .25.
Because 0.5−2 = ( 1 _ 2 ) −2 = 22 = 4.
Because a5 = a5.
You can use your calculator to find logarithms of
any base. Some calculators have a specific
log key for this function. Most calculators
also have separate buttons for logarithms to the
base 10 (usually written as log and logarithms
to the base e (usually written as ln ).
Example 9
Use your calculator to find the following logarithms to 3 decimal places.
a log3 40 b loge 8 c log10 75
a 3.358
b 2.
079
c 1.
875For part a use log .
For part b you can use either ln or log .
For part c you can use either log or log .
Exercise 14D
1 Rewrite using a logarithm.a
44 = 256 b 3−2 = 1 _ 9 c 106 = 1 000 000
d 111 = 11 e (0.2)3 = 0.008
2 Rewrite using a po
wer.
a log2 16 = 4 b log5 25 = 2 c log9 3 = 1 _ 2
d log5 0.2 = − 1 e log10 100 000 = 5 Log arithms to the base e are
typically called natural logarithms . This is why
the calculator key is labelled ln .Notation
Use the logarithm buttons on
yo
ur calculator.Online | [
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0.0270... |
321
Exponentials and logarithms
3 Without using a calcula
tor, find the value of
a log2 8 b log5 25 c log10 10 000 000 d log12 12
e log3 729 f log10 √ ___ 10 g log4 (0.25) h log0.25 16
i loga (a10) j lo g 2 _ 3 ( 9 _ 4 )
4 Without using a calcula
tor, find the value of x for which
a log5 x = 4 b logx 81 = 2 c log7 x = 1
d log2 (x − 1) = 3 e log3 (4x + 1) = 4 f logx (2x) = 2
5 Use your calcula
tor to evaluate these logarithms to three decimal places.
a log9 230 b log5 33 c log10 1020 d loge 3
6 a Without using a calcula
tor, justify why the value
of log2 50 must be between 5 and 6.
b Use a calculator to find the e
xact value of
log2 50 to 4 significant figures.
7 a Find the values of:
i log2 2 ii log3 3 iii log17 17
b Explain why lo
ga a has the same va lue for all positive values of a (a ≠ 1).
8 a Find the values of:i
log2 1 ii log3 1 iii log17 1
b Explain why lo
ga 1 has the same value f or all positive values of a (a ≠ 1).P Use corresponding
sta
tements involving powers of 2.Hint
14.5 Laws of logarithms
Expressions involving more than one logarithm can often be rearranged or simplified. For instance:
loga x = m and loga y = n
x = am and y = an
xy = am × an = am + n
loga xy = m + n = loga x + loga y
This result is one of the laws of logarithms.
You can use similar methods to prove two further laws.
■ The laws o
f logarithms:
• loga x + loga y = loga xy (the multiplication law)
• loga x − loga y = loga ( x __ y ) (the division law)
• loga (xk) = k loga x (the power la w)
■ You should also l
earn to recognise the following special cases:
• loga ( 1 __ x ) = loga (x−1) = −loga x (the power la w when k = −1)
• loga a = 1 (a > 0, a ≠ 1)
• loga 1 = 0 (a > 0, a ≠ 1) You n eed to learn
these three laws of logarithms,
and the special cases below.Watch outTake two logarithms with the same base
Rewrite these expressions using powersMultiply these powersRewrite your result using logarithms | [
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322
Chapter 14
Example 10
Write as a single logarithm.
a log3 6 + log3 7 b log2 15 − log2 3 c 2log5 3 + 3log5 2 d log10 3 − 4log10 ( 1 _ 2 )
a log3 (6 × 7)
= l
og3 42
b log2 (15 ÷ 3)
=
log2 5
c 2 lo
g5 3 = log5 (32) = log5 9
3 lo
g5 2 = log5 (23) = log5 8
log5 9 + log5 8 = log5 72
d 4 lo
g10 ( 1 __ 2 ) = log10 ( 1 __ 2 ) 4
= log10 ( 1 ___ 16 )
log10 3 − log10 ( 1 ___ 16 ) = log10 (3 ÷ 1 ___ 16 )
= log10 48Use the multiplication law.
Use the division law.
First apply the power law to both parts of the
expression.Then use the multiplication law.
Use the power law first.Then use the division law.
Example 11
Write in terms of loga x, loga y and loga z.
a loga (x2yz3) b loga ( x __ y3 ) c loga ( x √ __ y ____ z ) d loga ( x __ a4 )
a loga (x2yz3)
= loga (x2) + loga y + loga (z3)
= 2 loga x + loga y + 3 loga z
b loga ( x ___ y3 )
= loga x − loga (y3)
= loga x − 3 loga y
c loga ( x √ __ y ____ z )
= loga (x √ __ y ) − loga z
= loga x + loga √ __ y − loga z
= loga x + 1 __ 2 loga y − l oga z
d loga ( x ___ a4 )
= loga x − loga (a4)
= loga x − 4 loga a
= loga x − 4Use the power law ( √ __ y = y 1 _ 2 ).
loga a = 1. | [
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0.007... |
323
Exponentials and logarithms
Example 12
Solve the equation log10 4 + 2 log10 x = 2.
log10 4 + 2 log10 x = 2
log10 4 + log10 x2 = 2
log10 4x2 = 2
4x2 = 102
4x2 = 100
x2 = 25
x = 5Use the power law.
Use the multiplication law.
Rewrite the logarithm using powers.
log10 x is only defined for positive
v
alues of x , so x = − 5 cannot be a solution of the
equation. Watch out
Example 13
Solve the equation log3 (x + 11) − log3 (x − 5) = 2
log3 (x + 11) − log3 (x − 5) = 2
log3 ( x + 11 _______ x − 5 ) = 2
x +
11 _______ x − 5 = 32
x + 11 = 9 (x − 5)
x + 11 = 9 x − 45
56 = 8 x
x = 7Use the division law.
Rewrite the logarithm using powers.
Exercise 14E
1 Write as a single logarithm.
a log2 7 + log2 3 b log2 36 − log2 4 c 3 log5 2 + log5 10
d 2 log6 8 − 4 log6 3 e log10 5 + log10 6 − log10 ( 1 _ 4 )
2 Write as a single logarithm, then simplify y
our answer.
a log2 40 − log2 5 b log6 4 + log6 9 c 2 log12 3 + 4 log12 2
d log8 25 + log8 10 − 3 log8 5 e 2 log10 2 − (log10 5 + log10 8)
3 Write in terms of
loga x, loga y and loga z.
a loga (x3y4z) b loga ( x5
__ y2 ) c loga (a2x2)
d loga ( x ____ √ __
y z ) e loga √ ___ ax | [
0.09157591313123703,
0.06256301701068878,
-0.011818905360996723,
0.025537481531500816,
-0.05482561141252518,
-0.013487119227647781,
-0.01238163560628891,
-0.07130257785320282,
-0.010050013661384583,
0.12389561533927917,
-0.05206602066755295,
-0.07711072266101837,
-0.05300439894199371,
0.01... |
324
Chapter 14
4 Solve the follo
wing equations:
a log2 3 + log2 x = 2 b log6 12 − log6 x = 3
c 2 log5 x = 1 + log5 6 d 2 log9 (x + 1) = 2 log9 (2x − 3) + 1
5 a Given tha
t log3 (x + 1) = 1 + 2 log3 (x − 1), show that 3x2 − 7x + 2 = 0. (5 marks)
b Hence, or otherwise, solv
e log3 (x + 1) = 1 + 2 log3 (x − 1). (2 marks)
6 Given tha
t a and b are positive constants, and
that a > b, solve the simultaneous equations
a
+ b = 13
log6 a + log6 b = 2 Move the logarithms
on
to the same side if necessary
and use the division law.Hint
P
P
Pay careful attention to the conditions on
a and b given in the question.Problem-solving
By writing loga x = m and loga y = n, prove that loga x − loga y = loga ( x __ y ) .Challenge
14.6 Solving equations using logarithms
You can use logarithms and your calculator to solve equations of the form ax = b.
Example 14
Solve the following equations, giving your answers to 3 decimal places.
a 3x = 20 b 54x − 1 = 61
a 3x = 20,
so x = log3 20 = 2.727
b 54x − 1 = 61, so 4 x − 1 = log5 61
4x = log5 61 + 1
x = log5 61 + 1 ___________ 4
= 0.
889Use the log button on your calculator.
You can evaluate the final answer in one step on
your calculator.
Example 15
Solve the equation 52x − 12(5x) + 20 = 0, giving your answer to 3 significant figures.
52x − 12(5x) + 20 is a quadratic function of 5x
(5x − 10)(5x − 2) = 0
5x = 10 or 5x = 2
5x = 10 ⇒ x = log510 ⇒ x = 1.43
5x = 2 ⇒ x = log5 2 ⇒ x = 0.431An alternative method is to rewrite the equation using the substitution y = 5
x: y2 − 12 y + 20 = 0.
Sol ving the quadratic equation gives
you two possible values for 5x. Make sure you
calculate both corresponding values of x for your final answer.Watch out | [
0.01271229051053524,
0.13110095262527466,
-0.0567329004406929,
-0.0333261676132679,
0.014966117218136787,
0.03574111685156822,
-0.05198757350444794,
0.07292289286851883,
-0.007001521997153759,
0.059588730335235596,
0.029257820919156075,
-0.05803460627794266,
-0.05011863633990288,
0.0055902... |
325
Exponentials and logarithms
You can solve more complicated equations by ‘taking logs’ of both sides.
■ Whenever f( x) = g( x), loga f(x) = loga g(x)
Example 16
Find the solution to the equation 3x = 2x + 1, giving your answer to four decimal places.
3x = 2x + 1
log 3x = log 2x + 1
x log 3 = (x + 1) log 2
x lo
g 3 =
x log 2 +
log 2
x lo
g 3 −
x log 2 =
log 2
x (l
og 3 − l
og 2) = log 2
x = log 2 ____________ log 3 − l og 2 = 1.7095This step is called ‘taking logs of both sides’. The
logs on both sides must be to the same base .
Here ‘log’ is used to represent log10.
Move all the terms in x to one side then factorise.Use the power law.
Exercise 14F
1 Solve, giving your answers to 3 significant figures.
a 2x = 75 b 3x = 10 c 5x = 2 d 42x = 100
e 9x + 5 = 50 f 72x − 1 = 23 g 113x − 2 = 65 h 23 − 2x = 88
2 Solve, gi
ving your answers to 3 significant figures.
a 22x − 6(2x) + 5 = 0 b 32x − 15(3x) + 44 = 0
c 52x − 6(5x) − 7 = 0 d 32x + 3x + 1 − 10 = 0
e 72x + 12 = 7x + 1 f 22x + 3(2x) − 4 = 0
g 32x + 1 − 26(3x) − 9 = 0 h 4(32x + 1) + 17(3x) − 7 = 0
3 Solve the follo
wing equations, giving your answers to 3 significant figures where appropriate.
a 3x + 1 = 2000 (2 marks)
b log5 (x − 3) = −1 (2 marks)
4 a Sketch the gra
ph of y = 4x, stating the coordinates
of any points where the graph crosses the axes. (2 marks)
b Solve the equation 42x − 10(4x) + 16 = 0. (4 marks)
5 Solve the follo
wing equations, giving your answers to four decimal places.
a 5x = 2x + 1 b 3x + 5 = 6x c 7x + 1 = 3x + 2 3x + 1 = 3x × 31 = 3(3x) Hint
E
E/P
Attempt this question
wi
thout a calculator.Hint
Take logs of both sides. HintConsider these equations as functions
of functions. Part a is equivalent to u
2 − 6u + 5 = 0, with u = 2x.Problem-solving | [
0.08601774275302887,
0.13742302358150482,
0.025898804888129234,
0.019976604729890823,
-0.05205421894788742,
-0.01961282454431057,
-0.009186477400362492,
-0.053081266582012177,
0.00746889365836978,
0.06137632578611374,
-0.001987326890230179,
-0.10183223336935043,
-0.11204242706298828,
0.033... |
326
Chapter 14
14.7 Working with natural logarithms
■ The graph of y = ln x is a reflection of the graph y = ex in the line y = x.
The graph of y = ln x pa
sses through (1, 0) and does not
cro
ss the y-axis.
The y-axis is an asymptote of the graph y = ln x. This means
that ln x is only defined for po
sitive values of x.
As x increases, ln x gro
ws without limit, but relatively slowly.
You can also use the fact that logarithms are the inverses of
exponential functions to solve equations involving powers and logarithms.
■ eln x = ln (ex) = xy = ex y = x
y = ln x
1
1y
x O
Example 17
Solve these equations, giving your answers in exact form.
a e x = 5 b ln x = 3
a When e x = 5
ln (e x) = ln 5
x
= ln 5
b When ln x = 3
eln x = e3
x = e3The inverse operation of raising e to the power x is
taking natural logarithms (logarithms to the base
e) and vice versa.
You can write the natural logarithm on both sides. ln
(e x) = x
Leave your answer as a logarithm or a power of e so that it is exact.
Example 18
Solve these equations, giving your answers in exact form.
a e2x + 3 = 7 b 2 ln x + 1 = 5 c e2x + 5e x = 14
a e2x + 3 = 7
2x
+ 3 = ln 7
2x = l
n 7 − 3
x = 1 __ 2 ln 7 − 3 __ 2
b 2 ln x + 1 = 5
2 ln x = 4
ln x = 2
x = e2Take natural logarithms of both sides and use the
fact that the inverse of ex is ln x.
Rearrange to make ln x th e subject.
The inverse of ln x is ex. ln x = log ex Notation | [
0.07970160990953445,
0.049377817660570145,
0.04991794005036354,
0.06696084141731262,
-0.03220294788479805,
-0.046779483556747437,
-0.03368818014860153,
0.05588982626795769,
0.018349705263972282,
0.048857975751161575,
0.11100362986326218,
0.06473139673471451,
0.012540427967905998,
0.0770938... |
327
Exponentials and logarithms
c e2x + 5ex = 14
e2x + 5ex − 14 = 0
(ex + 7)(ex − 2) = 0
ex = −7 o r ex = 2
ex = 2
x = ln 2e2x = (ex)2, so this is a quadratic function of ex.
Start by setting the equation equal to 0 and
factorise. You could also use the substitution u = e
x and write the equation as u2 + 5u − 14 = 0.
ex is always positive, so you can’t
have ex = −7. You need to discard this solution.Watch out
Exercise 14G
1 Solve these equations, giving your answers in exact form.
a ex = 6 b e2x = 11 c e−x + 3 = 20
d 3e4x = 1 e e2x + 6 = 3 f e5 − x = 19
2 Solve these equations
, giving your answers in exact form.
a ln x =
2 b ln (4x
) = 1 c ln (2x
+ 3) = 4
d 2 ln (6
x − 2) = 5 e ln (18 −
x) = 1 _ 2 f ln (x2 − 7x + 11) = 0
3 Solve these equations
, giving your answers in exact form.
a e2x − 8ex + 12 = 0 b e4x − 3e2x = −2
c (ln x)2 + 2 ln x − 15 = 0 d ex − 5 + 4e−x = 0
e 3e2x + 5 = 16ex f (ln x)2 = 4(ln x + 3)
4 Find the exact solutions to the equation e x + 12e−x = 7. (4 marks)
5 Solve these equations
, giving your answers in exact form.
a ln (8x
− 3) = 2 b e5(x − 8) = 3 c e10x − 8e5x + 7 = 0
d (ln x −
1)2 = 4
6 Solve 3xe4x − 1 = 5, giving your answer in the form a + ln b _______ c + ln d
(5 marks)
7 Officials are testing a
thletes for doping at a sporting event. They model the concentration
of a particular drug in an athlete’s bloodstream using the equation D = 6 e −t __ 10 where D is the
concentration of the drug in mg/l and t is the time in hours since the athlete took the drug.
a Interpret the meaning of the constant 6 in this mode
l.
b Find the concentration of
the drug in the bloodstream after 2 hours.
c It is impossible to detect this drug in the b
loodstream if the concentration is lower than 3 mg/l.
Show tha
t this happens after t = −10 ln ( 1 _ 2 ) and convert this result into hours and minutes. All of the equations in question
3 ar
e quadratic equations in a
function of x .Hint
First in part d multiply each
te
rm by ex.Hint
E/P
E/P Take natural logarithms of both
si
des and then apply the laws of
logarithms.Hint
P | [
0.016947656869888306,
0.0563775897026062,
0.027611611410975456,
-0.055920276790857315,
-0.005833159200847149,
0.03826926648616791,
0.014503901824355125,
-0.012853797525167465,
-0.017352504655718803,
0.024254629388451576,
-0.01762135699391365,
-0.09600500762462616,
-0.04952447488903999,
0.0... |
328
Chapter 14
14.8 Logarithms and non-linear data
Logarithms can also be used to manage and explore non-linear trends in data.
■ If y = axn then the graph of log y against log x will be a straight line with gradient n and
vertical intercept log a.
log xOlog y
log a8 The graph of y = 3 + ln (4 − x) is shown to the right.
y = 3 + ln (4 – x)
BAy
x Oa State the exact coordinates of point A. (1 mark)
b Calculate the e
xact coordinates of point B. (3 marks)E/P
The graph of the function g( x) = A eBx + C passes through (0, 5) an d (6, 10) .
Given that the line y = 2 is an asymptote to the graph, show that B = 1 _ 6 ln ( 8 _ 3 ) .Challenge
Start with a non-linear relationship y = axn
Take logs of both sides (log = log10) log y = log axn
Use the multiplication law log y = log a + log xn
Use the power law log y = log a + n log x
Co
mpare this equation to the common form of a straight line, Y = MX + C .
log y
variable =n
constant
(gradient)log x
variable +log a
constant
(intercept)
Y
variable =M
constant
(gradient)X
variable +C
constant
(intercept)Case 1: y = ax n | [
0.04548950120806694,
0.07693807035684586,
-0.051012031733989716,
0.01486165076494217,
0.007791139185428619,
-0.034324489533901215,
-0.09149325639009476,
0.010571146383881569,
-0.00910615548491478,
0.0651603639125824,
0.05832807719707489,
-0.04007000848650932,
-0.022900497540831566,
0.05024... |
329
Exponentials and logarithms
Example 19
The table below gives the rank (by size) and population of the UK’s largest cities and districts
(London is ranked number 1 but has been excluded as an outlier).
City Birmingham Leeds Glasgow Sheffield Bradford
Rank, R 2 3 4 5 6
Population, P (2 s.f.) 1 000 000 730 000 620 000 530 000 480 000
The relationship between the rank and population can be modelled by the formula
R = aP n where a and n are constants.
a Draw a ta
ble giving values of log R and log P to 2 decimal places
.
b Plot a graph of
log R against lo
g P using the va
lues from your table and draw a line of best fit.
c Use your gra
ph to estimate the values of a and n to two significant figures.
alog R0.30 0.48 0.60 0.70 0.78
log P 6 5.86 5
.79 5.72 5.68
b
00.1 0.3 0.5 log Rlog P
0.7 0.9 0.2 0.4 0.6 0.85.05.25.45.65.86.06.26.4
c R = aP n
log R = log a(P n)
log R =
log a +
log( P n)
log R =
log a +
n log P
so t
he gradient is n and the intercept is log a
Rea
ding the gradient from the graph,
n = 5.68 − 6.16 ____________ 0.77 − 0.05 = −0.48 ______ 0.72 = −0.67
Reading the intercept from the graph,
log a =
6.2
a = 106.2 = 1 600 000 (2 s.f.).Start with the formula given in the question. Take
logs of both sides and use the laws of logarithms to rearrange it into a linear relationship between log
R an
d log P.
The gradient of the line of best fit will give you your value for n .
The y -intercept will give you the value of log a.
You n
eed to raise 10 to this power to find the
value of a . | [
0.12025459855794907,
0.04562569409608841,
-0.04132251441478729,
-0.015959633514285088,
-0.0355859249830246,
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-0.04174921661615372,
0.059487905353307724,
-0.015221820212900639,
0.0893004760146141,
-0.009510914795100689,
-0.005347810219973326,
-0.09322647005319595,
0.006... |
330
Chapter 14
■ If y = ab x then the graph of log y against x will be a straight line with gradient log b and
vertical intercept log a.
xlog y
log a
O For y = ab x you
need to plot log y ag
ainst x
to obtain a linear graph.
If you plot log y ag
ainst
log x yo
u will not get a
linear relationship.Watch out
Example 20
The graph represents the growth of a population of bacteria,
tlog P
O2
P, over t hours. The graph has a gradient of 0.6 and meets
the vertical axis at (0, 2) as shown.
A scientist suggests that this gr
owth can be modelled by
the equation P = abt, where a and b are constants to be found.
a Write down an equa
tion for the line.
b Using your answ
er to part a or otherwise, find the values
of a and b, giving them to 3 significant figures where necessary.
c Interpret the meaning of the constant
a in this model.
a log P = 0.6t + 2
b P =
100.6t + 2
P = 100.6t × 102
P = 102 × (100.6)t
P = 100 × 3.98t
a = 100, b = 3.98 (3 s.f.)
c The v
alue of a gives the initial size of the
bacteria population.log P = (gradient) × t + ( y-i ntercept)
Rewrite the logarithm as a power. An alternative
method would be to start with P = abt and take
logs of both sides, as in Example 19.
Rearrange the equation into the form abt. You can
use xmn = (xm)n to write 100.6t in the form bt.Start with a non-linear relationship y = ab x
Take logs of both sides (log = log10) log y = log ab x
Use the multiplication law log y = log a + log b x
Use the power law log y = log a + x log b
Co
mpare this equation to the common form of a straight line, Y = MX + C .
log y
variable =log b
constant
(gradient)x
variable +log a
constant
(intercept)
Y
variable =M
constant
(gradient)X
variable +C
constant
(intercept)Case 2: y = ab x | [
0.024222899228334427,
0.06282208859920502,
-0.035492926836013794,
-0.0064760856330394745,
0.0003259084769524634,
-0.028940461575984955,
-0.04683418199419975,
0.030342422425746918,
0.06507482379674911,
0.09648939967155457,
0.06238650530576706,
-0.03303369879722595,
-0.061348799616098404,
0.... |
331
Exponentials and logarithms
Exercise 14H
1 Two variables, S and x satisfy the formula S = 4 × 7x.
a Show that lo
g S =
log 4 +
x log 7.
b The straight line gra
ph of log S against
x is plotted. Write down the gradient and the value
of the intercept on the vertical axis.
2 Two v
ariables A and x satisfy the formula A = 6x 4.
a Show that lo
g A =
log 6 +
4 log x.
b The straight line gra
ph of log A against lo
g x is plotted. Write do
wn the gradient and the
value of the intercept on the vertical axis.
3 The data belo
w follows a trend of the form y = axn, where a and n are constants.
x 3 5 8 10 15
y 16.3 33.3 64.3 87.9 155.1
a Copy and complete the table of values of log x and log y, giving your ans wers to 2 decimal places.
log x 0.48 0.70 0.90 1 1.18
log y 1.21 2.19
b Plot a graph of log y against lo g x and dra w in a line of best fit.
c Use your gra
ph to estimate the values of a and n to one decimal place.
4 The data belo
w follows a trend of the form y = abx, where a and b are constants.
x 2 3 5 6.5 9
y 124.8 424.4 4097.0 30 763.6 655 743.5
a Copy and complete the table of values of x and log y, giving your ans wers to 2 decimal places.
x 2 3 5 6.5 9
log y 2.10
b Plot a graph of log y against x and draw in a line of best fit.
c Use your gra
ph to estimate the values of a and b to one decimal place.
5 Kleiber’s law is an empirica
l law in biology which connects the mass of an animal, m, to its
resting metabolic rate, R. The law follows the form R = amb, where a and b are constants.
The table below contains data on five animals.
Animal Mouse Guinea pig Rabbit Goat Cow
Mass, m (kg) 0.030 0.408 4.19 34.6 650
Metabolic rate
R (kcal per day) 4.2 32.3 195 760 7637
a Copy and complete this table giving values of log R and log m to 2 decimal places . (1 mark)
log m −1.52
log R 0.62 1.51 2.29 2.88 3.88E | [
0.04189661517739296,
0.09786305576562881,
-0.039515189826488495,
-0.004359736572951078,
-0.035148512572050095,
-0.006265555042773485,
-0.015506182797253132,
-0.0007047535036690533,
0.02532258629798889,
0.04892211779952049,
0.04438153654336929,
-0.06451766937971115,
-0.1034957692027092,
0.0... |
332
Chapter 14
b Plot a graph of
log R against lo
g m using the v
alues from your table and draw in a
line of best fit. (2 marks)
c Use your gra
ph to estimate the values of a and b to two significant figures. (4 marks)
d Using your va
lues of a and b, estimate the resting metabolic rate of a human male
with a mass of 80 kg. (1 mark)
6 Zipf’s la
w is an empirical law which relates how frequently a word is used, f, to its ranking in a
list of the most common words of a language, R. The law follows the form f = ARb, where A
and b are constants to be found.
The table below contains data on four words.
Word ‘the’ ‘it’ ‘well’ ‘detail’
Rank, R 1 10 100 1000
Frequency per
100 000 w
ords, f4897 861 92 9
a Copy and complete this table giving values of log f to 2 decimal places .
log R 0 1 2 3
log f 3.69
b Plot a graph of log f against lo g R using the values fr om your table and draw in a line of best fit.
c Use your gra
ph to estimate the value of A to two significant figures and the value of b to
one significant figure.
d The word ‘w
hen’ is the 57th most commonly used word in the English language. A trilogy of
novels contains 455 125 words
. Use your values of A and b to estimate the number of times
the word ‘when’ appears in the trilogy.
7 The table be
low shows the population of Mozambique between 1960 and 2010.
Year 1960 1970 1980 1990 2000 2010
Population, P (millions)7.6 9.5 12.1 13.6 18.3 23.4
This data can be modelled using an exponential function of the form P = abt, where t is the
time in years since 1960 and a and b are constants.
a Copy and complete the tab
le below.
Time in years
since 1960, t0 10 20 30 40 50
log P 0.88
b Show that P = abt can be rearranged into the form log P = log a + t log b.
c Plot a graph of
log P against t using the values from your table and draw in a line of
best fit.
d Use your gra
ph to estimate the values of a and b.
e Explain why an e
xponential model is often appropriate
for modelling population growth.P
For part e , thi nk about the
relationship between P and dP ___ dt .Hint | [
0.02005748450756073,
0.026336053386330605,
-0.040862325578927994,
-0.018776247277855873,
-0.01107512041926384,
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-0.009679335169494152,
0.09397448599338531,
0.009069261141121387,
0.09011393785476685,
0.015747489407658577,
-0.06631612777709961,
0.007764824200421572,
-0.... |
333
Exponentials and logarithms
8 A scientist is modelling the number of
people, N, who have fallen sick with a virus after t days.
tlog N
O1.6(10, 2.55)
From looking at this graph, the scientist suggests that the number of sick people can be
modelled by the equation N = abt, where a and b are constants to be found.
The graph passes through the points (0, 1.6) and (10, 2.55).
a Write down the equa
tion of the line. (2 marks)
b Using your answ
er to part a or otherwise, find the values of a and b, giving
them to 2 significant figures. (4 marks)
c Interpret the meaning of the constant
a in this model. (1 mark)
d Use your model to pr
edict the number of sick people after 30 days.
Give one reason why this might be an overestimate. (2 marks)
9 A student is investiga
ting a family of similar shapes. She measures the width, w, and the
area, A, of each shape. She suspects there is a formula of the form A = pw q, so she plots the
logarithms of her results.
log A
Olog w
–0.1049
The graph has a gradient of 2 and passes through −0.1049 on the vertical axis.a
Write down an equa
tion for the line.
b Starting with your answ
er to part a, or otherwise, find the exact value of q and the value of p
to 4 decimal places.
c Suggest the name of the family of
shapes that the
student is investigating, and justify your answer.E/P
P
Multiply p by 4 and think about
an
other name for ‘half the width’.Hint
Find a formula to describe the relationship between the data in
this table.
x 1 2 3 4
y 5.22 4.698 4.2282 3.805 38Challenge Sketch the graphs of log y
ag
ainst log x an
d log y ag
ainst x .
This will help you determine whether the relationship is of the form y = ax
n or y = abx.Hint | [
0.05219860374927521,
0.05805150419473648,
-0.04002410173416138,
0.02996273711323738,
0.013453750871121883,
-0.12710009515285492,
-0.011712493374943733,
0.023528827354311943,
0.04622252285480499,
0.016523826867341995,
0.01593511551618576,
-0.022541353479027748,
-0.049103908240795135,
0.0553... |
334
Chapter 14
1 Sketch each of the f
ollowing graphs, labelling all intersections and
asymptotes.
a y =
2−x b y = 5ex − 1 c y = ln x
2 a Express log a ( p2q) in terms of log a p and log a q.
b Given tha
t log a ( pq) = 5 and log a ( p2q) = 9, find the values of log a p and log a q.
3 Given tha
t p = logq 16, express in terms of p,
a logq 2
b logq (8q)
4 Solve these equations
, giving your answers to 3 significant figures.
a 4x = 23 b 72x + 1 = 1000 c 10x = 6x + 2
5 a Using the substitution u = 2x, show that the equation 4x − 2x + 1 − 15 = 0 can be written in the
form u2 − 2u − 15 = 0. (2 marks)
b Hence solve the equation 4x − 2x + 1 − 15 = 0, giving your answer to
2 decimal places. (3 marks)
6 Solve the equation lo
g2 (x + 10) − log2 (x − 5) = 4. (4 marks)
7 Differentiate each of
the following expressions with respect to x.
a e−x b e11x c 6e5x
8 Solve the following equations, giving exact solutions.a
ln (2x
− 5) = 8 b e4x = 5 c 24 − e−2x = 10
d ln x +
ln (x
− 3) = 0 e ex + e−x = 2 f ln 2 + ln x = 4
9 The price of a computer system can be modelled b
y the formula
P = 100 + 850 e − t _ 2
where
P is the price of the system in £s and t is the age of the computer in years after being
purchased.
a Calculate the ne
w price of the system.
b Calculate its price after 3 y
ears.
c When will it be worth less than £200?
d Find its price as t
→ ∞.
e Sketch the gra
ph showing P against t.
f Comment on the appropria
teness of this model. Recall that
2−x = (2−1)x = ( 1 _ 2 ) x
Hint
P
P
E/P
E
PMixed exercise 14 | [
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335
Exponentials and logarithms
10 The points P and
Q lie on the curve with equation y = e 1 _ 2 x .
The x-coordinates of P and Q are ln 4 and ln 16 respectiv
ely.
a Find an equation for the line
PQ.
b Show that this line passes thr
ough the origin O.
c Calculate the length, to 3 significant figur
es, of the line segment PQ.
11 The temperatur
e, T °C, of a cup of
tea is given by T = 55 e − t _ 8 + 20 t > 0
where t is the time in minutes since measurements began.
a Briefly explain w
hy t > 0. (1 mark)
b State the starting tempera
ture of the cup of tea. (1 mark)
c Find the time at which the temper
ature of the tea is 50 °C, giving y
our answer
to the nearest minute. (3 marks)
d By sketching a gra
ph or otherwise, explain why the temperature of the tea will
never fall below 20 °C. (2 marks)
12 The table be
low gives the surface area, S, and the volume, V of five different spheres, rounded
to 1 decimal place.
S 18.1 50.3 113.1 221.7 314.2
V 7.2 33.5 113.1 310.3 523.6
Given that S = aV b, where a and b are constants,
a show that lo
g S =
log a +
b log V. (2 marks)
b copy and complete the tab
le of values of log S and log V, giving y
our answers to
2 decimal places. (1 mark)
log S
log V 0.86
c plot a graph of log V against lo g S and dra w in a line of best fit. (2 marks)
d use your gra
ph to confirm that b = 1.5 and estimate the value of a to
one significant figure. (4 marks)
13 The radioactiv
e decay of a substance is modelled by the formula R = 140ekt t > 0
where
R is a measure of radioactivity (in counts per minute) at time t days, and k is a constant.
a Explain briefly wh
y k must be negative. (1 mark)
b Sketch the gra
ph of R against t. (2 marks)
After 30 days the radia
tion is measured at 70 counts per minute.
c Show that
k = c ln 2, stating the va
lue of the constant c. (3 marks)
14 The total number of
views (in millions) V of a viral video in x days is modelled by
V = e0.4x − 1
a Find the total number of
views after 5 days.
b Find dV ___ dx .P
E/P
E
E/P
P | [
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336
Chapter 14
c Find the rate of
increase of the number of views after 100 days, stating the units of your answer.
d Use your answ
er to part c to comment on the validity of the model after 100 days.
15 The moment magnitude scale is used b
y seismologists to express the sizes of earthquakes.
The scale is calculated using the formula
M = 2 _ 3 log10(S ) − 10.7
where
S is the seismic moment in dyne cm.
a Find the magnitude of an earthqua
ke with a seismic moment of 2.24 × 1022 dyne cm.
b Find the seismic moment of an earthquak
e with
i magnitude 6 ii magnitude 7
c Using your answ
ers to part b or otherwise, show that an earthquake of magnitude 7 is
approximately 32 times as powerful as an earthquake of magnitude 6.
16 A student is asked to solve the equa
tion
log2 x − 1 _ 2 log2 (x + 1) = 1
The student’s attempt is sho
wn
log2 x − log2 √ _____ x + 1 = 1
x
− √ _____ x + 1 = 21
x − 2 = √ _____ x + 1
(x
− 2)2 = x + 1
x2 − 5 x + 3 = 0
x
= 5 + √ ___ 13 ________ 2 x = 5 − √ ___ 13 ________ 2
a Identify the error made by the student. (1 mark)
b Solve the equation corr
ectly. (3 marks)P
E/P
1 For all real values of x:
• If f(x
) = ex then f9(x) = ex
• If y = ex then dy ___ dx = ex
2 For all real values of x and for any constant k:
• If f(x
) = ekx then f9(x) = kekx
• If y = ekx then dy ___ dx = kekxSummary of key pointsa Given that y = 9x, show that log3 y = 2 x.
b Henc
e deduce that log3 y = log9 y2.
c Use y
our answer to part b to solve the equation log3(2 − 3x) = log9(6x2 − 19x + 2)Challenge | [
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... |
337
Exponentials and logarithms
3 loga n = x is equivalent to ax = n (a ≠ 1)
4 The laws o
f logarithms:
• loga x + loga y = loga xy (the multiplication law)
• l
oga x − loga y = loga ( x __ y ) (the division law)
• l
oga (x k) = k loga x (the power la w)
5 You should also l
earn to recognise the following special cases:
• loga ( 1 __ x ) = loga (x −1) = −loga x (the power la w when k = −1)
• loga a = 1 (a > 0, a ≠ 1)
• loga 1 = 0 (a > 0, a ≠ 1)
6 Whenever f(x
) = g(x), loga f(x) = loga g(x)
7 The graph of
y = ln x is a refl
ection of the graph y = ex y = ex y = x
y = ln x
1
1 x O
in the line y = x.
8 eln x = ln (e x) = x
9 If y
= axn then the graph of log y against log x will be a log y
log xa
straight line with gradient n and vertical intercept log a.
10 If y
= ab x then the graph of log y against x will be a
xlog y
log a
O
straight line with gradient log b and ver
tical
intercept log a. | [
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-0... |
338Review exercise3
1 The vector 9i + qj is parallel to the vector
2i
− j. Find the value of the constant q.
(2 marks)
← Section 11.2
2 Given that |5i − kj| = |2ki + 2j| , find the exact
value of the positive constant k. (4 marks)
← Section 11.3
3 Given the four points X(9, 6), Y(13, −2),
Z(0,
−15), and C(1, −3),
a Show that | ⟶ CX | = | ⟶ CY | = | ⟶ CZ | . (3 marks)
b Using your answ
er to part a or
otherwise, find the equation of the circle
which passes through the points X, Y and Z.
(3 marks)
← Sections 6.2, 11.4
4 In the triangle ABC, ⟶ AB = 9 i + 2j and
⟶ AC = 7 i − 6j.
a Find ⟶ BC . (2 marks)
b Prov
e that the triangle ABC is
isosceles. (3 marks)
c Show that cos ∠ABC = 1 ___ √ __
5 (4 marks)
← Sections 9.1, 11.5
5 The vectors a, b and c are given as
a = ( 8 23 ) , b = ( −15 x ) and c = ( −13 2 ) , where x
is an integer. Given that a + b is parallel to
b − c, find the v
alue of x. (4 marks)
← Section 11.2
6 Two forces, F1 and F2, act on a particle.
F1 = 2i − 5j ne wtons
F2 = i + j newtons
The resultant force R acting on the particle is given by R = F
1 + F2.E
E/p
E/p
E/p
E/p
Ea Calculate the magnitude of R in newtons
. (3 marks)
A third for
ce, F3 begins to act on the
particle, where F3 = kj newtons and k is a
positive constant. The new resultant force is given by R
new = F1 + F2 + F3.
b Given tha
t the angle between the line
of action of Rnew and the vector i is 45
degrees, find the value of k. (3 marks)
← Section 11.6
7 A helicopter takes off from its starting position O
and travels 100 km on a bearing
of 060°. It then travels 30 km due east before
landing at point A . Gi
ven that the position
vector of A relative to O is (m i + nj) km, find
the exact values of m and n . (4 marks)
← Sections 10.2, 11.6
8 At the very end of a race, Boat A has a position vector of
(−65i + 180j) m and
Boa
t B has a position vector of (100i +
120j) m. The finish line has a position
vector of 10
i km.
a Show that Boa
t B is closer to the finish
line than Boat A. (2 marks)
Boat A
is travelling at a constant velocity
of (2.5i − 6j) m/s and Boat B
is travelling at
a constant velocity of (−3i − 4j) m/s.
b Calculate the speed of
each boat.
Hence, or otherwise, determine the result of the race.
(4 marks)
← Section 11.6
9 Prove, from first principles, that the deriv
ative of 5x2 is 10x. (4 marks)
← Section 12.2E/p
E/p
E/p | [
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-0... |
339
Review exercise 3
10 Given tha
t y = 4x3 − 1 + 2 x 1 _ 2 , x > 0,
find dy ___ dx . (2 marks)
← Section 12.5
11 The curve C has equation
y = 4x + 3 x 3 _ 2 − 2x2, x > 0.
a Find an expression f
or dy ___ dx (2 marks)
b Show that the point
P(4, 8) lies
on C. (1 mark)
c Show that an equa
tion of the normal to
C at point P is 3y = x + 20. (2 marks)
The normal to
C at P cuts the x-axis at
point Q.
d Find the length PQ, gi
ving your answer
in simplified surd form. (2 marks)
← Section 12.6
12 The curve C has equation
y = 4x2 + 5 − x _____ x , x ≠ 0. The point P on C
has x-coordinate 1.
a Show that the v
alue of dy ___ dx at P is 3.
(3 marks)
b Find an equation of the tangent to
C
at P. (3 marks)
This tangent meets the x-axis a
t the point
(k, 0).
c Find the value of
k. (1 mark)
← Section 12.6
13 f(x) = (2x
+ 1)(x + 4) _____________ √ __
x , x > 0.
a Show that f(
x) can be written in the
form P x 3 _ 2 + Q x 1 _ 2 + R x − 1 _ 2 , stating the
values of the constants P, Q and R.
(2 marks)
b Find f ′(
x). (3 marks)
c A curve has equation
y = f(x). Show
that the tangent to the curve at the
point where x = 1 is parallel to the line with equation 2y = 11x + 3.
(3 marks)
← Section 12.6
14 Prove that the function f(x ) = x3 − 12x2 + 48x
is increasing for all x ∈ ℝ . (3 marks)
← Section 12.7E
E/p
E/p
E/p
E/p15 The diagram shows part of the curve with
equation y = x + 2 __ x − 3 . The curve crosses
the
x-axis at A and B and the point C is
the minimum point of the curve.
B ACOy
x
a Find the coordinates of A and B.
(2 marks)
b Find the exact coordina
tes of C, giving
your answers in surd form. (4 marks)
← Section 12.9
16 A company mak es solid cylinders of
variable radius r cm and constant volume
128π cm3.
a Show that the surface ar
ea of the
cylinder is given by S = 256π _____ r + 2πr2.
(2 marks)
b Find the minium value f
or the surface
area of the cylinder. (4 marks)
← Section 12.11
17 Given that y = 3x2 + 4 √ __
x , x > 0, find
a dy ___ dx (2 marks)
b d2y ____ dx2 (2 marks)
c ∫ydx (3 marks)
← Sec tions 12.8, 13.2
18 The curve C with equation y = f(x) passes
through the point (5, 65).
Given that f ′(x
) = 6x2 − 10x − 12,
a use integration to find f(
x) (3 marks)
b hence show that f(
x) = x(2x + 3)(x − 4)
(2 marks)
c sketch C
, showing the coordinates of
the points where C crosses the x-axis.
(3 marks)
← Sections 4.1, 13.3
19 Use calculus to evaluate ∫
1 8
( x 1 _ 3 − x − 1 _ 3 ) dx.
← Section 13.4E/p
E/p
E
E | [
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340
Review exercise 3
20 Given tha
t ∫
0 6
(x 2 − kx) dx = 0 , find the
value of
the constant k. (3 marks)
← Section 13.4
21 The diagram shows a section of the curve
with equation
y = −x4 + 3x2 + 4. The curve
intersects the x -axis at points A and B . The
finite region R , which is shown shaded, is
bounded by the curve and the x -axis.
O AR
By
x
a Show that the equation −x
4 + 3x2 + 4 = 0 only has two solutions,
and hence or otherwise find the coordinates of A and B.
(3 marks)
b Find the area of the r
egion R.
(4 marks)
← Sections 4.2, 13.5
22 The diagram shows the shaded region T which is bounded b
y the curve
y = (x − 1)(x − 4) and the x-axis. Find the area of the shaded region T.
(4 marks)
O 14Ty
xy = (x – 1)(x – 4)
← Section 13.6
23 The diagram shows the curve with equation
y = 5 − x2 and the line with
equation y = 3 − x. The curve and the line intersect at the points P and Q.
OP
Qy
xE/p
E/p
E
E/pa Find the coordinates of P and Q.
(3 marks)
b Find the area of the finite r
egion
between PQ and the curve. (6 marks)
← Section 13.7
24 The graph of the function f(x) = 3e−x − 1,
x ∈ ℝ , has an asymptote y =
k, and
crosses the x and y axes at A and B respectively, as shown in the diagram.
A
BOy
x
y = k
a Write down the value of k and the y-coor
dinate of A. (2 marks)
b Find the exact va
lue of the
x-coordinate of B, giving your answer as simply as possible.
(2 marks)
← Sections 14.2, 14.7
25 A heated metal ball S is dropped into a liquid. As S
cools, its temperature, T °C,
t min
utes after it enters the liquid, is
given by
T = 400e−0.05t + 25, t > 0.
a Find the temperatur
e of S as it enters
the liquid. (1 mark)
b Find how long S
is in the liquid before
its temperature drops to 300 °C.
Giv
e your answer to 3 significant
figures. (3 marks)
c Find the rate
, dT ___ dt , in °C per minute
to 3 significant figures
, at which the
temperature of S is decreasing at the
instant t = 50. (3 marks)
d With refer
ence to the equation given
above, explain why the temperature of S can never drop to 20
°C. (2 marks)
← Sections 14.3, 14.7E
E/p | [
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341
Review exercise 3
26 a Find, to 3 significant figures, the v
alue
of x for which 5x = 0.75. (2 marks)
b Solve the equation 2lo
g5x − log53x = 1
(3 marks)
← Sections 14.5, 14.6
27 a Solve 32x − 1 = 10, giving your answer to
3 significant figures. (3 marks)
b Solve log2x + log2(9 − 2x) = 2
(3 marks)
← Sections 14.5, 14.6
28 a Express logp12 − ( 1 _ 2 logp9 + 1 _ 3 logp8) as a
single logarithm to base p. (3 marks)
b Find the value of
x in log4x = −1.5.
(2 marks)
← Sections 14.4, 14.5
29 Find the exact solutions to the equations
a ln x +
ln 3 =
ln 6 (2 marks)
b ex + 3e−x = 4 (4 marks)
← Section 14.7
30 The table below shows the population of
Angola between 1970 and 2010.
Year Population, P (millions)
1970 5.93
1980 7.64
1990 10.33
2000 13.92
2010 19.55
This data can be modelled using an exponential function of the form P = ab
t,
where t is the time in years since 1970 and a and b are constants.
a
Copy and complete the tab
le below,
giving your answers to 2 decimal
places. (1 mark)
Time in years since 1970, t log P
0 0.77
10
203040E
E
E/p
E/pb Plot a graph of log P against t using
the va
lues from your table and draw in
a line of best fit. (2 marks)
c By rearranging
P = abt, explain how
the graph you have just drawn supports the assumed model.
(3 marks)
d Use your gra
ph to estimate the values
of a and b to two significant figures.
(4 marks)
← Section 14.8
1 The position vector of a moving object is given
by (
cos θ)i +
(sin θ)j, whe
re 0 < θ < 90°.
a Fin
d the value of θ when the object has a
bearing of 090° from the origin.
b Cal
culate the magnitude of the position
vector. ← Sections 10.2, 10.3, 11.3, 11.4
2 The graph of the cubic function y = f( x) ha s
turning points at ( −3, 76) and (2, − 49).
a Sho
w that f ′(x) =
k(x2 + x − 6), where k is a
constant.
b Expr
ess f( x) in the form ax3 + bx2 + cx + d,
where a , b, c and d are real constants to be
found. ← Sections 12.9, 13.3
3 Given that ∫
0 9 f (x) dx = 24.2, sta te the value of
∫
0 9 (f(x) + 3) dx . ← Sections 4.5, 13.5
4 The functions f and g are defined as
f(x)
= x3 − kx + 1, where k is a constant, and
g(x) = e2x, x ∈ ℝ . The graphs of y = f( x) an d
y = g( x) intersect at the point P , where x = 0.
a Con
firm that f(0) = g(0) and hence state the
coordinates of P .
b Giv
en that the tangents to the graphs at P are
perpendicular, find the value of k .
← Sections 5.3, 14.3Challenge | [
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3423421 a Given tha
t 4 = 64n, find the value of n. (1)
b Write √ ___
50 in the form k √ __
2 where k is an integer to be determined. (1)
2 Find the equation of the line par
allel to 2x − 3y + 4 = 0 that passes through the point (5, 6).
Give your answer in the form y = ax + b where a and b are rational numbers. (3)
3 A student is asked to ev
aluate the integral ∫
1 2
(x4 − 3 ___ √ __
x + 2) dx
The student’s working is shown below
∫
1 2
(x4 − 3 ___ √ __
x + 2) dx = ∫
1 2
(x4 − 3 x 1 __ 2 + 2d x)
= [ x5 ___ 5 − 2 x 3 __ 2 + 2x] 2
1
= ( 1 __ 5 − 2 + 2) − ( 32 ___ 5 − 2 √ __
8 + 4)
= −4
.54 (3 s.f.)
a Identify two errors made b y the student. (2)
b Evalua
te the definite integral, giving your answer correct to 3 significant figures. (2)
4 Find all the solutions in the interva
l 0 < x < 180° of
2sin2(2x) − cos(2x) − 1 = 0
giving each solution in degrees. (7)Pearson Edexcel Level 3 GCE
Mathematics
Advanced Subsidiary
Paper 1: Pure Mathematics
Practice paper
Time: 2 hoursYou must have:
Mathematical Formulae and Statistical TablesCalculator | [
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343
Practice paper
5 A rectangular box has sides measuring
x cm, x
+ 3 cm and 2x cm.
2x cmx cm
x + 3 cm
Figure 1
a Write down an e
xpression for the volume of the box. (1)
Given tha
t the volume of the box is 980 cm3,
b Show that
x3 + 3x2 − 490 = 0. (2)
c Show that
x = 7 is a solution to this equation. (1)
d Prov
e that the equation has no other real solutions. (4)
6 f(x
) = x3 − 5x2 − 2 + 1 ___ x 2
The point P with x-coor
dinate −1 lies on the curve y = f(x). Find the equation of the normal to the
curve at P , giving your answer in the form ax + by + c = 0 where a , b and c are positive integers. (7)
7 The population, P
, of a colony of endangered Caledonian owlet-nightjars can be modelled
by the equation P = abt where a and b are constants and t is the time, in months, since the
population was first recorded.
log10P
t O(0, 2)(20, 2.2)
Figure 2
The line l shown in figure 2 shows the relationship between t and log10P for the population over
a period of 20 years.
a Write down an equa
tion of line l. (3)
b Work out the v
alue of a and interpret this value in the context of the model. (3)
c Work out the v
alue of b, giving your answer correct to 3 decimal places. (2)
d Find the population predicted b
y the model when t = 30. (1)
8 Prov
e that 1 + cos 4 x − sin 4 x 2 cos 2 x. (4)
9 Rela
tive to a fixed origin, point A has position vector 6i − 3j and point B has position
vector 4i + 2j.Find the magnitude of the vector
⟶ AB and the angle it makes with the unit v ector i. (5) | [
-0.015094109810888767,
0.09102751314640045,
-0.008386951871216297,
0.013399960473179817,
-0.07179321348667145,
0.018301865085959435,
-0.028097575530409813,
0.002638252917677164,
-0.0885985791683197,
0.03862061724066734,
0.036716800183057785,
-0.1241481751203537,
-0.01648656092584133,
0.023... |
344
Practice paper
10 A triangular lawn ABC
is shown in figure 3:
B
ACDiagram not
to scale
Figure 3
Given that AB = 7.5 m, BC
= 10.6 m and AC =
12.7 m,
a Find angle BAC . (3)
Grass seed costs £1.25 per square metr
e.
b Find the cost of seeding the whole la
wn. (5)
11 g(x
) = (x − 2)2(x + 1)(x − 7)
a Sketch the curve
y = g(x), showing the coordinates of any points where the curve meets or
cuts the coordinate axes. (4)
b Write down the r
oots of the equation g(x + 3) = 0. (1)
12 Given tha
t 92x = 27 x 2 −5 , find the possible values of x. (6)
13 f(x
) = (1 − 3x)5
a Expand f(x), in ascending po wers of x, up to the term in x2. Give each term in its
simplest form. (3)
b Hence find an appro
ximate value for 0.975. (2)
c State, with a r
eason, whether your approximation is greater or smaller than the true value. (2)
14 f ′(
x) = √ __
x − x 2 − 1 __________ x 2 , x > 0
a Show that f(
x) can be written as f(x) = − x 2 + 2 √ __
x − 1 ____________ x + c where c is a constant. (5)
Given tha
t f(x) passes through the point (3, −1),
b find the value of
c. Give your answer in the form p + q √ _
r where p, q and r are rational
numbers to be found. (4)
15 A circle,
C, has equation x2 + y2 − 4x + 6y = 12
a Show that the point
A(5, 1) lies on C and find the centre and radius of the circle. (5)
b Find the equation of the tangent to
C at point A. Give your answer in the form
y = ax + b where a and b are rational numbers. (4)
c The curve y
= x2 − 2 intersects this tangent at points P and Q. Given that O is the origin,
find, as a fraction in simplest form, the exact area of the triangle POQ. (7) | [
0.06017697602510452,
0.11352264881134033,
-0.0489020049571991,
0.011584916152060032,
-0.0008807019330561161,
0.058262042701244354,
-0.050780996680259705,
-0.03142072260379791,
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0.035160377621650696,
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0.0... |
345
Answers
345Answers
Answers
CHAPTER 1
Prior knowledge check
1 a 2m2n + 3mn2 b 6x2 − 12x − 10
2 a 28 b 24 c 26
3 a 3x + 12 b 10 − 15x c 12x − 30 y
4 a 8 b 2x c xy
5 a 2x b 10x c 5x ___ 3
Exercise 1A
1 a x7 b 6x5 c k d 2p2
e x f y10 g 5x2 h p2
i 2a3 j 2p k 6a9 l 3a2b3
m 27x8 n 24x11 o 63a12 p 32y6
q 4a6 r 6a12
2 a 9x − 18 b x2 + 9x
c −12y
+ 9y2 d xy + 5x
e −3x2 − 5x f −20x2 − 5x
g 4x2 + 5x h −15y + 6y3
i −10x2 + 8x j 3x3 − 5x2
k 4x − 1 l 2x − 4
m 9d2 − 2c n 13 − r2
o 3x3 − 2x2 + 5x p 14y2 − 35y3 + 21y4
q −10y2 + 14y3 − 6y4 r 4x + 10
s 11x − 6 t 7x2 − 3x + 7
u −2x2 + 26x v −9x3 + 23x2
3 a 3x3 + 5x5 b 3x4 − x6 c x3 __ 2 − x
d 4x2 + 5 __ 2 e 7x6 ____ 5 + x f 3x4 − 5x2 ____ 3
Exercise 1B
1 a x2 + 11x + 28
b x2 − x − 6
c x2 − 4x + 4
d 2x2 + 3x − 2xy − 3y
e 4x2 + 11xy − 3y2
f 6x2 − 10xy − 4y2
g 2x2 − 11x + 12
h 9x2 + 12xy + 4y2
i 4x2 + 6x + 16xy + 24y
j 2x2 + 3xy + 5x + 15y − 25
k 3x2 − 4xy − 8x + 4y + 5
l 2x2 + 5x − 7xy − 4y2 − 20y
m x2 + 2x + 2xy + 6y − 3
n 2x2 + 15x + 2xy + 12y + 18
o 13y − 4
x + 12 − 4y2 + xy
p 12xy − 4
y2 + 3y + 15x + 10
q 5xy − 20
y − 2x2 + 11x − 12
r 22y − 4
y2 − 5x + xy − 10
2 a 5x2 − 15x − 20
b 14x2 + 7x − 70
c 3x2 − 18x + 27
d x3 − xy2
e 6x3 + 8x2 + 3x2y + 4xy
f x2y − 4xy − 5y
g 12x2y + 6xy − 8xy2 − 4y2
h 19xy − 35 y − 2x2y
i 10x3 − 4x2 + 5x2y − 2xy
j x3 + 3x2y − 2x2 + 6xy − 8x k 2x2y + 9xy + xy2 + 5y2 − 5y
l 6x2y + 4xy2 + 2y2 − 3xy − 3y
m 2x3 + 2x2y − 7x2 + 3xy − 15x
n 24x3 − 6x2y − 26x2 + 2xy + 6x
o 6x3 + 15x2 − 3x2y − 18xy2 − 30xy
p x3 + 6x2 + 11x + 6
q x3 + x2 − 14x − 24
r x3 − 3x2 − 13x + 15
s x3 − 12x2 + 47x − 60
t 2x3 − x2 − 5x − 2
u 6x3 + 19x2 + 11x − 6
v 18x3 − 15x2 − 4x + 4
w x3 − xy2 − x2 + y2
x 8x3 − 36x2y + 54xy2 − 27y3
3 2x2 − xy + 29x − 7y + 24
4 4x3 + 12x2 + 5x − 6 cm3
5 a = 12, b = 32, c = 3, d = −5
Challenge
x4 + 4x3y + 6x2y2 + 4xy3 + y4
Exercise 1C
1 a 4(x + 2) b 6(x − 4)
c 5(4x
+ 3) d 2(x2 + 2)
e 4(x2 + 5) f 6x (x − 3)
g x(x
− 7) h 2x
(x + 2)
i x(3x
− 1) j 2x
(3x − 1)
k 5y
(2y − 1) l 7x
(5x − 4)
m x(x
+ 2) n y(3y
+ 2)
o 4x
(x + 3) p 5y
( y − 4)
q 3xy
(3y + 4x) r 2ab
(3 − b)
s 5x(
x − 5y) t 4xy
(3x + 2y)
u 5y
(3 − 4z2) v 6(2x2 − 5)
w xy( y −
x) x 4y
(3y − x)
2 a x(x
+ 4) b 2x
(x + 3)
c (x
+ 8)(x + 3) d (x
+ 6)(x + 2)
e (x
+ 8)(x − 5) f (x − 6)(
x − 2)
g (x
+ 2)(x + 3) h (x
− 6)(x + 4)
i (x
− 5)(x + 2) j (x
+ 5)(x − 4)
k (2x
+ 1)(x + 2) l (3x
− 2)(x + 4)
m (5x − 1)(
x − 3) n 2(3x
+ 2)(x − 2)
o (2x
− 3)(x + 5) p 2(x2 + 3)(x2 + 4)
q (x
+ 2)(x − 2) r (x
+ 7)(x − 7)
s (2x + 5)(2
x − 5) t (3x
+ 5y)(3x − 5y)
u 4(3x
+ 1)(3x − 1) v 2(x
+ 5)(x − 5)
w 2(3x − 2)(
x − 1) x 3(5x
− 1)(x + 3)
3 a x(x2 + 2) b x(x2 − x + 1)
c x(x2 − 5) d x(x + 3)(x − 3)
e x(x
− 4)(x + 3) f x(x
+ 5)(x + 6)
g x(x
− 1)(x − 6) h x(x
+ 8)(x − 8)
i x(2x
+ 1)(x − 3) j x(2x
+ 3)(x + 5)
k x(x
+ 2)(x − 2) l 3x(
x + 4)(x + 5)
4 (x2 + y2)(x + y)(x − y)
5 x(3x
+ 5)(2x − 1)
Challenge(x − 1)(x + 1)(2x + 3)(2x − 3)
Exercise 1D
1 a x5 b x−2 c x4 d x3
e x5 f 12x0 = 12 g 3 x 1 __ 2 h 5x
i 6x−1 j x 5 __ 6 k x 17 __ 6 l x 1 __ 6 | [
-0.09458763152360916,
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0.006152608431875706,
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0.01312678400427103,
... |
346
Answers
346 Full worked solutions are available in SolutionBank.
Online
2 a 5 b 729 c 3 d 1 __ 16
e 1 __ 3 f −1 ____ 125 g 1 h 216
i 125 ___ 64 j 9 __ 4 k 5 __ 6 l 64 __ 49
3 a 8x5 b 5 __ x2 − 2 __ x3 c 5x4
d 1 __ x2 + 4 e 2 __ x3 + 1 __ x2 f 8 ___ 27 x6
g 3 __ x − 5x2 h 1 ____ 3x2 + 1 ___ 5x
4 a 3 b 16 ___
3 √ __
x
5 a x __ 2 b 32 ___ x6
Exercise 1E
1 a 2 √ __
7 b 6 √ __
2 c 5 √ __
2 d 4 √ __
2
e 3 √ ___
10 f √ __
3 g √ __
3 h 6 √ __
5
i 7 √ __
2 j 12 √ __
7 k −3 √ __
7 l 9 √ __
5
m 23 √ __
5 n 2 o 19 √ __
3
2 a 2 √ __
3 + 3 b 3 √ __
5 − √ ___ 15
c 4 √ __
2 − √ ___ 10 d 6 + 2 √ __
5 − 3 √ __
2 − √ ___
10
e 6 − 2 √ __
7 − 3 √ __
3 + √ ___ 21 f 13 + 6 √ __
5
g 8 − 6 √ __
3 h 5 − 2 √ __
3
i 3 + 5 √ ___ 11
3 3 √ __
3
Exercise 1F
1 a √ __
5 ___ 5 b √ ___
11 ____ 11 c √ __
2 ___ 2
d √ __
5 ___ 5 e 1 __ 2 f 1 __ 4
g √ ___
13 ____ 13 h 1 __ 3
2 a 1 − √ __
3 _______ −2 b √ __
5 − 2 c 3 + √ __
7 _______ 2
d 3 + √ __
5 e √ __
5 + √ __
3 ________ 2 f (3 − √ __
2 )(4 + √ __
5 ) _______________ 11
g 5( √ __
5 − 2) h 5(4 + √ ___
14 ) i 11(3 − √ ___
11 ) ___________ −2
j 5 − √ ___
21 ________ −2 k 14 − √ ____ 187 __________ 3 l 35 + √ _____ 1189 ___________ 6
m −1
3 a 11 + 6 √ __
2 _________ 49 b 9 − 4 √ __
5 c 44 + 24 √ __
2 __________ 49
d 81 − 30 √ __
2 __________ 529 e 13 + 2 √ __
2 _________ 161 f 7 − 3 √ __
3 ________ 11
4 − 7 __ 4 + √ __
5 ___ 4
Mixed exercise
1 a y8 b 6x7 c 32x d 12b9
2 a x2 − 2x − 15 b 6x2 − 19x −7
c 6x2 − 2xy + 19x − 5y + 103 a x3 + 3x2 − 4x b x3 + 6x2 − 13x − 42
c 6x3 − 5x2 − 17x + 6
4 a 15y + 12 b 15x2 − 25x3 + 10x4
c 16x2 + 13x d 9x3 − 3x2 + 4x
5 a x(3x
+ 4) b 2y(2
y + 5)
c x(x
+ y + y2) d 2xy(4 y + 5x)
6 a (x + 1)(
x + 2) b 3x(
x + 2)
c (x − 7)(
x + 5) d (2x − 3)(
x + 1)
e (5x + 2)(
x − 3) f (1 − x)(6 +
x)
7 a 2x(
x2 + 3) b x(x + 6)(x − 6)
c x(2x
− 3)(x + 5)
8 a 3x6 b 2 c 6x2 d 1 _ 2 x − 1 _ 3
9 a 4 _ 9 b ± 3375 ___ 4913
10 a √ __
7 ___ 7 b 4 √ __
5
11 a 21 877
b (5x + 6)(7
x − 8)
When x = 25, 5x + 6 = 131 and 7x − 8 = 167; both
131 and 167 are prime numbers.
12 a 3 √ __
2 + √ ___ 10 b 10 + 2 √ __
3 − 5 √ __
5 − √ ___ 15
c 24 − 6 √ __
7 − 4 √ __
2 + √ ___ 14
13 a √ __
3 ___ 3 b √ __
2 + 1 c − 3 √ __
3 − 6
d 30 − √ ____ 851 __________ − 7 e 7 − 4 √ __
3 f 23 + 8 √ __
7 _________ 81
14 a b = −4 and c
= −5 b (x + 3)(
x − 5)(x + 1)
15 a 1 _ 4 x b 256x−3
16 5 __________
√ ___ 75 − √ ___ 50 = 1 _______
√ __
3 − √ __
2 = √ __
3 + √ __
2
17 −36 + 10 √ ___
11
18 x(1 + 8x
)(1 − 8x)
19 y = 6x
+ 3
20 4 √ __
3
21 3 − √ __
3 cm
22 4 − 4 x 1 _ 2 + x 1 ___________
x 1 _ 2 = 4 x − 1 _ 2 − 4 + x 1 _ 2
23 11 __ 2
24 4 x 5 _ 2 + x 2 , a = 5 __ 2 b = 2
Challenge
a a − b
b ( √ __
1 − √ __
2 ) + ( √ __
2 − √ __
3 ) + … + ( √ ___
24 − √ ___
25 ) _____________________________________ −1 = √ ___
25 − √ __
1 = 4
CHAPTER 2
Prior knowledge check
1 a x = −5 b x = 3
c x = 5 or x
= −5 d 16 or 0
2 a (x + 3)(
x + 5) b (x + 5)(
x − 2)
c (3x + 1)(
x − 5) d (x − 20)(
x + 20)
3 a y
x
–62O b y
x10
5 O | [
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0.019751090556383133,
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0.022827155888080597,
0.03088555298745632,
-0.09027034044265747,
-0.01064865943044424,
-0.03... |
347
Answers
3473 c y
x9
18 O d y
x O
4 a x < 3 b x > 9 c x < 2.5 d x > −7
Exercise 2A
1 a x = −1 or x = −2 b x = −1 or x = −4
c x = −5 or x
= −2 d x = 3 or x
= −2
e x = 3 or x
= 5 f x = 4 or x
= 5
g x = 6 or x
= −1 h x = 6 or x
= −2
2 a x = 0 or x
= 4 b x = 0 or x
= 25
c x = 0 or x
= 2 d x = 0 or x
= 6
e x = − 1 __ 2 or x = −3 f x = − 1 __ 3 or x = 3 __ 2
g x = − 2 __ 3 or x = 3 __ 2 h x = 3 __ 2 or x = 5 __ 2
3 a x = 1 __ 3 or x = −2 b x = 3 or x = 0
c x = 13 or x
= 1 d x = 2 or x
= −2
e x = ± √ __
5 __ 3 f x = 3 ± √ ___
13
g x = 1 ± √ ___
11 ________ 3 h x = 1 or x = − 7 __ 6
i x = − 1 __ 2 or x = 7 __ 3 j x = 0 or x = − 11 __ 6
4 x = 4
5 x = −1 or x
= − 2 __ 25
Exercise 2B
1 a x = 1 __ 2 (−3 ± √ __
5 ) b x = 1 __ 2 (3 ± √ ___
17 )
c x = −3 ± √ __
3 d x = 1 __ 2 (5 ± √ ___
33 )
e x = 1 __ 3 (−5 ± √ ___
31 ) f x = 1 __ 2 (1 ± √ __
2 )
g x = 2 or x
= − 1 __ 4 h x = 1 __ 11 (−1 ± √ ___
78 )
2 a x = −0.586 or x
= −3.41 b x = 7.87 or x
= 0.127
c x = 0.765 or x
= −11.8 d x = 8.91 or x
= −1.91
e x = 0.105 or x
= −1.90 f x = 3.84 or x
= −2.34
g x = 4.77 or x
= 0.558 h x = 4.89 or x
= −1.23
3 a x = −6 or x
= −2 b x = 1.09 or x
= −10.1
c x = 9.11 or x
= −0.110 d x = − 1 __ 2 or x = −2
e x = 1 or x
= −9 f x = 1
g x = 4.68 or x
= −1.18 h x = 3 or x
= 5
4 Area = 1 __ 2 (2x) (x + (x+10)) = 50 m2
So x2 + 5x − 25 = 0
Using the quadratic formula:
x = 1 __ 2 (−5 ± 5 √ __
5 )
Height = 2x = 5 ( √ __
5 − 1) m
Challenge
x = 13
Exercise 2C
1 a (x + 2)2 − 4 b (x − 3)2 − 9
c (x − 8)2 − 64 d (x + 1 __ 2 )2 − 1 __ 4
e (x − 7)2 − 49
2 a 2(x + 4)2 − 32 b 3(x − 4)2 − 48
c 5(x + 2)2 − 20 d 2(x − 5 __ 4 )2 − 25 __ 8
e −2(x − 2)2 + 8
3 a 2(x + 2)2 − 7 b 5(x − 3 __ 2 )2 − 33 __ 4
c 3(x + 1 __ 3 )2 − 4 __ 3 d −4 (x + 2)2 + 26
e −8(x − 1 __ 8 )2 + 81 __ 8 4 a = 3 __ 2 , b = 15 __ 4
5 A = 6,
B = 0.04, C = −10
Exercise 2D
1 a x = −3 ± 2 √ __
2 b x = −6 ± √ ___
33
c x = −2 ± √ __
6 d x = 5 ± √ ___
30
2 a x = 1 __ 2 (−3 ± √ ___
15 ) b x = 1 __ 5 (−4 ± √ ___
26 )
c x = 1 __ 8 (1 ± √ ____ 129 ) d x = 1 __ 2 (−3 ± √ ___
39 )
3 a p = −7,
q = −48
b (x − 7)2 = 48
x = 7 ± √ ___
48 = 7 ± 4 √ __
3
r = 7,
s = 4
4 x2 + 2bx + c = (x + b)2 − b2 + c
(x + b)2 = b2 − c
x = −b ± √ ______ b 2 − c
Challenge
a ax2 + 2bx + c = 0
x2 + 2b ___ a x + c __ a = 0
(x + b __ a ) 2
− b 2 ___ a 2 + c _ a = 0
(x + b __ a ) 2
= b 2 − ac _______ a 2
x = − b __ a ± √ _______
b 2 − ac _______ a2 b ax2 + bx + c = 0
x2 + b __ a x + c __ a = 0
(x + b __ 2a ) 2
− b 2 ___ 4 a 2 + c _ a = 0
(x + b __ 2a ) 2
= b 2 − 4ac ________ 4 a 2
x = − b ± √ _______ b 2 − 4ac ___________ 2a
Exercise 2E
1 a 8 b 7 c 3 d 10.5 e 0
f 0 g 25 h 2 i 7
2 a = 4 or a
= −2
3 a 2 __ 3 b 2 and −9 c −10 and 4
d 12 and −12 e 0, −5 and −7 f 0, 3 and −8
4 x = 3 and x
= 2
5 x = 0,
x = 2.5 and 6
6 a (x − 1)2 + 1
p = −1,
q = 1
b Squared terms are always >0,
so the minimum
value is 0 + 1 = 1
7 a −2 and −1 b 2, −2, 2 √ __
2 and −2 √ __
2
c −1 and 1 __ 3 d 1 __ 2 and 1
e 4 and 25 f 8 and −27
8 a (3x − 27)(3x − 1) b 0 and 3
Exercise 2F
1 a y
x O (4, 0)(2, 0)(0, 8)y = x2 – 6x + 8
Turning point: (3, −1)
Line of symmetry: x = 3 | [
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-0.04718343... |
348
Answers
348 Full worked solutions are available in SolutionBank.
Online
b y
x O
y = x2 + 2x – 15(3, 0)
(–5, 0)
(0, –15)
Turning point: (−1, −16)
Line of symmetry: x = −1
c y
x Oy = 25 – x2
(–5, 0)(0, 25)
(5, 0)
Turning point: (0, 25)
Line of symmetry: x = 0
d y
x Oy = x2 + 3x + 2
(–1, 0)
(–2, 0)(0, 2)
Turning point: (− 3 __ 2 , − 1 __ 4 )
Line of symmetry: x = − 3 __ 2
e y
x Oy = –x2 + 6x + 7
(–1, 0)(0, 7)
(7, 0)
Turning point: (3, 16)
Line of symmetry: x = 3
f y
x Oy = 2x2 + 4x + 10(0, 10)
Turning point: (−1, 8)
Line of symmetry: x = −1
g y
x O
y = 2x2 + 7x – 15( , 0)3
2 (–5, 0)
(0, –15)
Turning point: (− 7 __ 4 , − 169 ___ 8 )
Line of symmetry: x = − 7 __ 4 h y
x O
y = 6x2 – 19x + 10(0, 10)
( , 0)2
3( , 0)52
Turning point: ( 19 ___ 12 , − 121 ____ 24 )
Line of symmetry: x = 19 ___ 12
i y
x Oy = 4 – 7x – 2x2
(0, 4)
(–4, 0)( , 0)1
2
Turning point: (− 7 __ 4 , 81 ___ 8 )
Line of symmetry: x = − 7 __ 4
j y
x Oy = 0.5x2 + 0.2x + 0.02
(0, 0.02)
(–0.02, 0)
Turning point: (−0.2, 0)
Line of symmetry: x = −0.2
2 a a = 1,
b = −8, c = 15
b a = −1,
b = 3, c = 10
c a = 2,
b = 0, c = −18
d a = 1 __ 4 , b = − 3 __ 4 , c = −1
3 a = 3,
b = −30, c = 72
Exercise 2G
1 a i 52 ii −23 iii 37
iv 0 v −44
b i h(x) ii f(x) iii k(x)
iv j(x) v g(x)
2 k < 9
3 t = 9 __ 8
4 s = 4
5 k > 4 __ 3
6 a p = 6 b x = −9
7 a k2 + 16
b k2 is always positive so k2 + 16 > 0
Challenge
a Need b2 > 4ac. If a, c > 0 or a, c < 0, choose b such that
b > √ ____ 4ac . If a > 0 and c < 0 (or vice versa), then
4ac
< 0, so 4ac < b2 for all b.
b Not if one of a or
c are negative as this would require b
to be the square root of a negative number. Possible if
both negative or both positive. | [
-0.08171235769987106,
-0.0643116757273674,
0.034181348979473114,
-0.05814121663570404,
0.008347636088728905,
0.02351064421236515,
-0.005849470384418964,
-0.013341337442398071,
-0.10939627885818481,
-0.00675944285467267,
-0.020795000717043877,
-0.05500122159719467,
0.01088480930775404,
-0.0... |
349
Answers
349Exercise 2H
1 a The height of the bridge above ground level
b x = 1103 and x
= −1103
c 2206 m
2 a 21.8 mph and
75.7 mph
b A = 39.77,
B = 0.01, C = 48.75
c 48.75mph
d −11 mpg; a negative answer is impossible so this
model is not valid for very high speeds
3 a 6 tonnes
b 39.6 kilograms per hectare.
4 a M = 40 000
b r = 400 000 − 1000( p − 20)2
A = 400 000, B
= 1000, C = 20
c £20
Challenge
a a = 0.01,
b = 0.3, c = −4
b 36.2 mph
Mixed exercise
1 a y = −1 or −2 b x = 2 __ 3 or x = −5
c x = − 1 __ 5 or 3 d 5 ± √ __
7 _______ 2
2 a y
x (–4, 0)(–1, 0)(0, 4)
O
b y
x
(0, –3)(– , 0)(1, 0)
3
2O
c y
x(–3, 0)(0, 6)
( , 0)1
2
O
d y
x(0, 0)
(7 , 0)12 O
3 a k = 1 b x = 3 and x = −2
4 a k = 0.0902 or k
= −11.1
b t = 2.28 or t
= 0.219
c x = −2.30 or x
= 1.30
d x = 0.839 or x
= −0.239
5 a (x + 6)2 − 45; p = 1, q = 6, r = −45
b 5(x − 4)2 − 67; p = 5, q = −4, r = −67
c −2(x − 2)2 + 8; p = −2, q = −2, r = 8
d 2 (x − 1 __ 2 ) 2 − 3 __ 2 ; p = 2, q = − 1 __ 2 , r = − 3 __ 2 6 k = 1 __ 5
7 a p = 3,
q = 2, r = −7 b −2 ± √ __
7 __ 3
8 a f(x) = (2x − 16)(2x − 4) b 4 and 2
9 1 ± √ ___
13
10 x = −5 or x
= 4
11 a 10 m b 1.28 s
c h(t) = 10.625 − 10(
t − 0.25)2
A = 10.625, B = 10, C = 0.25
d 10.625 m at
0.25 s
12 a 16k2 + 4
b k2 > 0 for all k, so 16k2 + 4 > 0
c When k = 0,
f(x) = 2x + 1; this is a linear function
with only one root
13 1, −1, 2 and −2
14 a H = 10
b r = 1322.5 − 10(
p − 11.5)2
A = 1322.5, B = 10, C = 11.5
c Old revenue is 80 × £15 = £1200; new revenue is
£1322.50; difference is £122.50. The best selling price of a cushion is £11.50.
Challenge
a
a + b _____ a = a __ b
a2 − ba − b2 = 0
Using quadratic formula: a = b + √ ____
5b2 ________ 2
So a : b is b + √ ____
5b2 ________ 2 : b
Dividing by b
: 1 + √ __
5 _______ 2 : 1
b Let x = √ _____________________
1 + √ ________________ 1 + √ ___________ 1 + √ ______ 1 + …
So x = √ _____ 1 + x ⇒ x2 − x − 1 = 0
Using quadratic formula: x = 1 + √ __
5 _______ 2
CHAPTER 3
Prior knowledge check
1 a A ⋂ B = {1, 2, 4} b (A ⋃ B)9 = {7, 9, 11, 13}
2 a 5 √ __
3 c √5 + 2√2
3 a graph ii b graph iii c graph i
Exercise 3A
1 a x = 4, y = 2 b x = 1, y = 3
c x = 2,
y = −2 d x = 4 1 __ 2 , y = −3
e x = − 2 __ 3 , y = 2 f x = 3, y = 3
2 a x = 5,
y = 2 b x = 5 1 __ 2 , y = −6
c x = 1,
y = −4 d x = 1 3 __ 4 , y = 1 __ 4
3 a x = − 1,
y = 1 b x = 4,
y = −4
c x = 0.5,
y = −2.5
4 a 3x +
ky = 8 (1); x − 2ky = 5 (2)
(1) × 2: 6x + 2
ky = 16 (3)
(2) + (3) 7x = 21 so
x = 3
b −2
5 p = 3,
q = 1
Exercise 3B
1 a x = 5, y = 6 or x = 6, y = 5
b x = 0,
y = 1 or x = 4 __ 5 , y = − 3 __ 5
c x = −1,
y = −3 or x = 1, y = 3 | [
0.05560750886797905,
0.054463766515254974,
-0.010985961183905602,
-0.023514389991760254,
-0.05735262110829353,
-0.0852833092212677,
-0.0284983292222023,
0.01807441934943199,
-0.13538342714309692,
0.060148999094963074,
0.09056969732046127,
-0.0765831470489502,
0.007468268740922213,
-0.05170... |
350
Answers
350 Full worked solutions are available in SolutionBank.
Online
d a = 1, b = 5 or a = 3, b = −1
e u = 1 1 __ 2 , v = 4 or u = 2, v = 3
f x = −1 1 __ 2 , y = 5 3 __ 4 or x = 3, y = −1
2 a x = 3,
y = 1 __ 2 or x = 6 1 __ 3 , y = −2 5 __ 6
b x = 4 1 __ 2 , y = 4 1 __ 2 or x = 6, y = 3
c x = −19,
y = −15 or x = 6, y = 5
3 a x = 3 + √ ___
13 , y = −3 + √ ___
13 or x = 3 − √ ___
13 ,
y = −3 − √ ___
13
b x = 2 − 3 √ __
5 , y = 3 + 2 √ __
5 or x = 2 + 3 √ __
5 , y = 3 − 2 √ __
5
4 x = −5,
y = 8 or x = 2, y = 1
5 a 3x2 + x(2 − 4x) + 11 = 0
3x2 + 2x − 4x2 + 11 = 0
x2 − 2x − 11 = 0
b x = 1 + 2 √ __
3 , y = −2 − 8 √ __
3
x = 1 − 2 √ __
3 , y = −2 + 8 √ __
3
6 a k = 3,
p = −2
b x = −6,
y = −23 or x = 1, y = −2
Challenge
y = x + kx
2 + (x + k) 2 = 4
x2 + x2 + 2kx + k2 − 4 = 0
2x2 + 2kx + k2 − 4 = 0 for one solution b2 − 4ac = 0
4k2 − 4 × 2(k2 − 4) = 0
4k2 − 8k2 + 32 = 0 4k2 = 32 k2 = 8 k = ±2 √ __
2
Exercise 3C
1 a i y
x O
ii (2, 1)
b i y
x O
ii (3, −1)
c i y
x O
ii (−0.5, 0.5)
2 a y
x O
b (3.5, 9) and (−1.5, 4)3 a y
x O
b (−1, 8) and (3, 0)
4 a y
x O
b (6, 16) and (1, 1)
5 (−11, −15) and (3, −1)
6 (−1 1 __ 6 , −4 1 __ 2 ) and (2, 5)
7 a 2 points b 1 point c 0 points
8 a y = 2x
− 1
x2 + 4k(2x − 1) + 5k = 0
x2 + 8kx − 4k + 5k = 0 x2 + 8kx + k = 0
b k = 1 __ 16 c x = − 1 __ 4 , y = − 3 _ 2
9 If swimmer reaches the bottom of the pool
0.5x2 − 3x = 0.3x − 6
0.5x2 − 3.3x + 6 = 0
b2 − 4ac = (− 3.3)2 − 4 × 0.5 × 6 = −1.11
negative so no points of intersection and diver does not reach the bottom of the pool
Exercise 3D
1 a x < 4 b x > 7 c x > 2 1 __ 2 d x < −3
e x < 11 f x < 2 3 __ 5 g x > −12 h x < 1
i x < 8 j x > 1 1 __ 7
2 a x > 3 b x < 1 c x < −3 1 __ 4 d x < 18
e x > 3 f x > 4 2 __ 5 g x < 4 h x > −7
i x < − 1 __ 2 j x > 3 __ 4 k x > − 10 __ 3 l x > 9 __ 11
3 a {x:
x > 2 1 __ 2 } b {x: 2 < x < 4}
c {x: 2 1 __ 2 < x < 3} d No values
e x = 4 f {x: x < 1.2} ⋃ {x: x > 2.2}
g { x : x < − 2 __ 3 } ⋃ { x : x > 3 _ 2 }
Challenge
p = −1,
q = 4, r = 6
Exercise 3E
1 a 3 < x < 8 b −4 < x < 3
c x < −2,
x > 5 d x < −4,
x > −3
e − 1 __ 2 < x < 7 f x < −2, x > 2 1 __ 2
g 1 __ 2 < x < 1 1 __ 2 h x < 1 __ 3 , x > 2
i −3 < x < 3 j x < −2 1 __ 2 , x > 2 __ 3
k x < 0,
x > 5 l −1 1 __ 2 < x < 0
2 a −5 < x < 2 b x < −1,
x > 1
c 1 __ 2 < x < 1 d −3 < x < 1 __ 4
3 a {x: 2 <
x < 4} b {x:
x > 3}
c {x: − 1 __ 4 < x < 0} d No values | [
-0.06884238123893738,
-0.04596174135804176,
0.02920406311750412,
-0.09674413502216339,
0.007242766208946705,
0.037808068096637726,
-0.025201871991157532,
-0.03072923980653286,
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0.09661343693733215,
0.01819494366645813,
-0.03300723060965538,
0.018895741552114487,
0.026... |
351
Answers
351 e {x: −5 < x < −3} ⋃ {x: x > 4}
f {x: −1 <
x < 1} ⋃ {x: 2 < x < 3}
4 a x < 0 or x
> 2 b x < 0 or x
> 0.8
c x < −1 or x
> 0 d x < 0 or x
> 0.5
e x < − 1 __ 5 or x > 1 __ 5 f x < − 2 __ 3 or x > 3
5 a −2 < k < 6 b p < −8 or
p > 0
6 {x:
x < −2} ⋃ {x: x > 7}
7 a {x:
x < 2 _ 3 } b {x: − 1 __ 2 < x < 3}
c {x: − 1 __ 2 < x < 2 __ 3 }
8 x < 3 or x
> 5.5
9 No real roots b2 − 4ac < 0 (−2k)2 − 4 × k × 3 < 0
4k2 − 12k = 0 when k = 0 and k = 3
solution 0 < k
< 3
note when k = 0 equation gives 3 = 0
Exercise 3F
1 a P(3.2, −1.8) b x < 3.2
2 a i y
x O
ii (4, 5) iii x < 4
b i y
x O
ii (−3, 23) iii x > −3
c i y
x O
ii (−2, 9), (0, 5) iii −2 < x < 0
d i y
x O
ii (−5, −22), (3, −6) iii x < −5 or x > 3
e i y
x O
ii (−2, −1), (9, 76) iii −2 < x < 9 f i y
x O
ii (−5, −18), (3, −2) iii x < −5 or x > 3
3 a −1 < x < 2 b 0.5 < x < 3
c x < 0.5 or x
> 3 d x < 0 or x
> 2
e 1 < x < 3 f x < −1 or x
> −0.75
Challenge
a (−1.5, −3.75), (6,
0)
b {x: −1.5 <
x < 6}
Exercise 3G
1
xy
–4–2 246–22468
–4O 2
xy
–6 –2–4 246–22468
–4O
3
xy
–6 –2–4 246–22468
–4O 4
xy
–6 –2–4 246–2
–4246810
–6O
5
xy
–2 2468–2
–42468
O
6 a (1, 6), (3, 4), (1, 2)
b x >
1, y < 7 − x , y > x + 1
7 y < 2 − 5x
− x2, 2x + y > 0, x + y < 4
8 a
xy
–2–4–6 2–2
–4
–62468
O
b ( − 7 _ 6 , 17 __ 6 ), (2, 6), (2, −1), (−0.4, 1)
c (−0.4, 1) d 941 ___ 60 | [
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0.02979881316423416,
0.02298298478126526,
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0.0012667098781093955,
0.023505887016654015,
0.0549328587949276,
-0.0162173043936491,
-0.09046171605587006,
0.10245751589536667,
0.02741752192378044,
-0.10725383460521698,
0.04612688347697258,
-0.0096202... |
352
Answers
352 Full worked solutions are available in SolutionBank.
Online
Mixed exercise
1 a 4kx − 2 y = 8
4kx + 3y = −2
−5y = 10
y = −2
b x = 1 __ k
2 x = −4,
y = 3 1 __ 2
3 a Substitute x = 1 + 2
y into 3xy − y2 = 8
b (3, 1) and (− 11 __ 5 , − 8 __ 5 )
4 a Substitute y = 2 −
x into x2 + xy − y2 = 0
b x = 3 ± √ __
6 , y = −1 ± √ __
6
5 a 3x = (32)y − 1 = 32y − 2 ⇒ x = 2y − 2
b x = 4,
y = 3 and x = −2 2 __ 3 , y = − 1 __ 3
6 x = −1 1 __ 2 , y = 2 1 __ 4 and x = 4, y = − 1 __ 2
7 a k = −2 b (−1, 2)
8 Yes, the ball will hit the ceiling
9 a {x:
x > 10 1 __ 2 } b {x: x < −2} ⋃ {x: x > 7}
10 3 < x < 4
11 a x = −5,
x = 4 b {x:
x < −5} ⋃ {x: x > 4}
12 a x < 2 1 __ 2 b 1 __ 2 < x < 5
c 0 < x < 4 d 1 __ 2 < x < 2 1 __ 2
13 1 <
x < 8
14 k < 3 1 __ 5
15 b2 < 4ac so 16k2 < −40k
8k
(2k + 5) < 0 so − 5 __ 2 < k < 0
16 a y
x O
b (−7, 20), (3, 0) c x < −7, x > 3
17 1 __ 4 (−1 − √ ____ 185 ) < x < 1 __ 4 (−1 + √ ____ 185 )
18 y
x –8–10 –6–4–2 246–44
–8
–12
–16
–20O
19 a
xy
–4–3–2–1 12341234567
O b 9 __ 2
Challenge
1 0 < x < 1.6 2 −2 < k < 7CHAPTER 4
Prior knowledge check
1 a (x + 5)( x + 1) b (x − 3)( x − 1)
2 a y
x O–2 3 b y
x O–1 6
3 a x −2 −1.5 −1 −0.5 0
y −12 −6.875 −4 −2.625 −2
x 0.5 1 1.5 2
y−1.375 0 2.875 8
b y
x1 –1
–1010
–2 2 O
4 a x = 2, y = 4 b x = 2, y = 5
Exercise 4A
1 a
6
–1 23y
x O b
–6–2–3 1Oy
c y
6
x x –1–2–3O d y
x3
–1 1 –3 O
e
24y
234x O f y
x2 –1O
g
xy
1 –1 O h
xy
1 –1O | [
-0.07159418612718582,
0.039758432656526566,
0.029994383454322815,
-0.06266020983457565,
-0.06447648257017136,
0.011389545165002346,
-0.02702719159424305,
-0.03692164272069931,
-0.10028614103794098,
0.04789949208498001,
0.007884837687015533,
-0.05697328969836235,
0.030236465856432915,
0.000... |
353
Answers
353 i
xy
22
– 1
212O j
xy
–312O
2 a y
x–1
–11O b y
x–22
1O
c y
x–12
2O d y
x–1–22O
e y
x –2 O f y
x1O
g y
x13
–3O h y
x13
3O
i y
x2O j y
x2O
3 a y = x
(x + 2)(x − 1) b y = x
(x + 4)(x + 1)
y
x1 –2O
–1 –4y
x O
c y = x (x + 1)2 d y = x (x + 1)(3 − x)
–1y
x O
–1 3y
x O e y = x2(x − 1) f y = x (1 − x)(1 + x)
1y
x O
1 –1y
x O
g y = 3x
(2x − 1)(2x + 1) h y = x
(x + 1)(x − 2)
–12 12y
x O
–1 2y
x O
i y = x
(x − 3)(x + 3) j y = x2(x − 9)
–3 3y
x O
9y
x O
4 a y
x O
(0, –8)(2, 0) b
Oy
x(2, 0)(0, 8)
c
O
xy
(0, –1) (1, 0) d
O xy
(–2, 0)(0, 8)
e
Oxy
(–2, 0)
(0, –8) f
–327y
x O | [
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-0.009504067711532116,
0.02975899539887905,
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-0.05843671038746834,
0.06311101466417313,
0.0732831135392189,
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-0.013768315315246582,
-0.039649106562137604,
0.021541696041822433,
-0.07... |
354
Answers
354 Full worked solutions are available in SolutionBank.
Online
g
3
–27y
x O h
11y
x O
i
28y
x O j
1
218y
x O
5 a b = 4, c = 1, d = −6 b (0, −6)
6 a = 1 __ 3 , b = − 4 __ 3 , c = 1 __ 3 , d = 2
7 a x(x2 − 12x + 32) b x(x − 8)(x − 4)
c y
x O 84
Exercise 4B
1 a y
xO –4 –3 –2 –124 b y
x O1 –3 2
c y
x O–2–1 d y
x O–2 10.5
2–4
e y
x O –0.25 0.25 f y
x O 2
–644 g y
x O 3 –19 h y
x O 3 –2
–24
i y
x O0.5 –35 j y
xO –4256
2 a y
xO
–2 12–4 b y
x O –3 254
3
c y
x O 4480
56 d y
x O1152
–9 –4 48
3 a (0, 12)
b b = −2,
c = −7, d = 8, e = 12
4
–280–5 4y
x O
Challenge
a = 1 __ 3 , b = − 4 __ 3 , c = − 2 __ 3 , d = 4, e = 3 | [
-0.03521917387843132,
-0.032616157084703445,
0.033659614622592926,
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-0.04875417798757553,
0.04825134575366974,
-0.008693584240972996,
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0.05912495777010918,
0.04343504458665848,
-0.07272208482027054,
-0.05057069659233093,
-0.08... |
355
Answers
355Exercise 4C
1 a y
y =
x4
x
y =2
xO b
y =2
x
y = –2
xy
x O
c
y = –2
x
y = –4
xOy
x d
y =y
x8
x
y =3
xO
e
y = –3
x
y = –8
xOy
x
2 a
y =5
x2
y =2
x2y
x O b
y = –3
x2y
x Oy =3
x2
c y
x O
y = –6
x2y = –2
x2
Exercise 4D
1 a i y
xy = x2
y = x(x2 – 1)–1 1 O
ii 3 iii x2 = x(x2 − 1) b i y
xy = x(x + 2)
y = ––2
3
xO
ii 1 iii x(x
+ 2) = − 3 __ x
c i y
xy = (x + 1)(x – 1)2y = x2
–1 1O
ii 3 iii x2 = (x + 1)(x − 1)2
d i y
xy = x2 (1 – x)
y = 1
2
xO
ii 2 iii x2(1 − x) = − 2 __ x
e i y
x
y = x(x – 4) y = 4
1
xO
ii 1 iii x(x − 4) = 1 __ x
f i y
xy = x(x – 4)
y = – 4
1
xO
ii 3 iii x(x
− 4) = − 1 __ x | [
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0.027853356674313545,
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0.057712338864803314,
0.0009278937359340489,
-0.05335347354412079,
0.057499587535858154,
... |
356
Answers
356 Full worked solutions are available in SolutionBank.
Online
g i y
x
y = x(x – 4)y = (x – 2)3
42
O
ii 1 iii x(x − 4) = (x − 2)3
h i y
x
y = –x3y = –2
x
O
ii 2 iii −x3 = − 2 __ x
i i y
xy = x2
y = –x3O
ii 2 iii −x3 = x2
j i
y
x
y = –x3y = –x(x + 2)
–2O
ii 3 iii −x3 = −x(x + 2)
k i y
x O–2 1y = 4
y = x(x – 1)(x + 2)2
ii 2 iii x(x − 1)(x + 2)2 = 4 l i y
x Oy = x3
y = x2(x + 1)2
ii 1 iii x3 = x2(x + 1)2
2 a y
x 3y = x2(x – 3)y =2
x
O
b Only 2 intersections
3 a y
x –1 1y = 3x(x – 1)
y = (x + 1)3O
b Only 1 intersection
4 a y
x 11
x
y = –x(x – 1)2y =
O
b Graphs do not intersect
5 a y
x Oy = x2(x – a)
y =b
x
b 2; the graphs cross in two places so there are two
solutions. | [
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0.06201764568686485,
-0.026966454461216927,
-0.0669819563627243,
0.018533414229750633,
0.02798229642212391,
-0.06419938802719116,
-0.031203024089336395,
-0.06... |
357
Answers
3576 a
7y =y = 3x + 7
4
x2y
x O
b 3
c Expand brackets and rearrange.
d (−2, 1), (−1,
4), ( 2 __ 3 , 9)
7 a y
x–1
O4
y = x3 – 3x2 – 4xy = 6x
b (0, 0); (−2, −12); (5, 30)
8 a y
x2
–1 21y = 14x + 2
y = (x2 – 1)(x – 2)
O
b (0, 2); (−3, −40); (5, 72)
9 a y
x–2 2y = (x – 2)(x + 2)2
y = –x2 – 8O
b (0, −8); (1, −9); (−4, −24)
10 a y
x Oy = x2 + 1
1
1
–0.5y = x – 1
b Graphs do not intersect.
c a < − 7 __ 16 11 a y
x Oy = x2(x – 1)(x + 1)
–1 11
y = x3 + 11
3
b 2
Exercise 4E
1 a y y y
x x x O O O
(–2, 0), (0, 4)
(–2, 0), (0, 8) (0, ), x = –2, y = 012ii i iii
b
x
xy y y
x OO
O(0, 2)
(2 , 0), y = 2, x = 012(– 2, 0), (0, 2)ii i iii
3
c y y y
x x OO O x
(0, 1), (1, 0)
(0, –1), x = 1, y = 0 (0, –1), (1, 0)ii i iii
d y y y
xx x O O O
(–1, 0),
(0, –1),(1, 0) (1, 0),
y = –1, x = 0 (0, –1), (1, 0)ii i iii
e y y y
xx
xO
O
O
(– 3, 0),(0, –3),( 3, 0)( , 0),
y = –3, x = 0(0, –3),( 3, 0)1
3ii i iii
3 | [
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-0.04114476591348648,
-0.016668615862727165... |
358
Answers
358 Full worked solutions are available in SolutionBank.
Online
f y y y
xxx
OOO
(0, 9), (3, 0)
(0, –27), (3, 0)(0, – ), x = 3, y = 01
3ii i iii
2 a
Oy
x–2
–21y = f (x)
b i
Oy
x–44
–1y = f (x + 2) ii
Oy
x –1y = f (x) + 2
c f(x + 2) = (x + 1)(x + 4); (0, 4)
f(x
) + 2 = (x + 1)(x + 2) + 2; (0, 0)
3 a
Oy
x 1y = f(x) b
Oy
x–1y = f (x + 1)
c f(x + 1) = −x(x + 1)2; (0, 0)
4 a
Oy
x 2y = f (x) b
Oy
x2y = f (x) + 2
y = f (x + 2)–2 2
c f(x + 2) = (x + 2)x2; (0, 0); (−2, 0)
5 a
Oy
x4y = f(x) b
Oy
x4
–4–2 2
y = f (x + 2)y = f (x) + 4
c f(x + 2) = (x + 2)(x − 2); (2, 0); (−2, 0)
f(x
) + 4 = (x − 2)2; (2, 0)6 a y
x Oy = f(x)
12 b
y
x O
y = f(x) – 1y = f(x + 2)
–2–1–1
7 a (6, −1) b (4, 2)
8 y = 1 _____ x − 4
9 y
x O
y = (x – 2)3 – 5(x – 2)2 + 6(x – 2)42 5y = x3 – 5x2 + 6x
10 y
x O
y = x2(x – 3)(x + 2)y = (x + 2)2(x – 1)(x + 4)
–2 1 –4
11 a y
x Oy = x3 + 4x2 + 4x
–2
b −1 or 1
12 a y
x Oy = x(x + 1)(x + 3)2
–3 –1
b −2, −3 or −5
Challenge
1 (3, 2)
2 a (−7, −12) b f(x
− 2) + 1
Exercise 4F
1 a y
xi y
xii y
xiii f(2x) f(2x)
f(2x)f(x)f(x)f(x)
O
O O | [
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0.04570527374744415,
0.06389906257390976,
-0.05937379598617554,
-0.05886876583099365,
-0.028... |
359
Answers
359 b
OO Oy
xi y
xii y
xiii
f(–x) f(x) = f(– x)
f(–x)f(x)f(x)
c
OO Oy
xi y
xii y
xiii f(x)
f( x)f(x)
f(x) 1
2f( x)12f( x)12
d
OO Oy
xi y
xii y
xiii f(4x)
f(4x)
f(4x)f(x)
f(x)f(x)
e
OO Oy
xi y
xii y
xiii
f( x)f(x) f(x)
f(x)1
4f( x)14f( x)14
f
OO Oy
xi y
xii y
xiii f(x)
f(x)2f(x)
2f(x)2f(x)
f(x)
g
OO Oy
xi y
xii y
xiii
f(x)f(x)f(x)
–f(x)–f(x)–f(x)
h y
xi y
xii y
xiii
f(x) f(x)
4f(x)4f(x)4f(x)
f(x)OO O
i y
xi y
xii y
xiii
f(x)f(x)
f(x)f(x)
1
2f(x)12
f(x)12O OO j y
xi y
xii y
xiii f(x)f(x)
f(x)f(x)
14f(x)14
f(x)14O OO
2 a yy = f(x)
x –2
–42O b yy = f(4x)
x–
–412 12 O
Oyy = 3f(x)
x–2
–122
Oy
y = f(–x)
x–2
–42
Oy
y = –f(x)
x–24
2
3 a yy = f (x)
x–2 2O b
Oyy = f ( x)
x–4 412
yy = f (2x)
x–1 1O y
y = –f (x)x–2 2O
4 y
x Oy = x2(x – 3)
y = (2x)2(2x – 3)
y = –x2(x – 3)3 1.5 | [
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0.010177300311625004,
-... |
360
Answers
360 Full worked solutions are available in SolutionBank.
Online
5 y
x O5y = x2 + 3x – 4
y = x2 + 3x – 4
6 y = x2(x – 2)2
3y = –x2(x – 2)2y
x O 2
7 a (1, −3) b (2, −12)
8 (−4, 8)
9 a y
x O 32
y = (x – 2)(x – 3)2
b 2 and 3
Challenge
1 (2, −2)
2 1 __ 4 f ( 1 __ 2 x)
Exercise 4G
1 a y
x (0, 0)(3, 4)
(5, 0)(–1, 2)
O b y
x
(0, –2)(4, 0)
(1, –4)(6, –4)O
c y
x(–4, 2)
(–3, 0)(0, 4)
(2, 0)
O d y
x(3, 0)(2, 4)
(0, 2)
( , 0)1
2O
e y
x (1, 0)(6, 0)(4, 12)
(0, 6)
O f y
x(8, 4)
(0, 2)
(12, 0) (2, 0)O g
Oy
x(6, 0)(4, 2)
(1, 0)(0, 1) h
Oy
x (–1, 0)(–6, 0)(–4, 4)
(0, 2)
2 a y = 4, x = 1, (0, 2)
Oy
y = 4
x 12
b y = 2, x = 0, (−1, 0)
Oy
y = 2
x –1
c y = 4, x = 1, (0, 0)
Oy
y = 4
1 x
d y = 0, x = 1, (0, −2)
Oy
1 x
–2
e y = 2, x = 1 __ 2 , (0, 0)
Oy
xy = 2
1
2 | [
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0.... |
361
Answers
361 f y = 2, x = 2, (0, 0)
y
x O 2y = 2
g y = 1, x = 1, (0, 0)
Oy
x 1y = 1
h y = −2, x = 1, (0, 0)
Oy
x 1
y = –2
3 a A(−2, −6), B(0, 0), C(2, −3), D(6, 0)
Oy
x
(–2, –6)(2, –3)6
b A(−4, 0), B(−2, 6), C(0, 3), D(4, 6)
y
x(–2, 6)
–4(4, 6)
3
O
c A(−2, −6), B(−1, 0), C(0, −3), D(2, 0)
y
x
(–2, –6)–1
–32 O d A(−8, −6), B(−6, 0), C(−4, −3), D(0, 0)
y
x
(–8, –6)(–4, –3)–6
O
e A(−4, −3), B(−2, 3), C(0, 0), D(4, 3)
y
x(–2, 3)
(–4, –3)(4, 3)
O
f A(−4, −18), B(−2, 0), C(0, −9), D(4, 0)
y
x–2
4
(–4, –18)–9O
g A(−4, −2), B(−2, 0), C(0, −1), D(4, 0)
y
x–2
4
(–4, –2)–1O
h A(−16, −6), B(−8, 0), C(0, −3), D(16, 0)
y
x 16–8
(–16, –6)–3O
i A(−4, 6), B(−2, 0), C(0, 3), D(4, 0)
y
x –2 4(–4, 6)
3
O | [
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0.03353702649474144,
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0.039101533591747284,
0.03794702887535095,
-0.03324824944138527,
-0.011820523999631405,
-0... |
362
Answers
362 Full worked solutions are available in SolutionBank.
Online
j A(4, −6), B(2, 0), C(0, −3), D(−4, 0)
y
x –4
(4, –6)–32O
4 a i x = −2, y = 0, (0, 2)
Oy
–2 x2
ii x = −1, y = 0, (0, 1)
y
–1 x1
O
iii x = 0 y = 0
Oy
x
iv x = −2 y = −1 (0, 0)
Oy
y = –1x–2 v x = 2 y = 0 (0, 1)
Oy
x 21
vi x = −2 y = 0 (0, −1)
Oy
x –2–1
b f(x ) = 2 _____ x + 1
5 a 1 _ 2
b i (6, 1) ii (2, 3) iii (2, −3.5)
6 a A(−1, −2)
B(0, 0) C(1, 0) D(2, −2)
y
x OBC
ADy + 2 = f(x)
b A(−1, 0) B(0, 4) C(1, 4) D(2, 0)
OBC
ADy
xy = f(x)1
2
c A(−1, 3) B(0, 5) C(1, 5) D(2, 3)
OBC
ADy
xy – 3 = f(x)
d A(−1, 0)
B(0, 2 __ 3 ) C(1, 2 __ 3 ) D(2, 0)
OBC
ADy
x3y = f(x) | [
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0.00... |
363
Answers
363 e A(−1, 0.5) B(0, 1.5) C(1, 1.5) D(2, 0.5)
OBC
ADy
x2y – 1 = f(x)
Mixed exercise
1 a y
xy = x2(x – 2)
2x – x22 O
b x = 0, −1, 2; points (0, 0), (2, 0), (−1, −3)
2 a y
x
y = x2 + 2x – 5y = 1 + x
y =6
x1
AB
O
b A(−3, −2), B(2, 3)
c y = x2 + 2x − 5
3 a y
xy = 2
O B(0, 0)A( , 4)3
2
b y
xy = 1
B(0, 0)A(3, 2)
O
c y
x
y = 0 is asymptoteB(0, –2)A(3, 2)
O
d
Oy
x B(–3, 0)y = 2A(0, 4) e
Oy
x B(3, 0)y = 2A(6, 4)
f
Oy
xB(0, 1)y = 3A(3, 5)
4 a x = −1 at A , x = 3 at B
5 a,
b y
x O1 3y = x2(x – 1)(x – 3)
y = 2 – x
c 2 d (0, 2)
6 a y
x
(–2, 0) (2, 0)(0, 0) O
b y
x (2, 0)(0, 0)
O
7 a y = x2 − 4x + 3
b i y
x (1, 0) (–1, 0)
(0, –1)O
ii y
x( , 0)
(1, –1)32
( , 0)12 O
8 a (0, 2) b −2 c −1, 1, 2
9 a i ( 4 __ 3 , 3) ii (4, 6) iii (9, 3)
iv (4, −3) v (4, − 1 __ 2 )
b f(2x
), f(x + 2)
c i f(x
− 4) + 3 ii 2f( 1 __ 2 x) | [
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-... |
364
Answers
364 Full worked solutions are available in SolutionBank.
Online
10 a
y = x2(3x + b)y
x O( , 0) (0, 0)–b
3
y =a
x2
b 1; only one intersection of the two curves
11 a x(x
− 3)2
b
y = x(x – 3)2Oy
x 3
c −4 and −7
12 a
Oy
xy = x(x – 2)2
(2, 2)(0, 0)
b
Oy
x(0, k)y = x(x – 2)2 + x
13 a Asymptotes at x = 0 and
y = −2
Oy
x
y = –2y = f (x) – 2
b ( 1 __ 2 , 0)
c
Oy
xx = –3
y =1
x + 3
d Asymptotes at y = 0 and x = −3; intersection at (0, 1 __ 3 )Challenge
(6 − c, −4 − d)
Review exercise 1
1 a 2 b 1 _ 4
2 a 625 b 4 _ 3 x 2 _ 3
3 a 4 √ __
5 b 21 − 8 √ __
5
4 a 13 b 8 − 2 √ __
3
5 a 1 + 2 √ __
k b 1 + 6 √ __
k
6 a 25x−4 b x2
7 8 + 8 √ __
2
8 1 − 2 √ __
2
9 a (x − 8)(
x − 2) b y = 1,
y = 1 _ 3
10 a a = −4,
b = −45 b x = 4 ± 3 √ __
5
11 4.19 (3 s.f.)
12 a The height of the athlete’s shoulder is 1.7 m
b 2.16s (3 s.f.)
c 6.7 – 5(t − 1)2
d 6.7 m after 1 second
13 a (x – 3)2 + 9
b P is (0, 18),
Q is (3, 9)
c x = 3 + 4 √ __
2
14 a k = 2
b y
x O (– 2, 0)(0, 2)
15 a x3(x3 − 8)(x3 + 1) b −1, 0, 2
16 a a = 5,
b = 11
b discriminant < 0 so no real roots
c k = 25
d
xy
O25
–5
17 a a = 1, b = 2
b
xy
O3
c discriminant = −8
d −2 √ __
3 k < 2 √ __
3
18 a x2 + 4x – 8 = 0
b x = −2 ± 2 √ __
3 , y = −6 ± 2 √ __
3
19 a x > 1 _ 4
b x < 1 _ 2 or x > 3
c 1 _ 4 < x < 1 _ 2 or x > 3
20 −2(x
+ 1) = x2 − 5x + 2
x2 – 3x + 4 = 0
The discriminant of this is −7 < 0, so no real solutions. | [
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... |
365
Answers
36521 a x = 7 _ 2 , y = −2, x = −3, y = 11
b x < −3 or x
> 3 1 _ 2
22 a Different real roots, discriminant > 0
so k2 – 4k – 12 > 0
b k < −2 or k
> 6
23 −7 < x < 2
24
xy
–3 3914
(5, –16)(1, 8)y = g(x)
y = f(x)7
3
25 a x(x – 2)(x + 2)
b
xy
O –2 2
c
xy
–1 13O
26 a
xy
O 2(3, 2)
4
(2, 0) (4, 0) and (3, 2)
b
xy
O12
(1 , –2)12
(1, 0) (2, 0) and (1 1 _ 2 , −2)
27 a
xy
O 3
(0, 0) and (3, 0) b
xy
O 4 16
(1, 0) (4, 0) and (0, 6)
c
xy
O283
(2, 0) (8, 0) and (0, 3)
28 a
xy
O3
Asymptotes: y = 3 and x = 0
b (− 1 _ 3 , 0)
29 a 0.438, 1, 4,
4.56
b
xy
O 44.56y = t(x)
0.438
8
1
30 a (6, 8) b (9, −8) c (6, −4)
31 a y
x O(b, 0)(0, b2)c2
c1
b 1
32 a
xy
O
y = –4(– , 0)1
2 ( , 0)12
b − 1 __ 2 , 1 __ 2
Challenge
1 a x = 1,
x = 9 b 0, 2
2 √ __
2 cm, 3 √ __
2 cm
3 3x3 + x2 – x = 2x(x − 1)(x + 1)
3x3 + x2 – x = 2x3 – 2x
x3 + x2 + x = 0
x(x2 + x + 1) = 0
The discriminant of the bracket is −3 < 0 so this
contributes no real solutions.
The only solution is when x = 0 at (0,
0).
4 −3, 3 | [
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0.00444075558334589,
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-0.... |
366
Answers
366 Full worked solutions are available in SolutionBank.
Online
CHAPTER 5
Prior knowledge check
1 a (−2, −1) b ( 9 __ 19 , 26 __ 19 ) c (7, 3)
2 a 4 √ __
5 b 10 √ __
2 c 5 √ __
5
3 a y = 5 − 2x b y = 2 __ 5 x − 9 __ 5 c y = 3 __ 12 x + 12 __ 7
Exercise 5A
1 a 1 __ 2 b 1 __ 6 c − 3 __ 5 d 2
e −1 f 1 __ 2 g 1 __ 2 h 8
i 2 __ 3 j −4 k − 1 __ 3 l − 1 __ 2
m 1 n q2 − p2 _______ q − p = q + p
2 7
3 12
4 4 1 __ 3
5 2 1 __ 4
6 1 __ 4
7 26
8 −5
9 Gradient of AB = gradient of
BC = 0.5; point B is
common
10 Gradient of AB = gradient of
BC = −0.5; point B is
common
Exercise 5B
1 a −2 b −1 c 3 d 1 __ 3
e − 2 __ 3 f 5 __ 4 g 1 __ 2 h 2
i 1 __ 2 j 1 __ 2 k −2 l − 3 __ 2
2 a 4 b −5 c − 2 __ 3 d 0
e 7 __ 5 f 2 g 2 h −2
i 9 j −3 k 3 __ 2 l − 1 __ 2
3 a 4x −
y + 3 = 0 b 3x −
y − 2 = 0
c 6x +
y − 7 = 0 d 4x − 5
y − 30 = 0
e 5x − 3
y + 6 = 0 f 7x − 3
y = 0
g 14x − 7
y − 4 = 0 h 27x + 9
y − 2 = 0
i 18x + 3
y + 2 = 0 j 2x + 6
y − 3 = 0
k 4x − 6
y + 5 = 0 l 6x − 10
y + 5 = 0
4 (3, 0)
5 (0, 0)
6 (0, 5), (−4,
0)
7 a 1 __ 3 b x − 3y + 15 = 0
8 a − 2 __ 5 b 2x + 5 y − 10 = 0
9 ax + by
+ c = 0
by = −ax
− c
y = (− a __ b ) x − ( c __ b )
10 a = 6,
c = 10
11 P(3,0)
12 a −16 b −27
Challenge
Gradient = − a __ b ; y-intercept = a. So y = − a __ b x + a
Rearrange to give ax + by − ab = 0
Exercise 5C
1 a y = 2x + 1 b y = 3x + 7 c y = −x − 3
d y = −4x
− 11 e y = 1 __ 2 x + 12 f y = − 2 __ 3 x − 5
g y = 2x
h y = − 1 __ 2 x + 2b2 a y = 4x
− 4 b y = x
+ 2 c y = 2x
+ 4
d y = 4x
− 23 e y = x
− 4 f y = 1 __ 2 x + 1
g y = −4x
− 9 h y = −8x
− 33 i y = 6 __ 5 x
j y = 2 __ 7 x + 5 __ 14
3 5x +
y − 37 = 0
4 y = x
+ 2, y = − 1 __ 6 x − 1 __ 3 , y = −6x + 23
5 a = 3,
c = −27
6 a = −4,
b = 8
Challenge
a m = (y2 − y1) ________ (x2 − x1)
b y − y1 = (y2 − y1) ________ (x2 − y1) (x − x1)
(y −
y1) ________ (y2 − y1) = (x −
x1) ________ (x2 − x1)
c y = 3 _ 7 x + 52 __ 7
Exercise 5D
1 y = 3x − 6 2 y = 2x + 8
3 2x − 3
y + 24 = 0 4 − 1 __ 5
5 (−3, 0) 6 (0, 1)
7 (0, 3 1 __ 2 ) 8 y = 2 __ 5 x + 3
9 2x + 3
y − 12 = 0 10 8 __ 5
11 y = 4 __ 3 x − 4 12 6x + 15 y − 10 = 0
13 y = − 4 __ 5 + 4 14 x − y + 5 = 0
15 y = − 3 __ 8 x + 1 __ 2 16 y = 4x + 13
Exercise 5E
1 a Parallel b Not parallel c Not parallel
2 r: y
= 4 __ 5 x + 3.2, s: y = 4 __ 5 x − 7
Gradients equal therefore lines are parallel.
3 Gradient of AB = 3 __ 5 , gradient of BC = − 7 __ 2 , gradient of
CD = 3 __ 5 , gradient of AD = 10 __ 3 . The quadrilateral has
a pair of parallel sides, so it is a trapezium.
4 y = 5x
+ 3
5 2x + 5
y + 20 = 0
6 y = − 1 __ 2 x + 7
7 y = 2 __ 3 x
8 4x −
y + 15 = 0
Exercise 5F
1 a Perpendicular b Parallel
c Neither d Perpendicular
e Perpendicular f Parallel
g Parallel h Perpendicular
i Perpendicular j Parallel
k Neither l Perpendicular
2 y = − 1 __ 6 x + 1
3 y = 8 __ 3 x − 8
4 y = − 1 __ 3 x
5 y = − 1 __ 3 x + 13 __ 3
6 y = − 3 __ 2 x + 17 __ 2
7 3x + 2
y − 5 = 0
8 7x − 4
y + 2 = 0 | [
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-0.0... |
367
Answers
3679 l: y = − 1 __ 3 x − 1, n: y = 3x + 5. Gradients are negative
reciprocals,
therefore lines perpendicular.
10 AB: y
= − 1 __ 2 x + 4 1 __ 2 , CD: y = − 1 __ 2 x − 1 __ 2 , AD: y = 2x + 7,
BC: y
= 2x − 13. Two pairs of parallel sides and lines
with gradients 2 and − 1 __ 2 are perpendicular, so ABCD is
a rectangle.
11 a A( 7 __ 5 , 0) b 55x − 25 y − 77 = 0
12 − 9 __ 4
Exercise 5G
1 a 10 b 13 c 5 d √ __
5
e √ ____ 106 f √ ____ 113
2 Distance between A and
B = √ ___ 50 and distance between
B and C
= √ ___ 50 so the lines are congruent.
3 Distance between P and
Q = √ ___ 74 and distance between
Q and R
= √ ___ 73 so the lines are not congruent.
4 x = −8 or x
= 6
5 y = −2 or y
= 16
6 a Both lines have gradient 2.
b y = − 1 _ 2 x + 23 __ 2 or x + 2 y − 23 = 0
c ( 29 __ 5 , 43 __ 5 )
d 7 √ __
5 ____ 5
7 P (− 3 __ 5 , 29 __ 5 ) or P(3, −5)
8 a AB = √ ____ 178 , BC = 3 and AC = √ ____ 205 . All sides are
different lengths, therefore the triangle is a scalene
triangle.
b 39 __ 2 or 19.5
9 a A(2, 11)
b B ( 41 __ 4 , 0)
c 451 ___ 8
10 a ( 5 __ 2 , 0) b (−5, 0)
c (−10, −10) d 75 __ 2
11 a y = 1 __ 2 x − 9 __ 2 b y = −2x + 8
c T (0, 8) d RS = 2 √ __
5 and TR = 5 √ __
5
e 25
12 a x + 4y
− 52 = 0 b A(0,13)
c B(4, 12) d 26
Exercise 5H
1 a i k = 50 ii d = 50t
b i k = 0.3 or £0.30 ii C = 0.3t
c i k = 3 __ 5 ii p = 3 __ 5 t
2 a not linear
vp
10 0203040506070805
0101520253035404550 b linear
xy
5 01015202530354025
05075100125150175200
c not linear
wl
0.5 011.522.533.544.5510
02030405060
3 a
10 02030405060
kilowatt hours70809010011010
0203040506070Cost of electricity (£)
b The data forms a straight line, so a linear model is
appropriate.
c E = 0.12h
+ 45
d a = £0.12 = cost of 1 kilowatt hour of electricity,
b = £45 = fixed electricity costs (per month or per quarter)
e £52.80
4 a
1 023456
Time, t (seconds)78910100
0200300400500Distance, d (m)
b The data does not follow a straight line. There is a definite curve to the points on the graph.
5 a C = 350d
+ 5000
b a = 350 = daily fee charged by the website designer.
b = 5000 = initial cost charged by the website designer.
c 24 days | [
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-0.022240646183490753,
0.036309290677309036,
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0.06282756477594376,
0.028959106653928757,
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-0.03229277953505516,
0.0007258870173245668,
-0.0608617439866066,
0.03746010363101959,
-0.... |
368
Answers
368 Full worked solutions are available in SolutionBank.
Online
6 a F = 1.8C + 32 or F = 9 _ 5 C + 32
b a = 1.8 = increase in Fahrenheit temperature when
the Celsius temperature increases by 1°C. b
= 32 temperature in Fahrenheit when
temperature in Celsius is 0°.
c 38.5°C
d −40°C
7 a n = 750t
+ 17 500
b The increase in the number of homes receiving the internet will be the same each year.
8
a All the points lie close to the straight line shown.
b h = 4f
+ 69
c 175 cm
9 a
Price, P
Quality , QO135
supplyequilibrium
point
demand
b Q = 24, P = 17
Mixed exercise
1 a y = − 5 __ 12 x + 11 __ 6 b −22
2 a 2k − 2 _______ 8 − k = 1 __ 3 therefore 7k = 14, k = 2
b y = 1 __ 3 x + 1 __ 3
3 a L1 = y = 1 __ 7 x + 12 __ 7 , L2 = y = −x + 12
b (9, 3)
4 a y = 3 __ 2 x − 3 __ 2 b (3, 3)
5 11x − 10
y + 19 = 0
6 a y = − 1 __ 2 x + 3 b y = 1 __ 4 x + 9 __ 4
7 Gradient = 3 + 4 √ __
3 − 3 √ __
3 ______________
2 + √ __
3 − 1 = 3 + √ __
3 _______
1 + √ __
3 = √ __
3
y = √ __
3 x + c and A(1, 3 √ __
3 ), so c = 2 √ __
3
Equation of line is y = √ __
3 x + 2 √ __
3
When y = 0,
x = −2, so the line meets the x -axis at (−2, 0)
8 a y = −3x
+ 14 b (0, 14)
9 a y = − 1 _ 2 x + 4 b Students own work.
c (1, 1). Note: equation of line
n: y = − 1 _ 2 x + 3 _ 2
10 20
11 a 2x +
y = 20 b y = 1 __ 3 x + 4 __ 3
12 a 1 __ 2 b 6 c 2x + y − 16 = 0
d 10
13 a 7x + 5
y − 18 = 0 b 162 ___ 35
14 a y
x
(0, –3)(0, 0)l2
l1
O( , 0)3
2
b ( 4 __ 3 , − 1 __ 3 ) c 12x − 3 y − 17 = 015 a x + 2y
− 16 = 0
b y = − 2 _ 3 x
c C(−48, 32)
d Slope of OA is 3 __ 2 . Slope of OC is − 2 __ 3 .
Lines are perpendicular.
e OA = 2 √ ___ 13 and OC = 16 √ ___ 13
f Area = 208
16 a d = √ ____ 50 a 2 = 5a √ __
2 b 5 √ __
2
c 15 √ __
2 d 25 √ __
2
17 a d = √ _____________ 10 x 2 − 28x + 26
b B (− 6 __ 5 , − 18 __ 5 ) and C(4, 12)
c y = − 1 __ 3 x + 14 __ 5
d ( 7 __ 5 , 21 __ 5 )
e 20.8
18 a gradient = 10.5
b C = 10.5P
− 10751
c When the oil production increases by 1 million
tonnes, the carbon diox
ide emissions increase by
10.5 million tonnes.
d The model is not valid for small values of P,
as
it is not possible to have a negative amount of carbon dioxide emissions. It is always dangerous to extrapolate beyond the range on the model in this way.
Challenge1
130
2 ( 78 __ 19 , 140 ___ 19 )
3 (a, a(c − a) ________ b )
CHAPTER 6
Prior knowledge check
1 a (x + 5) 2 + 3 b (x − 3) 2 − 8
c (x − 6) 2 − 36 d (x + 7 __ 2 )2 − 49 __ 4
2 a y = 9 __ 4 x − 6 b y = − 1 __ 2 x − 3 __ 2
c y = 4 __ 3 x + 10 __ 3
3 a b 2 − 4ac = − 7 No real solutions
b b 2 − 4ac = 193 Two real solutions
c b 2 − 4ac = 0 One real solution
4 y = − 5 __ 6 x − 3 __ 2
Exercise 6A
1 a (5, 5) b (6, 4) c (−1, 4) d (0, 0)
e (2, 1) f (−8, 3 __ 2 ) g (4a, 0 ) h (− u __ 2 , −v)
i (2a,
a − b) j (3 √ __
2 , 4) k (2 √ __
2 , √ __
2 + 3 √ __
3 )
2 a = 10,
b = 1
3 ( 3 __ 2 , 7)
4 ( 3a ___ 5 , b __ 4 )
5 a ( 3 __ 2 , 3) or (1.5, 3) b y = 2x , 3 = 2 × 1.5
6 a ( 1 __ 8 , 5 __ 3 ) b 2 __ 3
7 Centre is (3, − 7 __ 2 ) . 3 − 2 (− 7 __ 2 ) − 10 = 0
8 (10, 5)
9 (−7a, 1
7a) | [
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0.022974004... |
369
Answers
36910 p = 8, q = 7
11 a = −2,
b = 4
Challenge
a p = 9,
q = −1
b y = −x
+ 13
c AC: y = −
x + 8. Lines have the same slope, so they are
parallel.
Exercise 6B
1 a y = 2x + 3 b y = − 1 __ 3 x + 47 __ 3 c y = 5 __ 2 x − 25
d y = 3 e y = − 3 __ 4 x + 37 __ 8 f x = 9
2 y = −x
+ 7
3 2x −
y − 8 = 0
4 a y = − 5 __ 3 x − 13 __ 3 b y = 3x − 8 c ( 11 __ 14 , 79 __ 14 )
5 q = − 5 __ 4 , b = − 189 ___ 8
Challenge
a PR: y
= − 5 __ 2 x + 9 __ 4
PQ: y
= − 1 __ 4 x + 33 __ 8
RQ: y = 2x + 6
b (− 5 __ 6 , 13 __ 3 )
Exercise 6C
1 a (x − 3)2 + ( y − 2)2 = 16
b (x + 4)2 + ( y − 5)2 = 36
c (x − 5)2 + ( y + 6)2 = 12
d (x − 2a)2 + ( y − 7a)2 = 25a2
e (x + 2 √ __
2 )2 + ( y + 3 √ __
2 )2 = 1
2 a (−5, 4), 9 b (7, 1), 4
c (−4, 0), 5 d (−4a, − a), 12a
e (3 √ __
5 , − √ __
5 ), 3 √ __
3
3 a (4 − 2)2 + (8 − 5)2 = 4 + 9 = 13
b (0 + 7)2 + (−2 − 2)2 = 49 + 16 = 65
c 72 + (−24)2 = 49 + 576 = 625 = 252
d (6a − 2 a)2 + (−3a + 5a)2 = 16a2 + 4a2 = 20a2
e ( √ __
5 − 3 √ __
5 )2 + (− √ __
5 − √ __
5 )2 = (−2 √ __
5 )2 + (−2 √ __
5 )2
= 20 + 20 = 40 = (2 √ ___
10 )2
4 (x − 8)2 + ( y − 1)2 = 25
5 (x − 3 __ 2 )2 + ( y − 4)2 = 65 __ 4
6 √ __
5
7 a r = 2
b Distance PQ =
PR = RQ = 2 √ __
3 , three equal length
sides triangle is equilateral.
8 a (x − 2)2 + y2 = 15
b Centre (2, 0) and radius = √ ___
15
9 a (x − 5)2 + ( y + 2)2 = 49
b Centre (5, −2) and radius = 7
10 a Centre (1, −4), radius 5
b Centre (−6, 2), radius 7
c Centre (11, −3), radius 3 √ ___
10
d 10 Centre (−2.5, 1.5), radius 5 √ __
2 ____ 2
e Centre (2 ,−2), radius
11 a Centre (−6, −1)
b k >
−37
12 Q(−13, 28)
13 k = −2 and k
= 8Challenge
1 k = 3, (
x − 3)2 + ( y − 2)2 = 50
k = 5,
(x − 5)2 + ( y − 2)2 = 50
2 (x
+ f )2 − f 2 + (y + g)2 − g2 + c = 0
So (x +
f )2 + (y + g)2 = f 2 + g2 − c
Circle with centre (−f, − g ) and radius √ __________ f 2 + g2 − c .
Exercise 6D
1 (7, 0), (−5, 0)
2 (0, 2), (0,
−8)
3 (6, 10), (2,
−2)
4 (4, −9), (−7,
2)
5 2x2 − 24x + 79 = 0 has no real solutions, therefore
lines do not intersect
6 a b 2 − 4ac = 64 − 4 × 1 × 16 = 0 . So there is only one
point of intersection.
b (4, 7)
7 a (0, −2), (4,
6) b midpoint of AB is (2,
2)
8 a 13 b p = 1 or 5
9 a A(5, 0) and
B(−3, −8) (or vice-versa)
b y = −x
− 3
c (4, −7) is a solution to y
= −x − 3.
d 20
10 a Substitute y =
kx to give
(k2 + 1)x2 − (12k + 10)x + 57 = 0
b2 − 4ac > 0, −84k2 + 240k − 128> 0,
21k2 − 60k + 32 < 0
b 0.71 < k < 2.15
Exact answer is 10 ___ 7 − 2 √ ___
57 _____ 21 < k < 10 ___ 7 + 2 √ ___
57 _____ 2
11 k < 8 __ 17
12 k = −20 ± 2 √ ____ 105
Exercise 6E
1 a 3 √ ___
10
b Gradient of radius = 3x,
gradient of line = − 1 __ 3 ,
gradients are negative reciprocals and therefore
perpendicular.
2 a (x − 4)2 + ( y − 6)2 = 73 b 3x + 8 y + 13 = 0
3 a y = −2x
− 1
b Centre of circle (1, −3) satisfies y
= −2x − 1.
4 a y = 1 __ 2 x − 3
b Centre of circle (2, −2) satisfies y
= 1 __ 2 x − 3
5 a (−7, −6) satisfies x2 + 18x + y2 − 2y + 29 = 0
b y = 2 __ 7 x − 4 c R(0, −4) d 53 __ 2
6 a (0, −17), (17,
0)
b 144.5
7 y = 2x
+ 27 and y = 2x − 13
8 a p = 4,
p = −6
b (3, 4) and (3, −6)
9 a (x − 11)2 + ( y + 5)2 = 100
b y = 3 __ 4 x − 3 __ 4
c A(8 − 4 √ __
3 , −1 − 3 √ __
3 ) and B(8 + 4 √ __
3 , −1 + 3 √ __
3 )
d 10 √ __
3
10 a y = 4x
− 22
b a = 5
c (x − 5)2 + (y + 2)2 = 34
d A(5 + √ __
2 ,−2 + 4 √ __
2 ) and B(5 − √ __
2 ,−2 − 4 √ __
2 ) | [
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-0.0... |
370
Answers
370 Full worked solutions are available in SolutionBank.
Online
11 a P(−2, 5) and Q(4, 7)
b y = 2x
+ 9 and y = − 1 __ 2 x + 9
c y = −3x
+ 9
d (0, 9)
Challenge
1 y = 1 __ 2 x − 2
2 a ∠ CPR = ∠ CQR = 90° (Angle between tangent and
radius)
CP =
CQ = √ ___
10 (Radii of circle)
CR = √ _______ (6 − 2)2 + (−1 − 1)2 = √ ___
20
So using Pythagoras’ Theorem,
PR = QR
= √ ________ 20 − 10 = √ ___
10
4 equal sides and two opposite right−angles,
so CPRQ
is a square
b y = 1 __ 3 x − 3 and y = −3x + 17
Exercise 6F
1 a WV 2 = WU 2 + UV 2
b (2, 3)
c (x − 2)2 + ( y − 3)2 = 41
2 a AC2 = AB2 + BC2
b (x − 5)2 + ( y − 2)2 = 25
c 15
3 a i y = 3 __ 2 x + 21 __ 2 ii y = − 2 __ 3 x + 4
b (−3, 6)
c (x + 3)2 + ( y − 6)2 = 169
4 a i y = 1 __ 3 x + 10 __ 3 ii x = −1
b (x + 1)2 + ( y − 3)2 = 125
5 (x − 3)2 + ( y + 4)2 = 50
6 a AB2 + BC 2 = AC 2
AB2 = 400, BC 2 = 100, AC 2 = 500
b (x + 2)2 + ( y − 5)2 = 125
c D (8, 0) satisfies the equation of the circle.
7 a AB = BC
= CD = DA = √ ___
50
b 50
c (3,6)
8 a DE 2 = b2 + 6b + 13
EF 2 = b2 + 10b + 169
DF 2 = 200
So b2 + 6b + 13 + b2 + 10b + 169 = 200
(b + 9)(b − 1) = 0; as b > 0, b = 1
b (x + 5)2 + ( y + 4)2 = 50
9 a Centre (−1, 12) and radius = 13
b Use distance formula to find AB = 26.
This is twice
radius, so AB is the diameter. Other methods
possible.
c C(−6, 0)
Mixed exercise
1 a C (3, 6)
b r = 10
c (x − 3)2 + ( y − 6)2 = 100
d P satisfies the equation of the circle.
2 (0 − 5)2 + (0 + 2)2 = 52 + 22 = 29 < 30 therefore point is
inside the circle
3 a Centre (0, −4) and radius = 3
b (0, −1) and (0, −7)
c Students’ own work. Equation x2 = −7 has no real
solutions.
4 a P (8, 8),
(8 + 1)2 + (8 − 3)2 = 92 + 52 = 81 + 25 = 106
b √ ____ 106 5 a All points satisfy x2 + y2 =1, therefore all lie on circle.
b AB = BC
= CA
6 a k = 1,
k = − 2 __ 5
b (x − 1)2 + ( y − 3)2 = 13
7 Substitute y = 3
x − 9 into the equation
x2 + px + y2 + 4y = 20
x2 + px + (3x − 9)2 + 4(3x − 9) = 20
10x2 + ( p − 42)x + 25 = 0
Using the discriminant: ( p − 42)2 − 1000 < 0
42 − 10 √ ___
10 < p < 42 + 10 √ ___
10
8 (x − 2)2 + ( y + 4)2 = 20
9 a 2 √ ___
29 b 12
10 (−1, 0), (11,
0)
11 The values of m and
n are 7 − √ ____ 105 and 7 + √ ____ 105 .
12 a a = 6 and b
= 8 b y = − 4 __ 3 x + 8 c 24
13 a p = 0,
q = 24 b (0, 49), (0,
−1)
14 x + y
+ 10 = 0
15 60
16 l1: y = −4x + 12 and l2: y = − 8 __ 19 x + 12
17 a y = 1 __ 3 x + 8 __ 3
b (x + 2)2 + ( y − 2)2 = 50
c 20
18 a P (−3, 1) and Q
(9, −7)
b y = 3 __ 2 x + 11 __ 2 and y = 3 __ 2 x − 41 __ 2
19 a y = −4x
+ 6 and y = 1 __ 4 x + 6
b P (−4, 5) and Q
(1, 2)
c 17
20 a P (5, 16) and Q
(13, 8)
b l2: y = 1 __ 7 x + 107 ___ 7 and l3: y = 7x − 83
c l4: y = x + 3
d All 3 equations have solution x = 15,
y = 18
so R(15, 18)
e 200 ___ 3
21 a (4,0), (0,12)
b (2,6)
c (x − 2)2 + ( y − 6)2 = 40
22 a q = 4
b (x + 5 __ 2 )2 + ( y − 2)2 = − 65 __ 4
23 a RS 2 + ST 2 = RT 2
b (x − 2)2 + ( y + 2)2 = 61
24 (x − 1)2 + ( y − 3)2 = 34
25 a i y = −4x
− 4
ii x = −2
b (x + 2)2 + ( y − 4)2 = 34
Challenge
a x + y
− 14 = 0
b P (7, 7) and
Q (9, 5)
c 10
CHAPTER 7
Prior knowledge check
1 a 15x7 b x ___ 3y
2 a (x − 6)(
x + 4) b (3x − 5)(
x − 4)
3 a 8567 b 1652
4 a y = 1 − 3x b y = 1 __ 2 x − 7
5 a (x − 1)2 − 21 b 2(x + 1)2 + 13
Exercise 7A
1 a 4x3 + 5x − 7 b 2x4 + 9x2 + x
c −x3 + 4x + 6 __ x d 7x4 − x2 − 4 __ x | [
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0.0090... |
371
Answers
371 e 4x3 − 2x2 + 3 f 3x − 4 x2 − 1
g 7x2 ____ 5 − x3 __ 5 − 2 ___ 5x h 2x − 3x3 + 1
i x 7 ___ 2 − 9 x 3 ____ 2 + 2 x 2 − 3 __ x j 3 x 8 + 2 x 5 − 4 x 3 ____ 3 + 2 ___ 3x
2 a x + 3 b x + 4 c x + 3
d x + 7 e x + 5 f x + 4
g x − 4 _____ x − 3 h x + 2 _____ x + 4 i x + 4 _____ x − 6
j 2x + 3 _______ x − 5 k 2x − 3 _______ x + 1 l x − 2 _____ x + 2
m 2x + 1 _______ x − 2 n x + 4 _______ 3x + 1 o 2x + 1 _______ 2x − 3
3 a = 1,
b = 4, c = −2
Exercise 7B
1 a (x + 1)( x2 + 5x + 3) b (x + 4)( x2 + 6x + 1)
c (x + 2)(
x2 − 3x + 7) d (x − 3)( x2 + 4x + 5)
e (x − 5)(
x2 − 3x − 2) f (x − 7)( x2 + 2x + 8)
2 a (x + 4)(6
x2 + 3x + 2) b (x + 2)(4 x2 + x − 5)
c (x + 3)(2
x2 − 2x − 3) d (x − 6)(2 x2 − 3x − 4)
e (x + 6)(−5
x2 + 3x + 5) f (x − 2)(−4 x2 + x − 1)
3 a x3 + 3x2 − 4x + 1 b 4x3 + 2x2 − 3x − 5
c −3x3 + 3x2 − 4x − 7 d −5x4 + 2x3 + 4x2 − 3x + 7
4 a x3 + 2x2 − 5x + 4 b x3 − x2 + 3x − 1
c 2x3 + 5x + 2 d 3x4 + 2x3− 5x2 + 3x + 6
e 2x4 − 2x3 + 3x2 + 4x − 7 f 4x4 − 3x3− 2x2 + 6x − 5
g 5x3 + 12x2 − 6x − 2 h 3x4 + 5x3 + 6
5 a x2 − 2x + 5 b 2x2 − 6x + 1
c −3x2 − 12x + 2
6 a x2 + 4x + 12 b 2x2 − x + 5
c −3x2 + 5x + 10
7 f(−2) = −8 + 8 + 10 − 10 = 0, so (x
+ 2) is a factor of
x3 + 2x2 − 5x − 10. Divide x3 + 2x2 − 5x − 10 by (x + 2)
to give (x2 − 5). So x3 + 2x2 − 5x − 10 = (x + 2)(x2 − 5).
8 a −8 b −7 c −12
9 f(1) = 3 − 2 + 4 = 5
10 f(−1) = 3 + 8 + 10 + 3 − 25 = −1
11 (x + 4)(5
x2 − 20x + 7)
12 3x2 + 6x + 4
13 x2 + x + 1
14 x3 − 2x2 + 4x − 8
15 14
16 a −200 b (x + 2)(
x − 7)(3x + 1)
17 a i 30 ii 0 b x = −3,
x = −4, x = 1
18 a a = 1,
b = 2, c = −3
b f(x) = (2
x − 1)(x + 3)(x − 1)
c x = 0.5,
x = −3, x = 1
19 a a = 3,
b = 2, c = 1
b Quadratic has no real solutions so only (4x − 1) is a
solution
Exercise 7C
1 a f(1) = 0 b f(−3) = 0 c f(4) = 0
2 (x − 1)( x + 3)(x + 4)
3 (x + 1)(
x + 7)(x − 5)
4 (x − 5)(
x − 4)(x + 2)
5 (x − 2)(2
x − 1)(x + 4)
6 a (x + 1)(
x − 5)(x − 6) b (x − 2)(
x + 1)(x + 2)
c (x − 5)(
x + 3)(x − 2)7 a i (x − 1)(
x + 3)(2x + 1) ii y
x O1 –3 –1
2
–3
b i (x − 3)( x − 5)(2x − 1) ii y
x O 3 5 12
–15
c i (x + 1)( x + 2)(3x − 1) ii y
x O –2–112
–2
d i (x + 2)(2 x − 1)(3x + 1) ii
xO
–2 12–13
–2y
e i (x − 2)(2 x − 5)(2x + 3) ii y
x O 230
52–32
8 2
9 −16
10 p = 3,
q = 7
11 c = 2,
d = 3
12 g = 3,
h = −7
13 a f(4) = 0
b f(x) = (
x − 4)(3x2 + 6)
For 3x2 + 6 = 0, b2 − 4ac = −72 so there are no real
roots. Therefore, 4 is the only real root of f(x) = 0.
14 a f(−2) = 0 b (x + 2)(2
x + 1)(2x − 3)
c x = −2,
x = − 1 __ 2 and x = 1 1 __ 2
15 a f(2) = 0 b x = 0,
x = 2, x = − 1 __ 3 and x = 1 __ 3
Challenge
a f(1) = 2 − 5 − 42 − 9 + 54 = 0
f(−3) = 162 + 135 − 378 + 27 + 54 = 0
b 2x4 − 5x3 − 42x2 − 9x + 54
= (x − 1)(x + 3)(x − 6)(2x + 3)
x = 1 x
= −3, x = 6, x = −1.5 | [
-0.016584021970629692,
0.026711642742156982,
0.0020620261784642935,
-0.04096461087465286,
-0.0318741649389267,
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0.04342261701822281,
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0.061253614723682404,
0.012651643715798855,
-0.07829251140356064,
-0.007970002479851246,
-0.... |
372
Answers
372 Full worked solutions are available in SolutionBank.
Online
Exercise 7D
1 n 2 − n = n(n − 1)
If n is even,
n − 1 is odd and even × odd = even
If n is odd,
n − 1 is even and odd × even = even
2 x ________
(1 + √ __
2 ) × (1 − √ __
2 ) ________
(1 − √ __
2 ) = x (1 − √ __
2 ) _________ (1 − 2 ) = x − x √ __
2 ________ − 1 = x √ __
2 − x
3 (x + √ __ y )(x − √ __ y ) = x 2 − x √ __ y + x √ __ y − y = x 2 − y
4 (2x − 1)(
x + 6)(x − 5) = (2x − 1)(
x2 + x − 30)
= 2x3 + x2 − 61x + 30
5 LHS = x2 + bx, using completing the square,
(x + b __ 2 ) 2
− ( b __ 2 ) 2
6 x2 + 2bx + c = 0, using completing the square
(x +
b) 2 + c − b 2 = 0
(x +
b) 2 = b 2 − c
x + b
= ± √ _____ b 2 − c
x = − b
± √ _____ b 2 − c
7 (x − 2 __ x ) 3
= (x − 2 __ x ) ( x 2 − 4 + 4 ___ x 2 ) = x 3 − 6x + 12 ___ x − 8 ___ x 3
8 ( x 3 − 1 __ x ) ( x 3 _ 2 + x −5 ___ 2 ) = x 9 _ 2 + x 1 _ 2 − x 1 _ 2 − x − 7 _ 2 = x 9 _ 2 − x − 7 _ 2
= x 1 _ 2 ( x 4 − 1 ___ x 4 )
9 3n2 − 4n + 10 = 3 [ n 2 − 4 _ 3 n + 10 __ 3 ] = 3 [ (n − 2 _ 3 ) 2 + 10 __ 3 − 4 _ 9 ]
= 3 (n − 2 _ 3 ) 2 + 26 __ 3
The minimum value is 26 __ 3 so 3n2 − 4n + 10 is always
positive.
10 −n2 − 2n − 3 = − [ n 2 + 2n + 3 ] = − [ (n + 1 ) 2 + 3 − 1 ]
= − (n + 1 ) 2 − 2
The maximum value is −2 so −
n2 − 2n − 3 is always
negative.
11 x2 + 8x + 20 = (x + 4)2 + 4
The minimum value is 4 so x2 + 8x + 20 is always
greater than or equal to 4.
12 k x 2 + 5kx + 3 = 0 , b2 − 4ac < 0, 25k2 − 12k < 0,
k(25k − 12) < 0, 0 < k < 12 __ 25 .
When k = 0 there are no real roots,
so 0 < k < 12 __ 25
13 p x 2 − 5x − 6 = 0 , b2 − 4ac > 0, 25 + 24p > 0, p > − 25 __ 24
14 Gradient AB = − 1 __ 2 , gradient BC = 2,
Gradient AB × gradient BC = − 1 __ 2 × 2 = −1,
so AB and BC are perpendicular.
15 Gradient AB = 3,
gradient BC = 1 __ 4 , gradient CD = 3,
gradient AD = 1 __ 4
Gradient AB = gradient
CD so AB and CD are parallel.
Gradient BC = gradient
AD so BC and AD are parallel.
16 Gradient AB = 1 __ 3 , gradient BC = 3, gradient CD = 1 __ 3 ,
gradient AD = 3
Gradient AB = gradient
CD so AB and CD are parallel.
Gradient BC = gradient
AD so BC and AD are parallel.
Length AB = √ ___ 10 , BC = √ ___ 10 , CD = √ ___ 10 and AD = √ ___ 10 ,
all four sides are equal
17 Gradient AB = −3,
gradient BC = 1 __ 3 ,
Gradient AB × gradient
BC = −3 × 1 __ 3 = −1, so AB and BC
are perpendicular
Length AB = √ ___ 40 , BC = √ ___ 40 , AB = BC
18 (x − 1)2 + y2 = k, y = ax, ( x − 1)2 + a2x2 = k,
x2(1 + a2) − 2x + 1 − k = 0
b2 − 4ac > 0, k > a 2 ______ 1 + a 2 .19 x = 2.
There is only one solution so the line
4y − 3x + 26 = 0 only touches the circle in one place so
is the tangent to the circle.
20 Area of square = (a +
b) 2 = a 2 + 2ab + b 2
Shaded area = 4 ( 1 __ 2 ab)
Area of smaller square: a 2 + 2ab + b 2 − 2ab
= a 2 + b 2 = c 2
Challenge
1 The equation of the circle is (x − 3) 2 + (y − 5) 2 = 25 and
all four points satisfy this equation.
2 ( 1 _ 2 (p + 1) ) 2 − ( 1 _ 2 (p − 1) ) 2 = 1 _ 4 ( (p + 1) 2 − (p − 1) 2 ) = 1 _ 4 (4p) = p
Exercise 7E
1 3, 4, 5, 6, 7 and 8 are not divisible by 10
2 3, 5, 7,
11, 13, 17, 19, 23 are prime numbers. 9, 15, 21,
25, are the product of two prime numbers.
3 2 2 + 3 2 = odd, 3 2 + 4 2 = odd, 4 2 + 5 2 = odd, 5 2 + 6 2 = odd,
6 2 + 7 2 = odd
4 (3n) 3 = 27 n 3 = 9n(3 n 2 ) which is a multiple of 9 | [
0.06108298525214195,
0.06655146181583405,
0.03833566606044769,
0.005557755474001169,
-0.05973406136035919,
0.0749589279294014,
0.03334863483905792,
-0.0715133547782898,
-0.06544898450374603,
-0.1278267204761505,
-0.04508297145366669,
-0.019872965291142464,
0.013491598889231682,
0.038086693... |
(3n + 1) 3 = 27 n 3 + 27 n 2 + 9n + 1 = 9n(3 n 2 + 3n + 1) + 1
which is one more than a multiple of 9
(3n + 2) 3 = 27 n 3 + 54 n 2 + 36n + 8 = 9n(3 n 2 + 6n + 4) + 8
which is one less than a multiple of 9
5 a For example, when n
= 2, 24 − 2 = 14, 14 is not
divisible by 4.
b Any square number
c For example,
when n = 1 __ 2
d For example,
when n = 1
6 a Assuming that x and
y are positive
b e.g. x
= 0, y = 0
7 (x + 5)2 > 0 for all real values of x, and
(x + 5)2 + 2x + 11 = (x + 6)2, so (x + 6)2 > 2x + 11
8 If a2 + 1 > 2a (a is positive, so multiplying both sides
by a does not reverse the inequality), then
a2 − 2a + 1 > 0, and (a − 1)2 > 0, which we know is
true.
9 a ( p + q
)2 = p2 + 2pq + q2 = ( p + q )2 + 4pq
( p − q
)2 > 0 since it is a square, so ( p + q )2 > 4pq
p > 0, q > 0 ⇒ p + q > 0 ⇒ p + q > √ ____ 4pq
b e.g. p
= q = −1: p + q = −2, √ ____ 4pq = 2
10 a Starts by assuming the inequality is true:
i.e. negative
> positive
b e.g. x
= y = −1: x + y = −2, √ _______ x2 + y2 = √ __
2
c (x
+ y)2 = x2 + 2xy + y2 > x2 + y2 since x > 0,
y > 0 ⇒ 2xy > 0
As x + y > 0, can take square roots: x + y > √ _______ x2 + y2
Mixed exercise
1 a x3 − 7 b x + 4 _____ x − 1 c 2x − 1 _______ 2x + 1
2 3x2 + 5
3 2x2 − 2x + 5
4 a When x = 3,
2x3 − 2x2 − 17x + 15 = 0
b A = 2,
B = 4, C = −5
5 a When x = 2,
x3 + 4x2 − 3x − 18 = 0
b p = 1,
q = 3
6 (x − 2)(
x + 4)(2x − 1)
7 7
8 a p = 1,
q = −15 b (x + 3)(2
x − 5)
9 a r = 3,
s = 0 b x(x
+ 1)(x + 3)
10 a (x − 1)(
x + 5)(2x + 1) b −5, − 1 __ 2 , 1 | [
0.021066656336188316,
0.09007084369659424,
0.00018930229998659343,
-0.00882311724126339,
0.04175085946917534,
0.015868905931711197,
-0.02084236964583397,
-0.048021845519542694,
-0.02944907732307911,
-0.009839308448135853,
0.0021875877864658833,
0.00882814172655344,
0.131444051861763,
0.094... |
373
Answers
37311 a When x = 2, x3 + x2 − 5x − 2 = 0
b 2, − 3 __ 2 ± √ __
5 ___ 2
12 1 __ 2 , 3
13 a When x = −4,
f(x) = 0
b (x + 4)(x − 5)(x − 1)
14 a f( 2 __ 3 ) = 0, therefore (3x − 2) is a factor of f(x)
a = 2, b = 7 and c = 3
b (3x − 2)(2x + 1)(x + 3)
c x = 2 __ 3 , − 1 __ 2 , −3
15 x − y ________ ( √ __ x − √ __ y ) × ( √ __ x + √ __ y ) ________ ( √ __ x + √ __ y ) = x √ __ x + x √ __ y − y √ __ x − y √ __ y _____________________ x − y = √ __ x + √ __ y
16 n2 − 8n + 20 = (n − 4) 2 + 4 , 4 is the minimum value so
n2 − 8n + 20 is always positive
17 Gradient AB = 1 __ 2 , gradient BC = −2, gradient CD = 1 __ 2 ,
gradient AD = −2
AB and BC
, BC and CD, CD and AD and AB and AD are
all perpendicular
Length AB = √ __
5 , BC = √ __
5 , CD = √ __
5 and AD = √ __
5 , all four
sides are equal
18 1 + 3 = even, 3 + 5 = even, 5 + 7 = even,
7 + 9 = even
19 For example when n = 6
20 (x − 1 __ x ) ( x 4 _ 3 + x −2 ___ 3 ) = x 7 _ 3 + x 1 _ 3 − x 1 _ 3 − x − 5 _ 3 = x 1 _ 3 ( x 2 − 1 ___ x 2 )
21 RHS = (x + 4)(x − 5)(2x + 3) = (x + 4)(2 x 2 − 7x − 15)
= 2 x 3 + x 2 − 43x − 60 = LHS
22 x 2 − kx + k = 0 , b 2 − 4ac = 0, k 2 − 4k = 0 , k( k − 4) = 0,
k = 4.
23 The distance between opposite edges
= 2 ( ( √ __
3 ) 2 − ( √ __
3 ___ 2 ) 2
) = 2 (3 − 3 _ 4 ) = 9 _ 2 which is rational.
24 a (2n + 2) 2 − (2n) 2 = 8n + 4 = 4(2n + 1) is always
divisible by 4.
b Yes, (2n + 1) 2 − (2n − 1) 2 = 8n which is always
divisible by 4.
25 a The assumption is that x is positive
b x = 0
Challenge
1 a Perimeter of inside square = 4 ( √ ________ ( 1 _ 2 ) 2 + ( 1 _ 2 ) 2 ) = 4 ___
√ __
2 = 2 √ __
2
Perimeter of outside square = 4,
therefore 2 √ __
2 < π < 4.
b Perimeter of inside hexagon = 3
Perimeter of outside hexagon = 6 × √ __
3 ___ 3 = 2 √ __
3 ,
therefore 3 < π < 2 √ __
3
2 a x 3 + b x 2 + cx + d ÷ (x − p) = a x 2 + (b + ap)x
+ (c + bp + a p 2 ) with remainder d + cp + b p 2 + a p 3
f( p) = a p 3 + b p 2 + cp + d = 0 , which matches the
remainder,
so (x − p) is a factor of f(x).
CHAPTER 8
Prior knowledge check
1 a 4x2 − 12xy − 9y2 b x3 + 3x2y + 3xy2 + y3
c 8 + 12x + 6 x2 + x3
2 a − 8 x 3 b 1 _____ 81 x 4 c 4 ___ 25 x 2 d 27 ___ x 3
3 a 5 √ __ x b 1 ______
16 3 √ __
x 2 c 10 ____ 3 √ __ x d 16 ___ 81 3 √ __
x 4 Exercise 8A
1 a 4th row b 16th row
c (n + 1)th row d (n + 5)th row
2 a x4 + 4x3y + 6x2y2 + 4xy3 + y4
b p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
c a3 − 3a2b + 3ab2 − b3
d x3 + 12x2 + 48x + 64
e 16x4 − 96x3 + 216x2 − 216x + 81
f a5 + 10a4 + 40a3 + 80a2 + 80a + 32
g 81x4 − 432x3 + 864x2 − 768x + 256
h 16x4 − 96x3y + 216x2y2 − 216xy3 + 81y4
3 a 16 b −10 c 8 d 1280
e 160 f −2 g 40 h −96
4 1 + 9x + 30
x2 + 44x3 + 24x4
5 8 + 12y + 6 y2 + y3, 8 + 12x − 6x2 − 11x3 + 3x4 + 3x5−x6
6 ±3
7 5 __ 2 , −1
8 12p
9 500 + 25X + X 2 ___ 2
Challenge
3 __ 4
Exercise 8B
1 a 24 b 362 880 c 720 d 210
2 a 6 b 15 c 20 d 5
e 45 f 126
3 a 5005 b 120 c 184 756 d 1140
e 2002 f 8568
4 a = 4C1, b = 5C2, c = 6C2, d = 6C3
5 330
6 a 120, 210 b 960
7 a 286, 715 b 57 915
8 0.1762 to 4 decimal places. Whilst it seems a low
probability
, there is more chance of the coin landing on
10 heads than any other amount of heads.
9 a n C 1 = n ! _________ 1 !(n − 1) !
= 1 × 2 × … × (n − 2) × (n − 1) × n ____________________________________ 1 × 1 × 2 × … × (n − 3) × (n − 2) × (n − 1) = n
b n C 2 = n ! _________ 2 !(n − 2) ! | [
-0.0577564612030983,
0.11641733348369598,
0.06692392379045486,
-0.028106648474931717,
0.036220453679561615,
0.10602177679538727,
-0.003296572482213378,
-0.04312630370259285,
-0.08979640901088715,
0.031300514936447144,
-0.02728533372282982,
-0.12264680862426758,
-0.012337922118604183,
-0.04... |
= 1 × 2 × … × (n − 2) × (n − 1) × n _______________________________ 1 × 2 × 1 × 2 × … × (n − 3) × (n − 2) = n(n − 1) ________ 2
10 a = 37
11 p = 17
Challenge
a 10C3 = 10 ! ____ 3 !7 ! = 120 and 10C7 = 10 ! ____ 7 !3 ! = 120
b 14C5 = 14 ! ____ 5 !9 ! = 2002 and 14C9 = 14 ! ____ 9 !5 ! = 2002
c The two answers for part a are the same and the two
answers for part
b are the same.
d nCr = n ! ________ r !(n − r) ! and nCn − r = n ! ________ (n − r) !r ! , therefore nCr = nCn − r
Exercise 8C
1 a 1 + 4x + 6 x2 + 4x3 + x4
b 81 + 108x + 54 x2 + 12x3 + x4
c 256 − 256x + 96 x2 − 16x3 + x4 | [
0.012324175797402859,
0.08403260260820389,
0.029683727771043777,
-0.008278165012598038,
-0.0851149708032608,
0.07936793565750122,
0.031360380351543427,
-0.08045043796300888,
-0.07227443903684616,
-0.004326690454035997,
-0.09868813306093216,
-0.09718167036771774,
0.058278534561395645,
0.013... |
374
Answers
374 Full worked solutions are available in SolutionBank.
Online
d x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64
e 1 + 8x + 24
x2 + 32x3 + 16x4
f 1 − 2x + 3 _ 2 x2 − 1 _ 2 x3 + 1 __ 16 x4
2 a 1 + 10x + 45 x2 + 120x3
b 1 − 10x + 40 x2 − 80x3
c 1 + 18x + 135 x2 + 540x3
d 256 − 1024x + 1792 x2 − 1792x3
e 1024 − 2560x + 2880 x2 − 1920x3
f 2187 − 5103x + 5103 x2 − 2835x3
3 a 64x6 + 192x5y + 240x4y2 + 160x3y3
b 32x5 + 240x4y + 720x3y2 + 1080x2y3
c p8 − 8p7q + 28p6q2 − 56p5q3
d 729x6 − 1458x5y + 1215x4y2 − 540x3y3
e x8 + 16x7y + 112x6y2 + 448x5y3
f 512x9 − 6912x8y + 41 472x7y2 − 145 152x6y3
4 a 1 + 8x + 28 x2 + 56x3
b 1 − 12x + 60 x2 − 160x3
c 1 + 5x + 45 __ 4 x2 + 15x3
d 1 − 15x + 90 x2 − 270x3
e 128 + 448x + 672 x2 + 560x3
f 27 − 54x + + 36 x2 − 8x3
g 64 − 576x + 2160 x2 − 4320x3
h 256 + 256x + 96 x2 + 16x3
i 128 + 2240x + 16 800x2 + 70 000x3
5 64 − 192x + 240 x2
6 243 − 810x + 1080 x2
7 x5 + 5x3 + 10x + 10 ___ x + 5 __ x3 + 1 __ x5
Challenge
a (a +
b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a − b)4 = a4 − 4a3b + 6a2b2 − 4ab3 + b4
(a + b)4 − (a − b)4 = 8a3b + 8ab3 = 8ab(a2 + b2)
b 82 896 = 24 × 3 × 11 × 157
Exercise 8D
1 a 90 b 80 c −20
d 1080 e 120 f −4320
g 1140 h −241 920 i −2.5
j 354.375 k −224 l 3.90625
2 a = ± 1 __ 2
3 b = −2
4 1, 5 ± √105 _________ 8
5 a p = 5 b −10 c −80
6 a 5 30 + 5 29 × 30px + 5 28 × 435 p 2 x 2
b p = 10
7 a 1 + 10qx + 45 q 2 x 2 +120q3x3
b q = ±3
8 a 1 + 11px + 55
p2x2
b p = 7, q = 2695
9 a 1 + 15px + 105
p2x2
b p = − 5 _ 7 , q = 10 5 _ 7
10 q __ p = 2 . 1
Challenge
a 314 928 b 43 750
Exercise 8E
1 a 1 − 0.6x + 0.15 x2 − 0.02x3
b 0.941 482 a 1024 + 1024x + 460.8
x2 + 122.88x3
b 1666.56
3 (1 − 3x) 5 = 1 5 + ( 5 1 ) 1 4 (−3x) 1 + ( 5 2 ) 1 3 (−3x) 2 = 1 − 15x + 90 x 2
(2 + x) (1 − 3x) 5 = (2 + x) (1 − 15x + 90 x 2 )
= 2 − 30x + 180 x 2 + x − 15 x 2 + 90 x 3 ≈ 2 − 29x + 165 x 2
4 a = 162,
b = 135, c = 0
5 a 1 + 16x + 112
x2 + 448x3
b x = 0.01, 1.028 ≈ 1.171 648
6 a 1 − 150x + 10875
x2 − 507500x3
b 0.860 368
c 0.860 384,
0.0019%
7 a 59 049 −
39 366x
+ 11 809.8x2
b Substitute x = 0.1 into the expansion.
8 a 1 − 15x + 90
x2 − 270x3
b (1 + x)(1 − 3 x)5 ≈ (1 + x)(1 − 15x) ≈ 1 − 14x
9 a So that higher powers of p can be ignored as they
tend to 0
b 1 + 200p − 19 900p2
c p = 0.000417 (3 s.f.)
Mixed exercise
1 a 455, 1365 b 3640
2 a = 28
3 a 0.0148 b 0.000 000 000 034 9 c 0.166
4 a p = 16 b 270 c −1890
5 A = 8192,
B = −53 248, C
= 159 744
6 a 1 − 20x + 180
x2 − 960x3
b 0.817 04, x = 0.01
7 a 1024 − 153 60x + 103 680
x2 − 414 720x3
b 880.35
8 a 81 + 216x + 216
x2 + 96x3 + 16x4
b 81 − 216x + 216 x2 − 96x3 + 16x4
c 1154
9 a n = 8 b 35 __ 8
10 a 81 + 1080x + 5400
x2 + 12 000x3 + 10 000x4
b 1 012 054 108 081, x = 100
11 a 1 + 24x + 264
x2 + 1760x3 b 1.268 16
c 1.268 241 795 d 0.006 45% (3 sf )
12 x5 − 5x3 + 10x − 10 ___ x + 5 __ x3 − 1 __ x5
13 a ( n 2 ) (2k)n − 2 = ( n 3 ) (2k)n − 3
n!(2k)n − 2 _________ 2!(n − 2)! = n!(2k
)n − 3 _________ 3!(n − 3)!
2k _____ n − 2 = 1 __ 3
So n = 6
k + 2
b 4096 _____ 729 + 2048 _____ 81 x + 1280 _____ 27 x2 + 1280 _____ 27 x3
14 a 64 + 192x + 240 x2 + 160x3 + 60x4 + 12x5 + x6
b k = 1560
15 a k = 1.25 b 3500
16 a A = 64,
B = 160, C = 20 b x = ± √ __
3 __ 2
17 a p = 1.5 b 50.625
18 672
19 a 128 + 448px + 672
p2x2
b p = 5, q = 16 800
20 a 1 − 12px + 66
p2x2
b p = −1 1 __ 11 , q = 13 1 __ 11 | [
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-0.03071... |
375
Answers
37521 a 128 + 224x + 168 x2
b Substitute x = 0.1 into the expansion.
22 k = 1 __ 2
Challenge
1 540 − 405p = 0,
p = 4 _ 3
2 −4704
CHAPTER 9
Prior knowledge check
1 a 3.10 cm b 9.05 cm
2 a 25.8° b 77.2°
3 a graph of x2 + 3x b graph of (x + 2)2 + 3(x + 2)
c graph of x2 + 3x − 3 b graph of (0.5x)2 + 3(0.5x)
Exercise 9A
1 a 3.19 cm b 1.73 cm ( √ __
3 cm) c 9.85 cm
d 4.31 cm e 6.84 cm f 9.80 cm
2 a 108(.2)° b 90° c 60°
d 52.6° e 137° f 72.2°
3 192 km
4 11.2 km
5 128.5° or 031.5° (Angle BAC = 48.5°)
6 302 yards (301.5…)
7 Using the cosine rule 52 + 42 − 62 ___________ 2 × 5 × 4 = 1 __ 8
8 Using the cosine rule 22 + 32 − 42 ___________ 2 × 2 × 3 = − 1 __ 4
9 ACB = 22.3°
10 ABC = 108(.4)°
11 104° (104.48)°
12 4.4 cm
13 42 cm
14 a y2 = (5 − x)2 + (4 + x)2 − 2(5 − x)(4 + x) cos 120°
= 25 − 10
x + x2 + 16 + 8x + x2 − 2(20 + x − x2) (− 1 __ 2 )
= x2 − x + 61
b Minimum AC 2 = 60.75; it occurs for x = 1 __ 2
15 a cos ∠ABC = x2 + 52 − (10 − x)2 _________________ 2x × 5
= 20x −
75 _________ 10x = 4x −
15 ________ 2x
b 3.5
16 65.3°
17 a 28.7 km b 056.6°
Exercise 9B
1 a 15.2 cm b 9.57 cm c 8.97 cm d 4.61 cm
2 a x = 84°,
y = 6.32
b x = 13.5,
y = 16.6
c x = 85°,
y = 13.9
d x = 80°,
y = 6.22 (isosceles triangle)
e x = 6.27,
y = 7.16
f x = 4.49,
y = 7.49 (right-angled)
3 a 36.4° b 35.8° c 40.5° d 130°
4 a 48.1° b 45.6° c 14.8° d 48.7°
e 86.5° f 77.4°
5 a 1.41 cm ( √ __
2 cm) b 1.93 cm
6 QPR = 50.6°,
PQR = 54.4°
7 a x = 43.2°,
y = 5.02 cm b x = 101°,
y = 15.0 cm
c x = 6.58 cm,
y = 32.1° d x = 54.6°,
y = 10.3 cm
e x = 21.8°,
y = 3.01 f x = 45.9°,
y = 3.87°
8 a 6.52 km b 3.80 km
9 a 7.31 cm b 1.97 cm10 a 66.3° b 148 m
11 Using the sine rule, x
= 4 √ __
2 _______
2 + √ __
2 ; rationalising
x = 4 √ __
2 (2 − √ __
2 ___________ 2 = 4 √ __
2 − 4 = 4( √ __
2 −1).
12 a 36.5 m
b That the angles have been measured from ground
level
Exercise 9C
1 a 70.5°, 109° (109.5°)
b
45° 109.5° 70.5°4.5 cm4.5 cm6 cmC
A9 A B
2 a x = 74.6°, y = 65.4°
x = 105°,
y = 34.6°
b x = 59.8°,
y = 48.4 cm
x = 120°,
y = 27.3cm
c x = 56.8°,
y = 4.37 cm
x = 23.2°,
y = 2.06 cm
3 a 5 cm (
ACB = 90°) b 24.6°
c 45.6°, 134(.4)°
4 2.96 cm
5 In one triangle ABC = 101° (100.9°); in the other B
AC = 131° (130.9°)
6 a 62.0° b The swing is symmetrical
Exercise 9D
1 a 23.7 cm2 b 4.31 cm2 c 20.2 cm2
2 a x = 41.8° or 138(.2)°
b x = 26.7° or 153(.3)°
c x = 60° or 120°
3 275(.3) m (third side = 135.3 m)
4 3.58
5 a Area = 1 __ 2 (x + 2)(5 − x) sin 30°
= 1 __ 2 (10 + 3x − x2) × 1 __ 2
= 1 __ 4 (10 + 3x − x2)
b Maximum A = 3 1 __ 16 , when x = 1 1 __ 2
6 a 1 __ 2 x(5 + x ) sin 150° = 15 __ 4
1 __ 2 (5x + x2) × 1 __ 2 = 15 __ 4
5x +
x2 = 15
x2 + 5x − 15 = 0
b 2.11
Exercise 9E
1 a x = 37.7°, y = 86.3°, z = 6.86
b x = 48°,
y = 19.5, z = 14.6
c x = 30°,
y = 11.5, z = 11.5
d x = 21.0°,
y = 29.0°, z = 8.09
e x = 93.8°,
y = 56.3°, z = 29.9°
f x = 97.2°,
y = 41.4°, z = 41.4° | [
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376
Answers
376 Full worked solutions are available in SolutionBank.
Online
g x = 45.3°, y = 94.7°, z = 14.7
or x = 135°,
y = 5.27°, z = 1.36
h x = 7.07,
y = 73.7°, z = 61.3°
or x = 7.07,
y = 106°, z = 28.7°
i x = 49.8°,
y = 9.39, z = 37.0°
2 a ABC = 108°,
ACB = 32.4°, AC = 15.1 cm
Area = 41.2 cm2
b BAC = 41.5°, ABC = 28.5°, AB = 9.65 cm
Area = 15.7 cm2
3 a 8 km b 060°
4 107 km
5 12 km
6 a 5.44 b 7.95 c 36.8°
7 a AB + BC
> AC ⇒ x + 6 > 7 ⇒ x > 1;
AC + AB >
BC ⇒ 11 > x + 2 ⇒ x < 9
b i x = 6.08 from x2 = 37
Area = 14.0 cm2
ii x = 7.23 from x2 − 4( √ __
2 − 1)x − (29 + 8 √ __
2 ) = 0
Area = 13.1 cm2
8 a x = 4 b 4.68 cm2
9 AC = 1.93 cm
10 a AC 2 = (2 − x)2 + (x + 1)2 − 2(2 − x)(x + 1) cos 120°
=
(4 − 4x + x2) + (x2 + 2x + 1) − 2(−x2 + x + 2) (− 1 __ 2 )
= x2 − x + 7
b 1 __ 2
11 4 √ ___
10
12 AC = 1 2 __ 3 cm and BC = 6 cm
Area = 5.05 cm2
13 a 61.3° b 78.9 cm2
14 a DAB = 136.3°, BCD = 50.1°
b 13.1 m2
c 5.15 m
15 34.2 cm2
Exercise 9F
1 y
O1
–1θ 90°y = cos θ
–90° 180° –180°
2
θ Oy
–180° –90° 90° 180°y = tan θ 3 y
O1
–1θ 90° –90° 180° –180°y = sin θ
4 a −30°
b i −120° ii −60°, 120°
c i 135° ii −45°, −135°
Exercise 9G
1 a i 1, x = 0° ii −1, x = 180°
b i 4, x
= 90° ii −4, x
= 270°
c i 1, x
= 0° ii −1, x
= 180°
d i 4, x
= 90° ii 2, x
= 270°
e i 1, x
= 270° ii −1, x
= 90°
f i 1, x
= 30° ii −1, x
= 90°
2
Oyy = cos 3θ y = cos θ
θ 90° 180° 270° 360°1
–1
3 a The graph of y = −cos θ is the graph of y = cos θ
reflected in the θ
-axis
O θ 90° 180° 270° 360°y
1
–1y = –cos θ
Meets θ-axis at (90°,
0), (270°, 0)
Meets y-axis at (0°,
−1)
Maximum at (180°, 1)
Minimum at (0°, −1) and (360°, −1)
b The graph of y = 1 __ 3 sin θ is the graph of y = sin θ
stretched by a scale factor 1 __ 3 in the y direction.
O θ 90° 180° 270° 360°y
1
313
–13y = sin θ
Meets θ-axis at (0°,
0), (180°, 0), (360°, 0)
Meets y-axis at (0°,
0)
Maximum at (90°, 1 __ 3 )
Minimum at (270°, − 1 __ 3 ) | [
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377
Answers
377 c The graph of y = sin 1 __ 3 θ is the graph of y = sin θ
stretched by a scale factor 3 in the θ direction.
O90° 180° 270° 360° θy
1
–11
3y = sin θ
Only meets axis at origin
Maximum at (270°, 1)
d The graph of y = tan (
θ − 45°) is the graph of tan θ
translated by 45° in the positive θ
direction.
90° 180° 270° 360° θ Oy
–1y = tan (θ – 45°)
Meets θ-axis at (45°, 0), (225°, 0)
Meets y-axis at (0°,
−1)
(Asymptotes at θ = 135° and
θ = 315°)
4 a This is the graph of y = sin θ stretched by scale
factor −2 in the y
-direction (i.e. reflected in the
θ-axis and scaled by 2 in the y-direction).
Oy
y = –2 sin θθ 90° 180° –180° –90°2
–2
Meets θ-axis at (−180°, 0), (0, 0), (180, 0)
Maximum at (−90°, 2)
Minimum at (90°, −2).
b This is the graph of y = tan θ translated by 180° in
the negative θ
direction.
Oy
y = tan (θ + 180°)
θ 90° 180° –180° –90°
As tan θ has a period of 180°
tan (θ
+ 180°) = tan θ
Meets θ-axis at (−180°,
0), (0, 0), (180°, 0)
Meets y-axis at (0,
0) c This is the graph of y = cos θ stretched by scale
factor 1 __ 4 horizontally.
Oy
y = cos 4θ
θ90° 180° –180° –90°1
–1
Meets θ-axis at (−157 1 __ 2 °, 0), (−112 1 __ 2 °, 0), (−67 1 __ 2 °, 0),
(−22 1 __ 2 °, 0), (22 1 __ 2 °, 0), (67 1 __ 2 °, 0), (112 1 __ 2 °, 0), (157 1 __ 2 °, 0)
Meets y-axis at (0,
1)
Maxima at (−180°, 1), (−90°,
1), (0, 1), (90°, 1), (180°, 1)
Minima at (−135°, −1), (−45°,
−1), (45°, −1), (135°, −1)
d This is the graph of y = sin θ reflected in the y
-axis.
(This is the same as y = −sin θ.)
Oy
y = sin (– θ)θ90° 180° –180° –90°1
–1
Meets θ-axis at (−180°, 0), (0°, 0), (180°, 0)
Maximum at (−90°, 1)
Minimum at (90°, −1)
5 a Period = 720°
90° 180° 270° 360° –180° –270° –360° –90°Oy
θ1
–11
2y = sin θ
b Period = 360°
90° 180° 270° 360° –180° –270° –360° –90°Oy
θ
–1
2
121
2y = cos θ
c Period = 180°
90° 180° 270° 360° –180° –270° –360° –90°Oy
θ2
–2y = tan (θ – 90°) | [
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378
Answers
378 Full worked solutions are available in SolutionBank.
Online
d Period = 90°
90° 180° 270° 360° –180° –270° –360° –90°Oy
θ2
–2y = tan 2θ
6 a i y = cos (−θ) is a reflection of y = cos θ in the
y-axis, which is the same curve,
so cos θ = cos (−θ).
y = cos θ
Oy
θ–90° –180° –270° 90° 180° 270° 360° 450°
ii y = sin (−θ ) is a reflection of y = sin θ in the y -axis.
y = sin(– θ)
Oy
θ–180° 180° 360°
y = −sin (−θ) is a reflection of y = sin (−θ) in the
θ
-axis, which is the graph of y = sin θ,
so
−sin (−θ
) = sin θ.
y = sin(– θ)
Oy
θ–180° 180° 360°
iii y = sin (θ − 90°) is the graph of y = sin θ
translated by 90° to the right, which is the graph
of
y = −cos θ, so sin (θ − 90°) = −cos θ.
y = sin (θ – 90°)
Oy
θ–90° –180° 90° 180° 270° 360° b sin (90° − θ )
= −sin (−(90° − θ)) = −sin (θ − 90°)
using (a) (ii)
= −(−cos θ) using (a) (iii)
= cos θ
c Using (a)(i) cos (90° − θ) = cos (−(90° − θ))
= cos (θ − 90°),
but cos (θ − 90°) = sin θ,
so cos (90° − θ) = sin θ
7 a (−300°, 0), (−120°,
0), (60°, 0), (240°, 0)
b (0°, √ __
3 ___ 2 )
8 a y = sin (x + 60°)
b Yes − could also be a translation of the cos graph,
e.g.
y = cos (x − 30°)
9 a
Oy
t1 11
23456
1y = sin(30 t)°
b Between 1 pm and 5 pm
Mixed exercise
1 a 155° b 13.7 cm
2 a x = 49.5°, area = 1.37 cm2
b x = 55.2°, area = 10.6 cm2
c x = 117°, area = 6.66 cm2
3 6.50 cm2
4 a 36.1 cm2 b 12.0 cm2
5 a 5 b 25 √ __
3 _____ 2 cm2
6 area = 1 __ 2 ab sin C
1 = 1 __ 2 × 2 √ __
2 sin C
1 ___
√ __
2 = sin C ⇒ C
= 45°
Use the cosine rule to find the other side:
x2 − 22 + ( √ __
2 )2 − 2 × 2 √ __
2 cos C ⇒ x = √ __
2 cm
So the triangle is isosceles, with two 45° angles,
thus is
also right-angled.
7 a AC = √ __
5 , AB = √ ___
18 , BC = √ __
5
cos ∠AC
B = AC 2 + BC 2 − AB2 ________________ 2 × AC × BC
= 5 + 5 − 18 ___________
2 × √ __
5 × √ __
5
= − 8 ___ 10 = − 4 __ 5
b 1 1 __ 2 cm2
8 a 4 b 15 √ __
3 _____ 4 (6.50) cm2
9 a 1.50 km b 241° c 0.789 km2
10 359 m2
11 35.2 m
12 a A stretch of scale factor 2 in the x direction.
b A translation of +3 in the y direction. | [
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379
Answers
379 c A reflection in the x-axis.
d A translation of −20 in the x direction.
13 a
Oy
x45°1
90° 135° 180°2
–1
–2y = –2cos xy = tan(x – 45°)
b There are no solutions.
14 a 300 b (30, 1) c 60 d √ __
3 ___ 2
15 a p = 5
90° 180° 270° 360°y = f(x)
Oy
x1
–1
b 72°
16 a The four shaded regions are congruent.
Oyy = sin θ
θ1
–190°α 180°180° + α
180° – α360° – α
270° 360°
b sin α and sin (180° − α) have the same y value,
(call it k)
so sin α = sin (180° − α)
sin (180° − α) and sin (360° − α) have the same
y
value, (which will be −k)
so sin α = sin (180° − α)
= −sin (180° + α)
= −sin (360° − α)
17 a
Oy
y = cos θ
θ1
–190°α 180°
180° + α360° – α180° – α
270° 360° y
y = tan θ
θ 90°α 180°180° + α
360° – α270° 360°180° – α
O
b i From the graph of y = cos θ, which shows four
congruent shaded regions,
if the y value at α
is k, then y at 180° − α is −k, y at
180° − α = −k and y at 360° − α = +k
so cos α = −cos (180° − α)
= −cos (180° + α)
= cos (360° − α)
ii From the graph of y = tan θ, if the
y value at α
is k, then at 180° − α it is −k, at 180° + α it is +k and at 360° − α it is −k,
so tan α = −tan (180° − α)
= +tan (180° + α)
= −tan (360° − α)
18 a
61 21 82 4Oy
x1
–1y = sin (60x)°
b 4
c The dunes may not all be the same height.
ChallengeUsing the sine rule:
sin
(180° − ∠
ADB − ∠AEB) = 5 ( 1 ___
√ __
5 ) _______
√ ___
10 = 1 ___
√ __
2
180° − ∠ADB
− ∠AEB = 135° (obtuse)
so ∠ADB + ∠B = 45° = ∠ACB
CHAPTER 10
Prior knowledge check
1 a
Oy
θ180° 360° 540°1
–1
b 4
c 143.1°, 396.9°, 503.1°
2 a 57.7° b 73.0°
3 a x = 11 b x = 9 __ 4 c x = −44.2°
4 a x = 1 or x
= 3
b x = 1 or x
= −9
c x = 3 ± √ ___
65 ________ 4 | [
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0.02573... |
380
Answers
380 Full worked solutions are available in SolutionBank.
Online
Exercise 10A
1 a
O80°
–80°y
x
P b
OP
80°+100°y
x
c
O
P20°+200°y
x d
15°+165°y
x OP
e
35°
–145°y
x O
P f
45°+225°y
x O
P
g
O
P80°+280°y
x h
O
P30°+330°y
x
i
O 20°
–160° Py
x j
OP
80°–280°y
x
2 a First b Second c Second
d Third e Third
3 a −1 b 1 c 0 d −1 e −1
f 0 g 0 h 0 i 0 j 0
4 a −sin 60° b −sin 80° c sin 20°
d −sin 60° e sin 80 f −cos 70°
g −cos 80° h cos 50° i −cos 20°
j −cos 5° k −tan 80° l −tan 35°
m −tan 30° n tan 5° o tan 60°
5 a −sin θ b −sin θ c −sin θ
d sin θ e −sin θ f sin θ
g −sin θ h −sin θ i sin θ
6 a −cos θ b −cos θ c cos θ
d −cos θ e cos θ f −cos θ
g −tan θ h −tan θ i tan θ
j tan θ k −tan θ l tan θChallenge
a
180 ° – θθ
θθ y
xa
sin θ = sin (180° − θ) = a
b
θ θ
θ
– θ y
xb
cos θ = cos (−θ) = b
c
θθθ y
xa
b –b180 ° – θ
tan θ = a __ b ; tan (180° − θ) = a ___ −b = −tan θ
For tan θ = x __ y
Exercise 10B
1 a √ __
2 ___ 2 b − √ __
3 ___ 2 c − 1 __ 2 d √ __
3 ___ 2
e √ __
3 ___ 2 f − 1 __ 2 g 1 __ 2 h − √ __
2 ___ 2
i − √ __
3 ___ 2 j − √ __
2 ___ 2 k −1 l −1
m √ __
3 ___ 3 n − √ __
3 o √ __
3
Challenge
a i √ __
3 ii 2 iii √ _______ 2 + √ __
3 iv √ _______ 2 + √ __
3 − √ __
2
b 15°
c i √ _______ 2 + √ __
3 − √ __
2 _____________ 2 ii √ _______ 2 + √ __
3 ________ 2 | [
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381
Answers
381Answers
Exercise 10C
1 a sin2 θ __ 2 b 5 c −cos2 A
d cos θ e tan x f tan 3A
g 4 h sin2 θ i 1
2 1 1 __ 2
3 tan x − 3 tan y
4 a 1 − sin2 θ b sin2 θ _________ 1 − sin2 θ c sin θ
d 1 − sin2 θ _________ sin θ e 1 − 2 sin2 θ
5 (One outline example of a proof is given)
a LHS = sin2 θ + cos2 θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ
= RHS
b LHS = 1 − cos2 θ _________ cos θ = sin2 θ _____ cos θ = sin θ × sin θ _____ cos θ
= sin θ tan θ = RHS
c LHS = sin x _____ cos x + cos x _____ sin x = sin2 x + cos2 x _____________ sin x + cos x
= 1 ____________ sin x + cos x = RHS
d LHS = cos2 A − (1 − cos2 A) = 2 cos2 A − 1
= 2 (1 − sin2 A) − 1 = 1 − 2 sin2 A = RHS
e LHS = (4 sin2 θ − 4 sin θ cos θ + cos2 θ)
+ (sin2 θ + 4 sin θ cos θ + cos2 θ)
= 5 (sin2 θ + cos2 θ) = 5 = RHS
f LHS = 2 − (sin2 θ − 2 sin θ cos θ + cos2 θ)
=
2 (sin2 θ + cos2 θ) − (sin2 θ − 2 sin θ cos θ + cos2 θ)
= sin2 θ + 2 sin θ cos θ + cos2 θ
= (sin θ + cos θ)2 = RHS
g LHS = sin2 x (1 − sin2 y) − (1 − sin2 x) sin2 y
= sin2 x − sin2 y = RHS
6 a sin θ = 5 __ 13 , cos θ = 12 __ 13
b sin θ = 4 __ 5 , tan θ = − 4 __ 3
c cos θ = 24 __ 25 , tan θ = − 7 __ 24
7 a − √ __
5 ___ 3 b − 2 √ __
5 ____ 5
8 a − √ __
3 ___ 2 b 1 __ 2
9 a − √ __
7 ___ 4 b − √ __
7 ___ 3
10 a x2 + y2 = 1
b 4x2 + y2 = 4 (or x2 + y2
__ 4 = 1)
c x2 + y = 1
d x2 = y2 (1 − x2) (or x2 + x2 __ y2 = 1)
e x2 + y2 = 2 (or (x + y)2 _______ 4 + (x −
y)2 _______ 4 = 1)
11 a Using cosine rule: cos B = 82 + 122 − 102 _____________ 2 × 8 × 12 = 9 ___ 16
b √ ____ 175 _____ 16
12 a Using sine rule: sin Q = sin 30 ______ 6 × 8 = 2 __ 3
b − √ __
5 ___ 3 Exercise 10D
1 a −63.4° b 116.6°, 296.6°
2 a 66.4° b 66.4°, 113.6°, 246.4°,
293.6°
3 a 270° b 60°, 240°
c 60°, 300° d 15°, 165°
e 140°, 220° f 135°, 315°
g 90°, 270° h 230°, 310°
4 a 45.6°, 134.4° b 135°, 225°
c 132°, 228° d 229°, 311°
e 8.13°, 188° f 61.9°, 242°
g 105°, 285° h 41.8°, 318°
5 a 30°, 210° b 135°, 315°
c 53.1°, 233° d 56.3°, 236°
e 54.7°, 235° f 148°, 328°
6 a −120°, −60°, 240°,
300° b −171°, −8.63°
c −144°, 144° d −327°, −32.9°
e 150°, 330°, 510°,
690° f 251°, 431°
7 a tan x should be 2 __ 3
b Squaring both sides creates extra solutions
c −146.3°, 33.7°
8 a
90° 180° 270° 360°Oy
x12
–1
–2y = cos x
y = 2 sin x
b 2 c 26.6°, 206.6°
9 71.6°, 108.4°, 251.6°,
288.4°
10 a 4 sin2 x − 3(1 − sin2 x) = 2.
Rearrange to get 7 sin2 x = 5
b 57.7°, 122.3°, 237.7°,
302.3°
11 a 2 sin2 x + 5(1 − sin2 x) = 1.
Rearrange to get 3 sin2 x = 4
b sin x > 1
Exercise 10E
1 a 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°, 360°
b 60°, 180°, 300°
c 22 1 __ 2 °, 112 1 __ 2 °, 202 1 __ 2 °, 292 1 __ 2 °
d 30°, 150°, 210°,
330°
e 300°
f 225°, 315°
2 a 90°, 270° b 50°, 170° c 165°, 345°
d 250°, 310° e 16.9°, 123°
3 a 11.2°, 71.2°, 131.2° b 6.3°, 186.3°, 366.3°
c 37.0°, 127.0° d −150°, 30°
4 a 10°, 130° b 71.6°, 108.4°
5 a
120° 300°(30°, 1)
(210°, –1)Oy
x210°1
–130° | [
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0.06... |
382
Answers
382 Full worked solutions are available in SolutionBank.
Online
b (0°, √ __
3 ___ 2 ) , (120°, 0), (300°, 0)
c 86.6°, 333.4°
6 a 0.75
b 18.4°, 108.4°, 198.4°,
288.4°
7 a 2.5
b No: increasing k will bring another
‘branch’ of the
tan graph into place.
Challenge
25°, 65°, 145°
Exercise 10F
1 a 60°, 120°, 240°, 300°
b 45°, 135°, 225°,
315°
c 0°, 180°, 199°,
341°, 360°
d 77.0°, 113°, 257°,
293°
e 60°, 300°
f 204°, 336°
g 30°, 60°, 120°,
150°, 210°, 240°, 300°, 330°
2 a ±45°, ±135° b −180°, −117°, 0°,
63.4°, 180°
c ±114° d 0°, ±75.5°, ±180°
3 a 72.0°, 144° b 0°, 60°
c No solutions in range
4 a ±41.8°, ±138° b 38.2°, 142°
5 60°, 75.5°, 284.5°,
300°
6 48.2°, 131.8°, 228.2°,
311.8°
7 2 cos2 x + cos x − 6 = (2 cos x − 3)(cos x + 2)
There are no solutions to cos x = −2 or to cos x = 3 __ 2
8 a 1 − sin2 x = 2 − sin x
Rearrange to get sin2 x − sin x + 1 = 0
b The equation has no real roots as b2 − 4ac < 0
9 a p = 1,
q = 5
b 72.8°, 129.0°, 252.8°,
309.0°, 432.8°, 489.0°
Challenge
1 −180°, −60°, 60°,
180°
2 0°, 90°, 180°,
270°, 360°
Mixed Exercise
1 a −cos 57° b −sin 48° c +tan 10°
2 a 0 b − √ __
2 ___ 2 c −1
d √ __
3 e −1
3 Using sin2 A = 1 − cos2 A, sin2 A = 1 − (− √ ___
7 ___ 11 ) 2
= 4 ___ 11 .
Since angle A is obtuse,
it is in the second quadrant
and sin is positive, so sin A = 2 ____
√ ___
11 .
Then tan A = sin A _____ cos A = 2 ____
√ ___
11 × (− √ ___
11 ___ 7 ) = − 2 ___
√ __
7 = − 2 __ 7 √ __
7 .
4 a − √ ___
21 ____ 5 b − 2 __ 5
5 a cos2 θ − sin2 θ b sin4 3θ c 1
6 a 1 b tan y = 4 + tan x __________ 2 tan x − 3
7 a LHS = (1 + 2 sin θ + sin2 θ) + cos2 θ
= 1 + 2 sin θ + 1
= 2 + 2 sin θ
= 2 (1 + sin θ) = RHS b LHS = cos4 θ + sin2 θ
= (1 − sin2 θ)2 + sin2 θ
= 1 − 2 sin2 θ + sin4 θ + sin2 θ
= (1 − sin2 θ) + sin4 θ
= cos2 θ + sin4 θ = RHS
8 a No solutions: −1 < sin θ <
1
b 2 solutions: tan θ = −1 has two solutions in the
interval.
c No solutions: 2 sin θ + 3 cos θ > −5
so 2 sin θ + 3 cos θ + 6 can never be equal to 0.
d No solutions: tan2 θ = −1 has no real solutions.
9 a (4x −
y)( y + 1) b 14.0°, 180°, 194°
10 a 3 cos 3θ b 16.1, 104, 136,
224, 256, 344
11 a 2 sin 2θ
= cos 2θ
⇒ 2 sin 2θ _______ cos 2θ = 1
⇒ 2 tan 2θ
= 1 ⇒ tan 2θ
= 0.5
b 13.3°, 103.3°, 193.3°,
283.3°
12 a 225°, 345°
b 22.2°, 67.8°, 202.2°,
247.8°
13 30°, 150°, 210°
14 0°, 131.8°, 228.2°
15 a Found additional solutions after dividing by three
rather than before. Not applied the full interval for solutions.
b −350°, −310°, −230°,
−190°, −110°, −70°, 10°, 50°,
130°, 170°, 250°, 290°
16 a
90° 180° 270° 360°Oy
x123
–1
–2
–3y = 3 sin xy = 2 cos x
b 2 c 33.7°, 213.7°
17 a 9 ___ 11 b √ ___
40 ____ 11
18 a Using sine rule: sin Q = sin 45 × 6 __ 5 = √ __
2 ___ 2 × 6 __ 5 = 3 √ __
2 ____ 5
b − √ __
7 ___ 5
19 a 3 sin2 x − (1 − sin2 x) = 2.
Rearrange to give 4 sin2 x = 3.
b −120°, −60°, 60°,
120°
20 −318.2°, −221.8°, 41.8°,
138.2°
Challenge
45°, 54.7°, 125.3°, 135°, 225°, 234.7°, 305.3°, 315°
Review exercise 2
1 x + 3y – 22 = 0
2 x – 3y
– 21 = 0
3 4, −2.5
4 a 0.45
b l = 0.45h
c The model may not be valid for young people who are still growing.
5
a y = − 1 __ 3 x + 4 b C is (3, 3) c 15 | [
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... |
383
Answers
3836 3 √ __
5
7 (−6, 0)
8 (x +
3)2 + (y − 8)2 = 10
9 a (x −
3)2 + (y + 1)2 = 20 (a = 3, b = −1, r = √ ___ 20 )
b Centre (3, −1), radius √ ___ 20
10 a (3, 5) and (4, 2)
b √ ___ 10
11 0 < r < √ __
2 _ 5
12 a (x −
1)2 + (y − 5)2 = 58
b 7y −
3x + 26 = 0
13 a AB = √ ___ 32 ; BC = √ __
8 ; AC = √ ___ 40 ; AC 2 = AB2 + BC 2
b AC is a diameter of the circle.
c (x −
5)2 + (y − 2)2 = 10
14 a = 3,
b = −2, c = −8
15 a 2( 1 _ 2 )3 – 7( 1 _ 2 )2 – 17( 1 _ 2 ) + 10 = 0
b (2x
− 1)(x − 5)(x + 2)
c y
x O5 –210
1
2
16 a 24
b (x
− 3)(3x − 2)(x + 4)
17 a g(3) = 33 – 13(3) + 12 = 0
b (x
– 3)(x + 4)(x – 1)
18 a a = 0,
b = 0
b a > 0,
b > 0
19 a 52 = 24 + 1; 72 = 2(24) + 1; 112 = 5(24) + 1;
132 = 7(24) + 1; 172 = 12(24) + 1; 192 = 15(24) + 1
b 3(24) + 1 = 73 which is not a square of a prime
number
20 a (x −
5)2 + (y − 4)2 = 32
b √ ___ 41
c Sum of radii = 3 + 3 < √ ___ 41 so circles do not touch
21 a 1 − 20x + 180
x2 − 960x3 + …
b 0.817
22 a = 2,
b = 19, c = 70
23 4
24 √ ___ 10 cm
25 a cos 60˚ = 1 _ 2 = (52 + (2x − 3)2 – (x + 1)2) ÷ 2(5)(2x − 3)
5(2x
− 3) = (25 + 4x2 − 12x + 9 – x2 −2x − 1)
0 = 3x2 – 24x + 48
x2 – 8x + 16 = 0
b 4
c 10.8 cm2
26 a 11.93 km
b 100.9°
27 a AB = BC
= 10 cm, AC = 6 √ ___ 10 cm
b 143.1°
28 19.4 km2
29 a (x − 5)2 + (y − 2)2 = 25
b 6
c XY = √ ___ 90 ; YZ = √ ___ 20 ; XZ = √ ___ 98
cos XYZ = (20 + 90 – 98) ÷ (2 × √ ___ 20 × √ ___ 90 )
cos XYZ = 12 ÷ 60 √ __
2 = √ __
2 ÷ 1030 a
xy
Oy = sin x
y = tan(x – 90°)90° 180° 270° 360°
b 2
31 a (−225, 0), (−45,
0), (135, 0) and (315, 0)
b (0, √ __
2 ___ 2 )
32 Area of triangle = 1 _ 2 × s × s × sin 60° = √ __
3 ___ 4 s2
Area of square = s2
Total surface area = 4 × ( √ __
3 ___ 4 s2
) + s2 = ( √ __
3 + 1)s2 cm2
33 a 1
b 45˚, 225˚
34 30°, 150°, 210°,
330°
35 90°, 150°
36 a 2(1 − sin2 x) = 4 − 5 sin x
2 − 2 sin2 x = 4 − 5 sin x
2 sin2 x − 5 sin x + 2 = 0
b x = 30°, 150°
37 72.3˚, 147.5˚, 252.3˚,
327.5˚
38 0˚, 78.5˚, 281.5˚,
360˚
39 cos2x (tan2x + 1)
= cos 2 x ( sin 2 x ______ cos 2 x + 1)
= cos2x + sin2x = 1
Challenge
1 a 160
b (− 28 __ 3 , 0)
2 The second circle has the same centre but a larger
radius
3 ( n k ) + ( n k + 1 ) = n ! _________
k ! (n − k ) ! + n ! _________________
(k + 1 ) ! (n − k − 1 ) !
= n ! (k + 1 ) _____________
(k + 1 ) ! (n − k ) ! + n ! (n − k ) _____________
(k + 1 ) ! (n − k ) !
= n ! ( (k + 1 ) + (n − k ) ) _________________
(k + 1 ) ! (n − k ) !
= n ! (n + 1 ) _____________
(k + 1 ) ! (n − k ) !
= (n + 1) ! ______________ (k + 1) !(n − k )!
= ( n + 1 k + 1 )
4 0°, 30°, 150°,
180°, 270°, 360°
CHAPTER 11
Prior knowledge check
1 a ( 4 2 ) b ( 5 2 ) c ( −1 −3 )
2 a 7 __ 9 b 2 __ 9 c 7 __ 2
3 a 123.2° b 13.6 c 5.3 d 21.4° | [
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0... |
384
Answers
384 Full worked solutions are available in SolutionBank.
Online
Exercise 11A
1 a
aa + c
c b
–b
c
–d
cc – d d
b dc
b + c + d
e
3d2c
2c + 3d f
a
–2ba – 2b
g
abdca + b + c + d
2 a 2b b d c b
d 2b e d +
b f d +
b
g −2d h −b i 2d +
b
j −b + 2
d k −b +
d l −d −
b
3 a 2m b 2p c m
d m e p + m f p +
m
g p + 2
m h p −
m i −m −
p
j −2m +
p k −2p
+ m l −m − 2
p
4 a d −
a b a +
b + c
c a +
b − d d a +
b + c − d
5 a 2a + 2
b b a +
b c b −
a
6 a b b b − 3a c a −
b
d 2a
− b
7 a ⟶ OB = a + b b ⟶ OP = 5 __ 8 (a + b) c ⟶ AP = 5 __ 8 b − 3 __ 8 a
8 a Yes (λ
= 2) b Yes (λ
= 4) c No
d Yes (λ
= −1) e Yes (λ
= −3) f No
9 a i b − a ii 1 __ 2 a iii 1 __ 2 b iv 1 __ 2 b − 1 __ 2 a
b ⟶ BC = b − a, ⟶ PQ = 1 __ 2 (b − a) so PQ is parallel to BC.
10 a i 2b ii a − b
b ⟶ AB = 2b, ⟶ OC = 3b so AB is parallel to OC.
11 1.2
Exercise 11B
1 v1: 8i, ( 8 0 ) v2: 9i + 3j, ( 9 3 ) v3: −4i + 2j, ( −4 2 )
v4: 3i + 5j, ( 3 5 ) v5: −3i − 2j, ( −3 −2 ) v6: − 5j, ( 0 −5 ) 2 a 8i + 12
j b i + 1.5
j c −4i +
j
d 10i +
j e −2i + 11
j f −2i − 10
j
g 14i − 7
j h −8i + 9
j
3 a ( 45 35 ) b ( 4 0.5 ) c ( 12 3 )
d ( −1 16 ) e ( −21 −29 ) f ( 10 2 )
4 a λ = 5 b μ = − 3 __ 2
5 a λ = 1 __ 3 b μ = −1
c s = −1 d t = − 1 __ 17
6 i – j
7 a ⟶ AC = 5i − 4 j = ( 5 −4 ) b ⟶ OP = 5i + 8 _ 5 j = ( 5
8 __ 5 )
c ⟶ AP = 3i − 12 __ 5 j = ( 3
− 12 __ 5 )
8 j = 4,
k = 11
9 p = 3,
q = 2
10 a p = 5 b 8i − 12
j
Exercise 11C
1 a 5 b 10 c 13
d 4.47 (3 s.f.) e 5.83 (3 s.f.) f 8.06 (3 s.f.)
g 5.83 (3 s.f.) h 4.12 (3 s.f.)
2 a √ ___
26 b 5 √ __
2 c √ ____ 101
3 a 1 __ 5 ( 4 3 ) b 1 ___ 13 ( 5 −12 )
c 1 ___ 25 ( −7 24 ) d 1 ____
√ ___
10 ( 1 −3 )
4 a 53.1° above b 53.1° below
c 67.4° above d 63.4° above
5 a 149° to the right b 29.7° to the right
c 31.0° to the left d 104° to the left
6 a 15 √ __
2 _____ 2 i + 15 √ __
2 _____ 2 j, ⎛
⎜
⎝ 15 √ __
2 _____ 2
15 √ __
2 _____ 2 ⎞
⎟
⎠ b 7.52i + 2.74 j, ( 7.52 2.74 )
c 18.1i − 8.45
j, ( 18.1 −8.45 ) d 5 √ __
3 ____ 2 i − 2.5j,
( 5 √ __
3 ____ 2
−2.5
)
7 a |3i + 4
j| = 5, 53.1° above
34
b |2i − j| = √ __
5 , 26.6° below
12
c |−5i + 2 j| = √ ___
29 , 158.2° above
52
8 k = ±6
9 p = ±8,
q = 6 | [
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385
Answers
38510 a 36.9° b 33.7° c 70.6°
11 a 67.2° b 19.0
Challenge
Possible solution:
p xy
p + rrqrqr
qr
q
s
sq + spq1
2
pq12rs12
rs12
Area of parallelogram = area of large rectangle − 2(area of
small rectangle) − 2 (area triangle 1) − 2(area triangle 2)
Area of parallelogram = ( p + r
)(q + s) − 2qr − 2( 1 __ 2 pq)
− 2( 1 __ 2 rs) = ps − qr
Exercise 11D
1 a i ⟶ OA = 3i − j, ⟶ OB = 4i + 5 j, ⟶ OC = −2i + 6 j
ii i + 6
j iii −5i + 7
j
b i √ ___
40 = 2 √ ___
10 ii √ ___
37 iii √ ___
74
2 a −i + 5
j or ( −1 5 )
b i 5 ii √ ___
13 iii √ ___
26
3 a −i − 9
j or ( −1 −9 )
b i √ ___
82 ii 5 iii √ ___
61
4 a −2a + 2
b b −3a + 2
b c −2a +
b
5 ( 7 9 ) or ( 9 3 )
6 a 2i + 8
j b 2 √ ___
17
7 3 √ __
5 ____ 5
Challenge
⟶ OB = 2i + 3 j or ⟶ OB = 46 __ 13 i + 9 __ 13 j
Exercise 11E
1 ⟶ XY = b − a and ⟶ YZ = c − b, so b − a = c − b.
Hence a + c = 2b.
2 a i 2r ii r
b Sides of triangle OAB are twice the length of sides of
triangle
PAQ and angle A is common to both SAS.
3 a 2 __ 3 a + 1 __ 3 b
b ⟶ AN = 1 __ 3 (b − a), ⟶ AB = b − a, ⟶ NB = 2 __ 3 (b − a)
so AN : NB =
1 : 2.
4 a 3 __ 5 a + 2 __ 5 c
b ⟶ AP = −a + 3 __ 5 a + 2 __ 5 c = 2 __ 5 (c − a),
⟶ PC = c − ( 3 __ 5 a + 2 __ 5 c) = 3 __ 5 (c − a) so AP : PC = 2 : 3
5 a √ ___
26 b 2 √ __
2 c 3 √ __
2
d ∠ BAC = 56°, ∠ ABC = 34°, ∠ ACB = 90°
6 a ⟶ OR = a + 1 __ 3 (b − a) = 2 __ 3 a + 1 __ 3 b,
⟶ OS = 3 ⟶ OR = 3( 2 __ 3 a + 1 __ 3 b) = 2a + b b ⟶ TP = a + b, ⟶ PS = 1 __ 3 (b − a) + 2 __ 3 (2a + b) = a + b
⟶ TP is parallel (and equal) to ⟶ PS and they have a point,
P,
in common so T, P and S lie on a straight line.
Challenge:
a ⟶ PR = b − a, ⟶ PX = j (b − a) = −ja + j b
b ⟶ ON = a + 1 __ 2 b, ⟶ PX = −a + k(a + 1 __ 2 b) = (k − 1)a + 1 __ 2 kb
c Coefficients of a and
b must be the same in both
expressions for ⟶ PX
Coefficients of a:
k − 1 = −j; Coefficients of b: j = 1 __ 2 k
d Solving simultaneously gives j = 1 __ 3 and k = 2 __ 3
e ⟶ PX = 1 __ 3 ⟶ PR .
By symmetry, ⟶ PX = ⟶ YR = ⟶ XY , so ON and OM divide PR
into 3 equal parts.
Exercise 11F
1 a 5 m s−1 b 25 km h−1
c 5.39 m s−1 d 8.06 cm s−1
2 a 50 km b 51.0 m
c 4.74 km d 967 cm
3 a 5 m s−1, 75 m b 5.39 m s−1, 16.2 m
c 5.39 km h−1, 16.2 km d 13 km h−1, 6.5 km
4 (2.8i − 1.6j) m s−2
5 a 54.5° b 0.3 √ ___ 74 Newtons
6 a 26.6° below i
b R = (3 + p
)i + (q − 4)j, 3 + p = 2λ and
q − 4 = − λ ⇒ λ = 4 −
q
3 + p = 2(4 −
q) ⇒ 3 + p = 8 − 2
q so p + 2q = 5
c |R| = 2 √ __
5 newtons
7 a 10i − 100
j b 109.4° c 1700 m2
8 a √ ___
41 b 303.7°
c ⟶ AB = 4i − 5 j, v = 2(4i − 5j) so the boat is travelling
directly towards the buoy.
d 2 √ ___
41 e 30 minutes
Mixed exercise
1 a 2 √ ___
10 newtons b 18°
2 a 108° b 9.48 km h−1
3 a 9.85 m s−1 b 59.1 m
c The model ignores friction and air resistance.
The model will become less accurate as t
increases.
4 a b − 3 __ 5 a b b − 4 a c 8 __ 5 a − b d 3a − b
5 1.25
6 a ( 12 −1 ) b ( −18 5 ) c ( 49 13 )
7 a 3i − 2
j b 32.5° c 10.5
8 a p = −1.5 b i − 1.5j
9 a i 1 __ 17 (8i + 15j) ii 61.9° above
b i 1 __ 25 (24i − 7j) ii 16.3° below
c i 1 __ 41 (−9i + 40j) ii 102.7° above
d i 1 ____
√ ___
13 (3i
− 2j) ii 33.7° below
10 p = 8.6,
q = 12.3
11 ±6
12 a 3 __ 5 a + 2 __ 5 b b 2 __ 5 b
c ⟶ AB = b − a, ⟶ AN = 2 __ 5 (b − a) so AN : NB = 2 : 3 | [
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386
Answers
386 Full worked solutions are available in SolutionBank.
Online
13 a 18.4° below
b R = (4 + p
)i + (5 − q)j, 4 + p = 3λ and 5 − q = −λ
4 + p = 3(
q − 5) so p + 3q = 11
c 2 √ ___
10 newtons
14 √ ____ 193 _____ 2
Challenge
⟶ OB = 3 __ 2 i + 5 __ 2 j or 99 __ 34 i + 5 __ 34 j
CHAPTER 12
Prior knowledge check
1 a 5 b − 2 __ 3 c 1 __ 3
2 a x10 b x 2 __ 3 c x−1 d x 3 __ 4
3 a y = 1 __ 2 x − 2 b y = − 1 __ 2 x + 8 1 __ 2 c y = − 1 __ 4 x + 7 1 __ 2
4 y = − 1 __ 2 x or y = − 1 __ 3 x + 1 2 __ 3
Exercise 12A
1 a x-coordinate −1 0 1 2 3
Estimate for gradient
of curve−4 −2 0 2 4
b Gradient = 2p − 2 c 1
2 a √ ________ 1 − 0.62 = √ _____ 0.64 = 0.8
b Gradient = −0.75
c i −1.21 (3 s.f.) ii −1 iii −0.859 (3 s.f.)
d As other point moves closer to A,
gradient tends to
−0.75.
3 a i 7 ii 6.5 iii 6.1
iv 6.01 v h + 6
b 6
4 a i 9 ii 8.5 iii 8.1
iv 8.01 v 8 + h
b 8
Exercise 12B
1 a f9(2) = lim
h→0 f(2 + h) − f(2) _____________ h = lim
h→0 (2 + h)2 − 22 ____________ h
= lim
h→0 4h + h2 _______ h = lim
h→0 (4 + h ) = 4
b f9(−3) = lim
h→0 f(−3 + h) − f(−3) _______________ h = lim
h→0 (−3 + h)2 − 32 _____________ h
= lim
h→0 −6h + h2 _________ h = lim
h→0 (−6 + h ) = −6
c f9(0) = lim
h→0 f(0 + h) − f(0) _____________ h = lim
h→0 h2 − 02 _______ h = lim
h→0 h = 0
d f9(50) = lim
h→0 f(50 + h) − f(50) _______________ h = lim
h→0 (50 + h)2 − 502 ______________ h
= lim
h→0 100h + h2 __________ h = lim
h→0 (100 + h ) = 100
2 a f9(
x) = lim
h→0 f(x + h) − f(x) _____________ h = lim
h→0 (x + h)2 − x2 ____________ h
= lim
h→0 2xh + h2 ________ h = lim
h→0 (2x + h)
b As h
→ 0, f 9(x) = lim
h→0 (2x + h) = 2x3 a g = lim
h→0 (−2 + h)3 − (−2)3 _______________ h
= lim
h→0 −8 + 3(−2)2h + 3(−2)h2 + h3 + 8 _____________________________ h
= lim
h→0 12h − 6h2 + h3 ______________ h = lim
h→0 (12 − 6h + h2)
b g = 12
4 a Gradient of AB = (−1 + h)3 − 5(−1 + h) − 4 _______________________ (−1 + h) − (−1)
= −1 + 3h
= 3h2 + h3 + 5 − 5h − 4 _____________________________ h
= h3 − 3h2 − 2h _____________ h = h2 − 3h − 2
b gradient = −2
5 dy ___ dx = lim
h→0 6(x + h) − 6x _____________ h = lim
h→0 6h ___ h = 6
6 dy ___ dx = lim
h→0 4(x + h)2 − 4x2 ______________ h = lim
h→0 8xh + 4h2 __________ h
= lim
h→0 (8x + 4h) = 8x
7 dy ___ dx = lim
h→0 a(x + h)2 − ax2 ______________ h = lim
h→0 (a − a)x2 + 2axh + ah2 _____________________ h
= lim
h→0 2axh + ah2 ___________ h = lim
h→0 (2ax + ah) = 2ax
Challenge
a f9
(x) = lim
h→0 1 _____ x + h − 1 __ x _________ h = lim
h→0 x − (x + h) __________ xh(x + h) = lim
h→0 −1 ________ x(x + h)
= lim
h→0 −1 _______ x2 + xh
b f9
(x) = lim
h→0 −1 ________ x(x + h) = −1 _______ x2 + xh = −1 ______ x2 + 0 = − 1 __ x2
Exercise 12C
1 a 7x6 b 8x7 c 4x3 d 1 _ 3 x − 2 _ 3
e 1 _ 4 x − 3 _ 4 f 1 _ 3 x − 2 _ 3 g –3x–4 h –4x–5
i –2x–3 j –5x–6 k − 1 _ 2 x − 3 _ 2 l − 1 _ 3 x − 4 _ 3
m 9x8 n 5x4 o 3x2 p –2x–3
q 1 r 3x2
2 a 6x b 54x8 c 2x3
d 5 x − 3 _ 4
e 15 __ 2 x 1 _ 4 f −10x−2 g 6x2 h − 1 ____ 2 x 5
i x − 3 _ 2 j 15 __ 2 √ __
x
3 a 3 _ 4 b 1 _ 2 c 3 d 2
4 dy ___ dx = 3 __ 2 √ __
x __ 2
Exercise 12D | [
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1 a 4x – 6 b x + 12 c 8x d 16x + 7
e 4 – 10x
2 a 12 b 6 c 7 d 2 1 _ 2
e –2 f 4
3 4, 0
4 (–1, –8)
5 1, –1
6 6, –4 | [
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0... |
387
Answers
3877 a, b
O xy
–2 14
(1, –9)–2y = f (x) y = f9(x)
c At the turning point, the gradient of y = f(x) is zero,
i.e. f9(x) = 0.
Exercise 12E
1 a 4x3 – x−2 b 10x4 – 6x−3 c 9 x 1 _ 2 − x − 3 _ 2
2 a 0 b 11 1 _ 2
3 a (2 1 _ 2 , − 6 1 _ 4 ) b (4, –4) and (2, 0)
c (16, –31) d ( 1 _ 2 , 4) (– 1 _ 2 , –4)
4 a x − 1 _ 2 b –6x–3 c –x–4
d 4 __ 3 x3 − 2x2 e 1 _ 2 x − 1 _ 2 − 6x–4 f 1 _ 3 x − 2 _ 3 − 1 _ 2 x –2
g –3x–2 h 3 + 6x–2 i 5 x 3 _ 2 + 3 _ 2 x − 1 _ 2
j 3x2 – 2x + 2 k 12x3 + 18x2 l 24x – 8 + 2 x–2
5 a 1 b 2 _ 9 c –4 d 4
6 − 3 _ 4 √ __
2
7 a 512 − 2304x + 4608
x2
b f9 (x) ≈ d ___ dx (512 − 2304x + 4608x2)
= −2304 + 2 × 4608x
= 9216x − 2304
Exercise 12F
1 a y + 3x – 6 = 0 b 4y – 3 x – 4 = 0
c 3y – 2
x – 18 = 0 d y = x
e y = 12x
+ 14 f y = 16x
– 22
2 a 7y +
x – 48 = 0 b 17y + 2
x – 212 = 0
3 (1 2 _ 9 , 1 8 _ 9 )
4 y = –x
, 4y + x – 9 = 0; (–3, 3)
5 y = –8x
+ 10, 8y – x – 145 = 0
6 (− 3 _ 4 , 9 _ 8 )
Challenge
L has equation y = 12x − 8.
Exercise 12G
1 a x > – 4 _ 3 b x < 2 _ 3 c x < –2
d x <
2, x > 3 e x ∈
핉 f x ∈
핉
g x >
0 h x >
6
2 a x <
4.5 b x >
2.5 c x >
–1
d –1 <
x < 2 e –3 <
x < 3 f –5 <
x < 5
g 0 < x
< 9 h –2 <
x < 0
3 f9
(x) = −6x2 − 3
x2 > 0 for all x ∈ 핉 , so −6x2 − 3 ≤ 0 for all x ∈ 핉 .
∴
f is decreasing for all x ∈ 핉 .
4 a Any p
> 2
b No. Can be any p
> 2.Exercise 12H
1 a 24x + 3, 24
b 15 – 3x–2, 6x–3
c 9 _ 2 x − 1 _ 2 + 6x–3, – 9 _ 4 x − 3 _ 2 – 18x–4
d 30x + 2, 30
e –3x–2 – 16x–3, 6x–3 + 48x–4
2 Acceleration = 3 _ 4 t − 1 _ 2 + 3 _ 2 t − 5 _ 2
3 3 _ 2
4 − 1 _ 2
Exercise 12I
1 a –28 b –17 c – 1 _ 5
2 a 10 b 4 c 12.25
3 a (− 3 _ 4 , − 9 _ 4 ) minimum
b ( 1 _ 2 , 9 1 _ 4 ) maximum
c (− 1 _ 3 , 1 5 __ 27 ) maximum, (1, 0) minimum
d (3,–18) minimum, (– 1 _ 3 , 14 __ 27 ) maximum
e (1, 2) minimum, (–1,
–2) maximum
f (3, 27) minimum
g ( 9 _ 4 , − 9 _ 4 ) minimum
h (2, –4 √ __
2 ) minimum
i ( √ __
6 , –36) minimum, (– √ __
6 , –36) minimum,
(0,0) maximum
4 a
Oy
x
(– , – )y = 4x2 + 6x
3
494
b
Oy
x( , 9 )1
214
y = 9 + x – x2
c
Oy
x(– , 1 )13
(1, 0)5
27y = x3 – x2 – x + 1 | [
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-0.... |
388
Answers
388 Full worked solutions are available in SolutionBank.
Online
d
Oy
x(– , )1
31427
(3, –18)y = x(x2 – 4x – 3)
5 (1, 1) inflection (gradient is positive either side of point)
Oy
x(1, 1)y = x3 – 3x2 + 3x
6 Maximum value is 27; f(x) < 27
7 a (1, −3): minimum, (−3,
−35): minimum, (− 1 _ 4 , 357 ___ 256 ) :
maximum
b y
x O(– , )1
4
(1, –3)
(–3, –35)357256 y = f (x)
Exercise 12J
1 a y
x O(6, 0)
(0, 0) (–9, 0)y = f9(x)
b y
x Oy = f9(x)
y = 0 c y
x Ox = –7
(4, 0)
y = f9(x)
d y
x O
y = f9(x)y = 0(–2, 0) (0, 0)
e y
x Oy = f9(x)x = 6
f y
x Oy = 0
y = f9(x)
2 a y
x O(–1, 0)16
(4, 0)y = f(x)
b y
x O 4y = f9(x)
8
23
c f(x ) = x3 − 7x2 + 8x + 16
f9
(x) = 3x2 − 14x + 8 = (3x − 2)(x − 4)
d (4, 0), ( 2 _ 3 , 0) and (0, 8) | [
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