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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) _______...
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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 ...
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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 respec...
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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...
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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(...
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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 ...
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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 ) − (−...
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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 ...
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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 boun...
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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 ...
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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, factor...
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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 ...
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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 ...
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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 diag...
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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 th...
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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 diagr...
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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 ...
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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 o...
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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,...
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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 functio...
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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 _ ...
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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 t...
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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 grad...
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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 ...
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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 tr...
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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 m...
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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 ...
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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 Demons...
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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 yo...
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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 lo...
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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 )...
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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 , s...
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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 ...
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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...
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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 v...
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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 − ...
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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...
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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 ...
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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 o...
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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 f...
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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 m...
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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...
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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...
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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 ...
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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...
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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 sp...
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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 ...
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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 ...
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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 ...
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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) ← Section...
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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 n...
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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...
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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 ...
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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...
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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 __...
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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 ...
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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, ...
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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 hec...
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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 ...
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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 < ...
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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 x...
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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 = ...
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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...
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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)...
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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...
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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,...
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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; (...
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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(...
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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)...
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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 ...
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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) ...
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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...
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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 ...
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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...
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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 Exerci...
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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 − ...
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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 = 7...
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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 = −...
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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)...
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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 _____...
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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 ...
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(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 squ...
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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 , − ...
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= 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 ...
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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 + 1...
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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...
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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 ...
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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 p...
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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° ...
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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 ...
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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 Thir...
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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 = sin...
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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...
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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...
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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 + ...
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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( ...
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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 ...
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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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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) ...
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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 ...
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