SkillFi/llm-documents · Datasets at Hugging Face

document stringlengths 121 3.99k	embedding listlengths 384 384
389 Answers 389Exercise 12K 1 2t – 3 2 2π 3 – 4 _ 3 4 48π 5 18 6 a Let x = width of garden. x + 2y = 80 A = xy = 2y b 20 m × 40 m, 800 m2 7 a 2π r2 + 2πrh = 600π ⇒ h = 300 – r2 ________ r V = π r2h = πr (300 − r2) = 300πr − πr3 b 2000π cm3 8 a Let θ = angle of sector . πr2 × θ ____ 360 = 100 ⇒ θ = 36 000 _______ πr2 P = 2r + 2πr × θ ____ 360 = 2r + 200πr ______ πr2 = 2r + 100 ____ r θ < 2π ⇒ Area < πr2, so πr2 > 100 ∴ r > √ ____ 100 ____ π b 40 cm 9 a Let h = height of rectangle. P = πr + 2r + 2h = 40 ⇒ 2h = 40 – 2r – πr A = π __ 2 r2 + 2rh = π __ 2 r2 + r (40 – 2r – πr) = 40r – 2r2 – π __ 2 r2 b 800 _____ 4 + π cm2 10 a 18x + 14y = 1512 ⇒ y = 1512 – 18x ___________ 14 A = 12xy = 12x ( 1512 – 18x ___________ 14 ) = 1296x − 108x2 ______ 7 b 27 216 mm2 Mixed exercise 1 f9(x) = lim h→0 10(x + h)2 − 10x2 ________________ h = lim h→0 20xh + 10h2 ____________ h = lim h→0 (20x + 10 h) = 20x 2 a y-coordinate of B = (δx)3 + 3(δx)2 + 6δx + 4 Gradient = ((δx )3 + 3(δx)2 + 6δx + 4) − 4 __________________________ (1 + δx) − 1 = (δx )3 + 3(δx)2 + 6δx __________________ (δx) = (δx)2 + 3δx + 6 b 6 3 4, 11 3 _ 4 , 17 25 __ 27 4 2, 2 2 _ 3 5 (2, –13) and (–2, 15) 6 a 1 – 9 __ x2 b x = ±3 7 3 _ 2 x − 1 _ 2 + 2 x − 3 _ 2 8 a dy ___ dx = 6 x − 1 _ 2 − 3 _ 2 x 1 _ 2 = 3 _ 2 x − 1 _ 2 (4 − x ) b (4, 16) 9 a x + x 3 _ 2 − x − 1 _ 2 − 1 b 1 + 3 _ 2 x 1 _ 2 + 1 _ 2 x − 3 _ 2 c 4 1 __ 16 10 6x2 + 1 _ 2 x − 1 _ 2 – 2x–211 a = 1, b = –4, c = 5 12 a 3x2 – 10x + 5 b i 1 _ 3 ii y = 2x – 7 iii 7 _ 2 √ __ 5 13 y = 9x – 4 and 9y + x = 128 14 a ( 4 _ 5 , − 2 _ 5 ) b 1 _ 5 15 P is (0, −1), dy ___ dx = 3x2 − 4x – 4 Gradient at P = −4, so L is y = −4x − 1. −4x −1 = x3 − 2x2 − 4x − 1 ⇒ x2(x − 2) = 0 x = 2 ⇒ y = −9, so Q is (2, −9) Distance PQ = √ ___________________ (2 − 0 ) 2 + (−9 − (−1) ) 2 = √ ___ 68 = 2 √ ___ 17 16 a x = 4, y = 20 b d2y ____ dx2 = 3 __ 4 x − 1 __ 2 + 96x3 At x = 4, d2y ____ dx2 = 15 ___ 8 > 0 (4, 20) is a local minimum. 17 (1, −11) and ( 7 _ 3 , − 329 ___ 27 ) 18 a 7 31 __ 32 b f9( x) = (x − 1 __ x ) 2 > 0 for all values of x 19 (1, 4) 20 a (1, 33) maximum, (2, 28) and (−1, 1) minimum b 20y x O(2, 28)(1, 33) (–1, 1)y = f(x) 21 a 250 ____ x 2 – 2x b (5, 125) 22 a P (x, 5 − 1 __ 2 x2) OP2 = (x – 0)2 + (5 – 1 __ 2 x2 − 0 ) 2 = 1 __ 4 x4 – 4x2 + 25 b x = ±2 √ __ 2 or x = 0 c OP = 3; f9 (x) > 0 so minimum when x = ±2 √ __ 2 , maximum when x = 0 23 a 3 + 5(3) + 32 − 33 = 0 therefore C on curve b A is (–1, 0); B is ( 5 _ 3 , 9 13 __ 27 ) 24 1.2 2.1 0, 5Velocity (cm/s)OTime (s) 25 10 ___ 3 , 2300π _______ 27 26 dA ___ dx = 4πx − 2000 _____ x 2 dA ___ dx = 0: 4πx = 2000 _____ x 2 → x 3 = 2000 _____ 4π = 500 ____ π 27 a y = 1 – x __ 2 − πx ___ 4	[ 0.0785108432173729, 0.05477064475417137, 0.04917829483747482, 0.02535714954137802, -0.0881166160106659, -0.02878936380147934, 0.0010375514393672347, -0.004131839144974947, -0.08299031108617783, -0.005186629015952349, -0.005194454919546843, -0.0246596522629261, 0.026318367570638657, -0.0413...
390 Answers 390 Full worked solutions are available in SolutionBank. Online b R = xy + π __ 2 ( x __ 2 ) 2 = x (1 – x __ 2 – πx ___ 4 ) + πx2 ____ 8 = x – x2 __ 2 – πx2 ____ 4 + πx2 ____ 8 = x __ 8 (8 – 4x – πx) c 2 _____ 4 + π m2 (0.280 m2) 28 a πx2 + 2πx + πx2 + 2πxh = 80π h = 40 – x – x2 __________ x V = π x2h = πx2 ( 40 – x – x2 __________ x ) = π(40 x − x2 − x3) b 10 __ 3 c d2V ____ dx2 < 0 ∴ maximum d 2300π _______ 27 e 22 2 _ 9 % 29 a Length of short sides = x ___ √ __ 2 Area = 1 __ 2 × base × height = 1 __ 2 ( x2 __ 2 ) = 1 __ 4 x2 m2 b Let l be length of EF. 1 __ 4 x2 l = 4000 ⇒ l = 16 000 _______ x2 S = 2 ( 1 __ 4 x2 ) + 2x l ____ √ __ 2 = 1 __ 2 x2 + 32 000x ________ √ __ 2 x2 = x2 __ 2 + 16 000 √ __ 2 _________ x c x = 20 √ __ 2 , S = 1200 m2 d d2S ____ dx2 > 0 Challenge a x7 + 7x6h + 21x5h2 + 35x4h3 b d ___ dx (x7) = lim h→0 (x + h)7 − x7 ____________ h = lim h→0 7x6h + 21x5h2 + 35x4h3 ______________________ h = lim h→0 (7x6 + 21x5h + 35x4h2) = 7x6 CHAPTER 13 Prior knowledge check 1 a 5 x 5 _ 2 b 2 x 3 _ 2 c x 5 _ 2 − √ __ x d x – 3 _ 2 + 4x 2 a 6x2 + 3 b x − 1 c 3x2 + 2x d − 1 ___ x 2 − 3x2 3 a O xy –1 3 –3 b Oy x –15 –5 Exercise 13A 1 a y = 1 _ 6 x6 + c b y = 2x5 + c c y = x−1 + c d y = 2x−2 + c e y = 3 _ 5 x 5 _ 3 + c f y = 8 _ 3 x 3 _ 2 + c g y = − 2 _ 7 x7 + c h y = 2 x 1 _ 2 + c i y = −10 x − 1 _ 2 + c j y = 9 _ 2 x 4 _ 3 + c k y = 3x12 + c l y = 2x−7 + c m y = −9 x 1 _ 3 + c n y = −5x + c o y = 3x2 + c p y = 10 __ 3 x 0.6 + c 2 a y = 1 _ 4 x4 − 3 x 1 _ 2 + 6x−1 + c b y = x4 + 3 x 1 _ 3 + x−1 + c c y = 4x + 4x−3 + 4 x 1 _ 2 + c d y = 3 x 5 _ 3 − 2x5 − 1 _ 2 x−2 + c e y = 4 x − 1 _ 3 − 3x + 4x2 + c f y = x5 + 2 x − 1 _ 2 + 3x−4 + c 3 a f(x) = 6 x2 − 3 x − 1 _ 2 + 5x + c b f(x) = x6 − x−6 + x – 1 _ 6 + c c f(x) = x 1 _ 2 + x − 1 _ 2 + c d f(x) = 2 x5 − 4x−2 + c e f(x) = 3 x 2 _ 3 − 6 x − 2 _ 3 + c f f(x) = 3 x3 − 2x−2 + 1 _ 2 x 1 _ 2 + c 4 y = 4 x 3 ____ 3 + 6x2 + 9x + c 5 f(x ) = −3x−1 + 4 x 3 _ 2 + x 2 ___ 2 − 4x + c Challenge y = − 12 ____ 7 x 7 _ 2 − 4 ____ 5 x 5 _ 2 + 3 ____ 2 x 2 + 1 __ x + c Exercise 13B 1 a x 4 ___ 4 + c b x 8 ___ 8 + c c −x−3 + c d 5 x 3 ____ 3 + c 2 a 1 __ 5 x5 + 1 __ 2 x4 + c b x 4 ___ 2 − x 3 ___ 3 + 5 x 2 ____ 2 + c c 2 x 5 _ 2 − x3 + c 3 a −4x−1 + 6 x 1 _ 2 + c b −6x−1 − 2 _ 5 x 5 _ 2 + c c −4 x − 1 _ 2 + x 3 ___ 3 − 2 x 1 _ 2 + c 4 a x4 + x−3 + rx + c b 1 _ 2 x 2 + 2 x 1 _ 2 − 2 x − 1 _ 2 + c c px5 ____ 5 + 2tx − 3x−1 + c 5 a t3 + t −1 + c b 2 _ 3 t3 + 6 t − 1 _ 2 + t + c c p __ 4 t4 + q2t + px3t + c 6 a 2x − 3 __ x + c b 4 _ 3 x3 + 6x2 + 9x + c c 4 _ 5 x 5 _ 2 + 2 x 3 _ 2 + c	[ 0.003952407278120518, -0.01145996805280447, 0.0033628090750426054, -0.014501330442726612, -0.044275134801864624, -0.017787249758839607, 0.03490917384624481, 0.04907870665192604, -0.0640309751033783, 0.025924302637577057, 0.05210623890161514, -0.07937609404325485, 0.0049435351975262165, -0....
391 Answers 3917 a 1 _ 3 x3 + 2x − 1 __ x + c b 1 _ 2 x2 + 8 _ 3 x 3 _ 2 + 4x + c c 2 x 1 _ 2 + 4 _ 3 x 3 _ 2 + c 8 a 3 _ 5 x 5 _ 3 − 2 ___ x 2 + c b − 1 __ x 2 − 1 _ x + 3x + c c 1 __ x x4 − 1 _ 3 x3 + 3 _ 2 x2 − 3x + c d 8 _ 5 x 5 _ 2 + 8 _ 3 x 3 _ 2 + 2 x 1 _ 2 + c e 3x + 2 x 1 _ 2 + 2x3 + c f 2 _ 5 x 5 _ 2 + 3x2 + 6 x 3 _ 2 + c 9 a − A __ x − 3x + c b 2 _ 3 √ __ P x 3 _ 2 – 1 ___ x 2 + c c − p __ x + 2 qx 3 _ 2 _____ 3 + rx + c 10 − 6 __ x + 8 x 3 _ 2 ____ 3 − 3 x 2 ____ 2 + 2x + c 11 2x4 + 3x2 − 6 x 1 _ 2 + c 12 a (2 + 5 √ __ x )2 = 4 + 10 √ __ x + 10 √ __ x + 25x = 4 + 20 √ __ x + 25x b 4x + 40 x 3 _ 2 _____ 3 + 25 x 2 _____ 2 + c 13 x 6 ___ 2 − 8 x 1 _ 2 + c 14 p = −4, q = −2.5 15 a 1024 − 5120x + 11 520x2 b 1024x − 2560x2 + 3840x3 + c Exercise 13C 1 a y = x3 + x2 − 2 b y = x4 − 1 ___ x 2 + 3x + 1 c y = 2 _ 3 x 3 _ 2 + 1 __ 12 x 3 + 1 _ 3 d y = 6 √ __ x − 1 _ 2 x 2 − 4 e y = 1 _ 3 x3 + 2x2 + 4x + 2 _ 3 f y = 2 _ 5 x 5 _ 2 + 6 x 1 _ 2 + 1 2 f(x ) = 1 _ 2 x 4 + 1 __ x + 1 _ 2 3 y = 1 − 2 ___ √ __ x − 3 __ x 4 f(x ) = 3x3 + 2x2 − 3x − 2 5 y = 6 x 1 _ 2 − 4 x 5 _ 2 ____ 5 + 118 ___ 5 6 a p = 1 _ 2 , q = 1 b y = 4 x 3 _ 2 + 5 x 2 ____ 2 − 421 ___ 2 7 a f(t) = 10 t − 5 t 2 ___ 2 b 7 1 _ 2 8 a f(t) = −4.9 t2 + 35 b 23.975 m c 35 m d 2.67 seconds e e.g. the ground is flat Challenge 1 f2(x) = x 3 ___ 3 ; f4(x) = x 4 ___ 12 b x n+1 _____________________ 3 × 4 × 5 × … × (n + 1 ) 2 f2(x) = x + 1; f3(x) = 1 _ 2 x2 + x + 1; f4(x) = 1 _ 6 x 3 + 1 _ 2 x 2 + x + 1 Exercise 13D 1 a 152 1 _ 4 b 48 2 _ 5 c 5 1 _ 3 d 2 2 a 5 1 _ 4 b 10 c 11 5 _ 6 d 60 1 _ 2 3 a 16 2 _ 3 b 46 1 _ 2 c 11 __ 14 d 2 1 _ 2 4 A = −7 or 4 5 28 6 −8 + 8 √ __ 3 7 k = 25 __ 4 8 450 mChallenge k = 2 Exercise 13E 1 a 22 b 36 2 _ 3 c 48 8 __ 15 d 6 2 4 3 6 4 10 2 _ 3 5 21 1 _ 3 6 4 __ 81 7 k = 2 8 a (−1, 0) and (3, 0) b 10 2 _ 3 9 1 1 _ 3 Exercise 13F 1 a 1 1 _ 3 b 20 5 _ 6 y xO –2 y x O 4 –1 c 40 1 _ 2 d 1 1 _ 3 y x O–3 3 2y x O e 21 1 __ 12 y x O 5 2 2 a (−3, 0) and (2, 0) b 21 1 __ 12 3 a f(−3) = 0 b f(x ) = (x + 3)(−x2 + 7x − 10) c f(x ) = (x + 3)(x − 5)(2 − x) d (−3, 0), (2, 0) and (5, 0) e 143 5 _ 6 Challenge 1 a 4 1 _ 2 b 9 c 9a ___ 2 d 4 1 _ 2 e 9 ___ 2a 2 a B has x -coordinate 1 ∫ 1 0 (x3 + x2 − 2x)dx = [ 1 __ 4 x4 + 1 __ 3 x3 − x2] 1 0 = 1 __ 4 + 1 __ 3 − 1 = − 5 __ 12 So area under x-axis is 5 __ 12 Area above x-axis is ( 1 __ 4 04 + 1 __ 3 03 − 02) − ( 1 __ 4 x4 + 1 __ 3 x3 − x2) = 5 __ 12 So the x-coordinate of a satisfies 3x4 + 4x3 − 12x2 + 5 = 0 Then use the factor theorem twice to get (x − 1)2(3x2 + 10x + 5) = 0	[ -0.037734612822532654, 0.024762989953160286, 0.018300039693713188, -0.032828524708747864, -0.023333560675382614, 0.1281108409166336, -0.005333567038178444, -0.016568941995501518, -0.08189444243907928, 0.044049493968486786, -0.04301578179001808, -0.12091062217950821, 0.06332910805940628, -0...
392 Answers 392 Full worked solutions are available in SolutionBank. Online b A has coordinates ( −5 + √ ___ 10 _________ 3 , −80 + 37 √ ___ 10 _____________ 27 ) The roots at 1 correspond to point B. The root −5 − √ ___ 10 _________ 3 gives a point on the curve to the left of −2 below the x-axis, so cannot be A. Exercise 13G 1 a A(−2, 6), B(2, 6) b 10 2 _ 3 2 a A(1, 3), B(3, 3) b 1 1 _ 3 3 6 2 _ 3 4 4.5 5 a (2, 12) b 13 1 _ 3 6 a 20 5 _ 6 b 17 1 _ 6 7 a, b Substitute into equation for y c y = x − 4 d 8 3 _ 5 8 3 3 _ 8 9 a Substitute x = 4 into both equations b 7.2 10 a 21 1 _ 3 b 2 5 _ 9 11 a (−1, 11) and (3, 7) b 21 1 _ 3 Mixed exercise 1 a 2 _ 3 x 3 − 3 _ 2 x 2 − 5x + c b 3 _ 4 x 4 _ 3 + 3 _ 2 x 2 _ 3 + c 2 1 _ 3 x 3 − 3 _ 2 x 2 + 2 _ x + 1 _ 6 3 a 2x4 − 2x3 + 5x + c b 2 x 5 _ 2 + 4 _ 3 x 3 _ 2 + c 4 4 _ 5 x 5 _ 2 − 2 _ 3 x 3 _ 2 – 6 x 1 _ 2 + c 5 x = 1 _ 3 t3 + t2 + t − 8 2 _ 3 ; x = 12 1 _ 3 6 a A = 6, B = 9 b 3 _ 5 x 5 _ 3 + 9 _ 2 x 4 _ 3 + 9x + c 7 a 9 _ 2 x − 1 _ 2 − 8 x − 3 _ 2 b 6 x 3 _ 2 + 32 x 1 _ 2 − 24x + c 8 a = 4, b = −3.5 9 25.9 m 10 a f(t) = 5 t + t2 b 7.8 seconds 11 a −1,3 b 10 2 _ 3 12 a − 2 x 3 _ 2 ____ 3 + 5x − 8 √ __ x + c b 7 __ 3 13 a (3, 0) b (1, 4) c 6 3 _ 4 14 a 3 _ 2 x − 1 _ 2 + 2 x − 3 _ 2 b 2 x 3 _ 2 − 8 x 1 _ 2 + c c A = 6, B = −2 15 a dy ___ dx = 6 x − 1 __ 2 − 3 __ 2 x 1 __ 2 = 3 __ 2 x − 1 __ 2 (4 − x) b (4,16) c 133 (3 sf ) 16 a (6,12) b 13 1 _ 3 17 a A(1, 0), B(5, 0), C (6, 5) b 10 1 _ 6 18 a q = − 2 b C (6,17) c 1 1 _ 3 19 – 9 __ x − 16 x 3 _ 2 _____ 3 + 2 x 2 − 5x + c 20 A = −6 or 1 21 a f9( x) = (2 − x2)(4 − 4x2 + x4) ___________________ x2 = 8x−2 − 12 + 6x2 − x4 b f 0( x) = −16x−3 + 12x − 4x3 c f(x) = − 8 __ x − 12x + 2 x 3 − x 5 ___ 5 − 47 ___ 5 22 a (−3, 0) and ( 1 _ 2 , 0) b 14 7 __ 24 23 a ( − 3 _ 2 , 0) and (4, 0) b 55 11 __ 24 24 a −2 and 3 b 21 1 __ 12 Challenge 10 5 __ 12 CHAPTER 14 Prior knowledge check 1 a 125 b 1 __ 3 c 32 d 49 e 1 2 a 66 b y21 c 26 d x4 3 gradient 1.5, intercept 4.1 Exercise 14A 1 a 2 O13y 4 x –1–2–3–4 2 2.634 1y = (1.7)x y = 4 b x ≈ 2.6 2 a 2 O13y 4 x–1–2–3–4 2 –1.434 1y = (0.6)x y = 2 b x ≈ −1.4 3 2 O13y 4 x–1–2–3–4 234 1y = 1x 4 a True, because a0 = 1 whenever a is positive b False, for example when a = 1 __ 2 c True, because when a is positive, ax > 0 for all values of x 5 2ba cd O1y –3x	[ -0.06452642381191254, 0.08347923308610916, -0.035933539271354675, -0.05223415419459343, 0.03906994313001633, 0.023860573768615723, -0.07203859835863113, -0.028486385941505432, -0.04397943243384361, 0.010822592303156853, -0.005721673369407654, -0.04940275102853775, -0.013422289863228798, -0...
393 Answers 3936 k = 3, a = 2 7 a As x increases, y decreases b p = 1.2, q = 0.2 Challenge Oy y = 2x – 2 + 5 x21 4 Exercise 14B 1 a 2.7183 b 54.5982 c 0.0000 d 1.2214 2 a Oy x1y = ex e = 2.71828… e3 = 20.08553… 3 a Oy x2y = ex + 1 y = 1 b y xy = 4e–2x O4 c y x y = –3–1y = 2ex – 3 O d y x4 y = 4 – ex3 e y xy = 6 + 10e 16 61 2 Ox f y xy = 100e–x + 10110 10 O 4 a A = 1, C = 5, b is positive b A = 4, C = 0, b is negative c A = 6, C = 2, b is positive 5 A = e2, b = 3 Oy = f(x) e2y x 6 a 6e6x b − 1 _ 3 e − 1 _ 3 x c 14e2x d 2e0.4x e 3e3x + 2ex f 2e2x + ex 7 a 3e6 b 3 c 3e−1.5 8 f9(x) = 0.2e0.2x The gradient of the tangent when x = 5 is f9(5) = 0.2e1 = 0.2e. The equation of the tangent is therefore e = 0.2e × 5 + c, so c = 0. Exercise 14C 1 a £20 000 b £14 331 c t20 000V (£) OV = 20 000e–t 12	[ -0.002270190045237541, 0.034995514899492264, 0.013202124275267124, -0.004304564092308283, -0.03082462027668953, 0.04877760633826256, 0.02412908524274826, 0.030379317700862885, -0.06413812935352325, 0.01448080874979496, 0.032256487756967545, -0.06143585965037346, 0.010788286104798317, -0.05...
394 Answers 394 Full worked solutions are available in SolutionBank. Online 2 a 30 000 b 38 221 c tP (thousands) 30 10094 OP = 20 + et 50 d Model predicts population of the country to be over 200 million, this is highly unlikely and by 2500 new factors are likely to affect population growth. Model not valid for predictions that far into the future. 3 a 200 b Disease will infect up to 300 people. c N t200N = 300 – 100e–0.5t300 O 4 a i 15 rabbits ii 132 rabbits b The initial number of rabbits c dR ____ dm = 2.4 e 0.2m When m = 6, dR ____ dm = 7.97 ≈ 8 d The rabbits may begin to run out of food or space 5 a 0.565 bars b dp ___ dh = –0.13 e −0.13h = − 0.13p , k = −0.13 c The atmospheric pressure decreases exponentially as the altitude increases d 12% 6 a Model 1: £15 733 Model 2: £15 723 Similar results b Model 1: £1814 Model 2: £2484 Model 2 predicts a larger value c 20k OModel 1 Model 2T t d In Model 2 the tractor will always be worth at least £1000. This could be the value of the tractor as scrap metal. Exercise 14D 1 a log4 256 = 4 b log3 1 _ 9 = –2 c log10 1 000 000 = 6 d log11 11 = 1 e log0.2 0.008 = 3 2 a 24 = 16 b 52 = 25 c 9 1 _ 2 = 3 d 5–1 = 0.2 e 105 = 100 0003 a 3 b 2 c 7 d 1 e 6 f 1 _ 2 g –1 h –2 i 10 j –2 4 a 625 b 9 c 7 d 9 e 20 f 2 5 a 2.475 b 2.173 c 3.009 d 1.099 6 a 5 = log2 32 < log2 50 < log2 64 = 6 b 5.644 7 a i 1 ii 1 iii 1 b a1 ≡ a 8 a i 0 ii 0 iii 0 b a0 ≡ 1 Exercise 14E 1 a log2 21 b log2 9 c log5 80 d log6 ( 64 __ 81 ) e log10 120 2 a log2 8 = 3 b log6 36 = 2 c log12 144 = 2 d log8 2 = 1 _ 3 e log10 10 = 1 3 a 3 loga x + 4 loga y + loga z b 5loga x – 2loga y c 2 + 2loga x d logax – 1 _ 2 loga y – loga z e 1 _ 2 + 1 _ 2 loga x 4 a 4 _ 3 b 1 __ 18 c √ ___ 30 d 2 5 a log3 (x + 1) – 2 log3 (x – 1) = 1 log3 ( x + 1 _______ (x − 1 ) 2 ) = 1 x + 1 _______ (x − 1 ) 2 = 3 x + 1 = 3(x –1)2 x + 1 = 3(x2 – 2x + 1) 3x2 – 7x + 2 = 0 b x = 2 6 a = 9, b = 4 Challenge loga x = m and loga y = n x = am and y = an x ÷ y = am ÷ an = am − n loga ( x __ y ) = m − n = loga x − loga y Exercise 14F 1 a 6.23 b 2.10 c 0.431 d 1.66 e –3.22 f 1.31 g 1.25 h –1.73 2 a 0, 2.32 b 1.26, 2.18 c 1.21 d 0.631 e 0.565, 0.712 f 0 g 2 h –1 3 a 5.92 b 3.2 4 a (0, 1) 1y y = 4x x O b 1 _ 2 , 3 _ 2 5 a 0.7565 b 7.9248 c 0.2966	[ 0.02497393637895584, 0.003860347904264927, 0.006339541636407375, -0.011927502229809761, -0.010583660565316677, -0.05687012895941734, -0.03460678830742836, 0.041691116988658905, -0.009282687678933144, 0.08540212363004684, 0.03840861842036247, -0.043896712362766266, -0.015999415889382362, -0...
395 Answers 395Exercise 14G 1 a ln 6 b 1 _ 2 ln 11 c 3 – ln 20 d 1 _ 4 ln ( 1 _ 3 ) e 1 _ 2 ln 3 – 3 f 5 – ln 19 2 a e2 b e _ 4 c 1 _ 2 e4 – 3 _ 2 d 1 _ 6 ( e 5 _ 2 + 2) e 18 – e 1 __ 2 f 2, 5 3 a ln 2, ln 6 b 1 _ 2 ln 2, 0 c e3, e−5 d ln 4, 0 e ln 5, ln ( 1 _ 3 ) f e6, e−2 4 ln 3, 2 ln 2 5 a 1 _ 8 (e2 + 3) b 1 _ 5 (ln 3 + 40) c 1 _ 5 ln 7, 0 d e3, e−1 6 1 + ln 5 ________ 4 + ln 3 7 a The initial concentration of the drug in mg/l b 4.91 mg/l c 3 = 6 e − t __ 10 1 _ 2 = e − t __ 10 ln ( 1 _ 2 ) = – t ___ 10 t = –10 ln ( 1 _ 2 ) = 6.931… = 6 hours 56 minutes 8 a (0, 3 + ln 4) b (4 – e–3) Challenge As y = 2 is an asymptote, C = 2.Substituting (0, 5) gives 5 = A e0 + 2, so A is 3. Substituting (6, 10) gives 10 = 3e6B + 2. Rearranging this gives B = 1 _ 6 ln ( 8 _ 3 ) . Exercise 14H 1 a log S = log (4 × 7x) log S = log 4 + log 7x log S = log 4 + x log 7 b gradient log 7, intercept log 4 2 a log A = log (6x4) log A = log 6 + log x4 log A = log 6 + 4 log x b gradient 4, intercept log 6 3 a Missing values 1.52, 1.81, 1.94 b log xlog y 0.20 0.40.60.811.21.40.5 011.522.5 c a = 3.5, n = 1.4 4 a Missing values 2.63, 3.61, 4.49, 5.82 b x 2 0 4 6 8 101 0234567log y c b = 3.4, a = 105 a Missing values −0.39, 0.62, 1.54, 2.81 b O 1 –1 –2 2 31234log R log m c a = 60, b = 0.75 d 1,600 kcal per day (2 s.f.) 6 a Missing values 2.94, 1.96, 0.95 b log Rlog f 0.50 11.522.533.50.5 011.522.533.54 c A = 5800, b = −0.9 d 694 times 7 a Missing values 0.98, 1.08, 1.13, 1.26, 1.37 b P = abt log P = log (abt) log P = log a + log bt log P = log a + t log b c tlog P 50 10152025303540450.2 00.40.60.811.21.4 d a = 7.6, b = 1.0 e The rate of growth is often proportional to the size of the population 8 a log N = 0.095t + 1.6 b a = 40, b = 1.2 c The initial number of sick people d 9500 people. After 30 days people may start to recover , or the disease may stop spreading as quickly. 9 a log A = 2 log w – 0.1049 b q = 2, p = 0.7854 c Circles: p is approximately one quarter π, and the width is twice the radius, so A = π __ 4 w 2 = π __ 4 (2r) 2 = π r 2 . Challenge y = 5.8 × 0.9x	[ -0.00033812926267273724, 0.025742139667272568, -0.04405737668275833, -0.0063821119256317616, -0.04907136783003807, 0.06211070716381073, -0.03236329182982445, -0.04145098850131035, -0.07328147441148758, 0.022982964292168617, 0.008300061337649822, -0.057222574949264526, -0.0568169429898262, ...
396 Answers 396 Full worked solutions are available in SolutionBank. Online Mixed exercise 1 a Oy x1y = 2–x y = 0 b Oy x4y = 5ex – 1 y = –1 c Oy x1y = ln x x = 0 2 a 2 loga p + loga q b loga p = 4, loga q = 1 3 a 1 _ 4 p b 3 _ 4 p + 1 4 a 2.26 b 1.27 c 7.02 5 a 4x – 2x + 1 – 15 = 0 22x – 2 × 2x – 15 = 0 (2x)2 – 2 × 2x – 15 = 0 u2 – 2u – 15 = 0 b 2.32 6 x = 6 7 a –e–x b 11e11x c 30e5x 8 a e 8 + 5 ______ 2 b ln 5 ____ 4 c – 1 _ 2 ln 14 d 3 + √ ___ 13 ________ 2 e 0 f e 4 ___ 2 9 a £950 b £290 c 4.28 years d £100 e OP t100950 P = 100 + 850e–t 2 f A good model. The computer will always be worth something10 a y = ( 2 ____ ln 4 ) x b (0, 0) satisfies the equation of the line. c 2.43 11 a We cannot go backwards in time b 75°C c 5 minutes d The exponential term will always be positive, so the overall temperature will be greater than 20°C. 12 a S = aV b log S = log (aVb) log S = log a + log (V b) log S = log a + b log V blog S 1.26 1.70 2.05 2.35 2.50 log V 0.86 1.53 2.05 2.49 2.72 c log Slog V 0.50 0.00 1.00 1.50 2.00 2.50 3.000.50 0.001.001.502.002.503.00 d The gradient is approximately 1.5; a = 0.09 13 a The model concerns decay, not growth b OR t140 R = 140ekt c 70 = 140e30k 1 _ 2 = e30k ln ( 1 _ 2 ) = 30k k = 1 __ 30 ln ( 1 _ 2 ) k = – 1 __ 30 ln (2), so c = – 1 __ 30 14 a 6.3 million views b dV ___ dx = 0.4e0.4x c 9.42 × 1016 new views per day d This is too big, so the model is not valid after 100 days 15 a 4.2 b i 1.12 × 1025 dyne cm ii 3.55 × 1026 dyne cm c divide b ii by b i 16 a They exponentiated the two terms on RHS separately rather than combining them first. b x = 2 ± √ __ 5	[ -0.004907224327325821, -0.007413041777908802, 0.018521111458539963, 0.023648159578442574, -0.0245670173317194, 0.0364726223051548, 0.01137244701385498, -0.03667840734124184, -0.06409579515457153, 0.0992368832230568, -0.006522845476865768, -0.03199135139584541, -0.07769443094730377, -0.0009...
397 Answers 397Challenge a y = 9x = 32x, log3(y) = 2x b y2 = (9x)2 = 92x, log9(y2) = 2x c x = − 1 __ 3 or x = −2 Review exercise 3 1 −4.5 2 √ __ 7 3 a All equal to √ ____ 145 b (x − 1)2 + (y + 3)2 = 145 4 a −2i − 8j b \| ⟶ AB \| = \| ⟶ AC \| = √ ___ 85 c \| ⟶ BC \| = √ ___ 68 cos ∠ABC = 85 + 68 − 85 _____________ 2 × √ ___ 85 × √ ___ 68 = 1 ___ √ __ 5 5 12 6 a 5N b 7 7 m = 50 √ __ 3 + 30, n = 50 8 a √ _____________ (−75) 2 + 180 2 = 195 > 150 = √ _________ 90 2 + 120 2 b Boat A: 6.5 m/s; Boat B: 5 m/s; Both boats arrive at the same time – it is a tie. 9 lim h→0 5 (x + h) 2 − 5 x 2 _____________ h = lim h→0 5x 2 + 10xh + 5 h 2 − 5 x 2 _____________________ h = lim h→0 10xh + 5 h 2 ___________ h = lim h→0 10x + 5h = 10x 10 dy ___ dx = 12x2 + x − 1 __ 2 11 a dy ___ dx = 4 + 9 __ 2 x 1 __ 2 − 4x b Substitute x = 4 into equation for C c Gradient of tangent = −3 so gradient of normal = 1 __ 3 Substitute (4, 8) into y = 1 __ 3 x + c Rearrange y = 1 __ 3 x + 20 __ 3 d PQ = 8 √ ___ 10 12 a dy ___ dx = 8x − 5x−2, at P this is 3 b y = 3x + 5 c k = − 5 __ 3 13 a P = 2, Q = 9, R = 4 b 3 x 1 __ 2 + 9 __ 2 x − 1 __ 2 − 2 x − 3 __ 2 c When x = 1, f9(x) = 5 1 __ 2 , gradient of 2y = 11x + 3 is 5 1 __ 2 , so it is parallel with tangent 14 f9 (x) = 3x2 – 24x + 48 = 3(x − 4)2 > 0 15 a A (1,0) and B (2,0) b ( √ __ 2 , 2 √ __ 2 – 3) 16 a V = πr2h = 128π, so h = 128 ____ r 2 S = 2π rh + 2πr2 = 256π _____ r + 2πr2 b 96π cm217 a dy ___ dx = 6x + 2 x − 1 __ 2 b d2y ____ dx2 = 6 − x − 3 __ 2 c x3 + 8 __ 3 x 3 __ 2 + c 18 a 2x3 − 5x2 − 12x b x(2x + 3)(x − 4) c xy O 4 –3 2 19 6 3 _ 4 20 4 21 a −x4 + 3x2 + 4 = (−x2 + 4)(x2 + 1); x2 + 1 = 0 has no real solutions; so solutions are A(−2, 0), B(2, 0) b 19.2 22 4 1 _ 2 square units 23 a P(−1, 4), Q(2, 1) b 4.5 24 a k = −1, A(0, 2) b ln 3 25 a 425 °C b 7.49 minutes c 1.64 °C/minute d The temperature can never go below 25 °C, so cannot reach 20 °C. 26 a −0.179 b x = 15 27 a x = 1.55 b x = 4 or x = 1 __ 2 28 a logp 2 b 0.125 29 a x = 2 b x = ln 3 or x = ln 1 = 0 30 a Missing values 0.88, 1.01, 1.14 and 1.29 b tlog P 01 0 20 30 400.2 00.40.60.811.21.4 c P = abt log P = log (abt) = log a + t log b This is a linear relationship. The gradient is log b and the intercept is log a. d a = 5.9, b = 1.0 Challenge 1 a 0 b 1 2 a f9 (−3) = f9(2) = 0, so f9(x) = k(x + 3)(x − 2) = k(x2 + x − 6); there are no other factors as f(x) is cubic b 2x3 + 3x2 − 36x – 5 3 51.2 4 a f(0) = 03 – k(0) + 1 = 1; g(0) = e2(0) = 1; P(0, 1) b 1 _ 2	[ 0.06604847311973572, 0.08798037469387054, -0.024763712659478188, 0.008190706372261047, -0.005782369524240494, 0.08301936089992523, -0.01188302505761385, -0.00824385229498148, -0.09283727407455444, 0.08265465497970581, -0.035439666360616684, -0.026332475244998932, 0.017151208594441414, -0.0...
398 Answers 398 Full worked solutions are available in SolutionBank. Online Practice paper 1 a 1 __ 3 b 5 √ __ 2 2 y = 2 __ 3 x + 8 __ 3 3 a error 1: = − 3 ___ √ __ x = −3 x − 1 _ 2 , not −3 x 1 _ 2 error 2: [ x5 __ 5 − 2 x 3 __ 2 + 2x] 2 1 = ( 32 ___ 5 − 2 √ __ 8 + 4) − ( 1 __ 5 − 2 + 2) not ( 1 __ 5 − 2 + 2) − ( 32 ___ 5 − 2 √ __ 8 + 4) b 5.71 (3 s.f.) 4 x = 30°, 90°, 150° 5 a 2x2(x + 3) b 2x2(x + 3) = 980 ⇒ 2x3 + 6x2 − 980 = 0 ⇒ x3 + 3x2 − 490 = 0 c For f( x) = x3 + 3x2 − 490 = 0, f(7) = 0, so x − 7 is a factor of f(x) and x = 7 is a solution. d Equation becomes (x − 7)( x2 + 10x +70) = 0 Quadratic has discriminant 102 − 4 × 1 × 70 = −180 So the quadratic has no real roots, and the equation has no more real solutions. 6 x + 15y + 106 = 0 7 a log10 P = 0.01t + 2 b 100, initial population c 1.023 d Accept answers from 195 to 200 8 1 + cos4 x − sin4 x ≡ 1 + (cos2x + sin2x)(cos2 − sin2x) ≡ (1 − sin2x) + cos2x ≡ 2cos2x 9 Magnitude = √ ___ 29 , angle = 112° (3 s.f.) 10 a 56.5° (3 s.f.) b £49.63 11 a O 27–28–1y = g(x) b −4, −1, 4 12 x = 3 or − 5 __ 3 13 a 1 − 15x + 90 x2 b 0.859 c Greater: The next term will be subtracting from this, and future positive terms will be smaller . 14 a f(x ) = ∫ ( x − 3 _ 2 − 1 − x −2 ) dx = −2 x − 1 _ 2 − x + x −1 + c = − x 2 + 2 √ __ x − 1 ____________ x + c b c = 5 __ 3 + 2 __ 3 √ __ 3 15 a 52 + 12 − 4 × 5 + 6 × 1 = 12, so (5, 1) lies on C. Centre = (2, −3), radius = 5 b y = − 3 __ 4 x + 19 __ 4 c 12 15 __ 32	[ -0.03663749247789383, 0.007427210453897715, 0.061256539076566696, -0.04550020024180412, 0.024715406820178032, 0.04097634181380272, 0.008829600177705288, -0.026126131415367126, -0.06632319837808609, 0.05424593761563301, 0.04923490807414055, -0.04833931848406792, 0.09628164023160934, 0.01978...
399 Index 399elimination 39–40 equations circle 117–119 normal 268–269 solving using logarithms 324–325 straight line 90, 93–95 tangent 268–269 equilater al triangles 208 exhaustion, proof by 150 expanding brack ets 4–6, 163–164 exponential functions 312–318 gradient functions of 314–316 graphs 312–313, 326 inverses 326 modelling with 317 exponents 2–3, 312 expressions expanding 4–6 factorising 6–8 simplifying 2–3 factor theorem 143–144 factorial notation 161 factorising expressions 6–8 polynomials 143–144 quadratic equa tions 19–21 force 248 for mula completing the square 22–23 distance between points 100–101 gradient of str aight line 91–92 quadratic 21 fractions , simplifying algebraic 138–139 function notation first order deriv ative 260, 289 second order deriv ative 271 functions 25–26 cubic 60–62 decreasing 270–271 domain 25 finding 90–91 gradient see gradient functions increasing 270–271 quadra tic see quadratic functions quartic 64–65 range 25 reciproca l 66–67 roots 25, 30–31 transfor ming 79–80, 194–197 fundamental theorem of calculus 296 geometric prob lems, solving with vectors 244–246 gradient curve 256–257 normal 268–269 paralle l lines 97 perpendicular lines 98–99 straight line 91–92 tangent 256, 268–269 zero 264, 273 gradient functions 260–261 of exponentia l functions 314–316 rate of change of 271–272, 274 sketching 277–278 gra phs 59–80 cosine 192–193, 195–197, 203 cubic 60–62, 68 exponential functions 312–313, 326 inequalities on 51–52 intersection points 42–43, 68–69 logarithms 326 periodic 192 quadratic functions 27–29 quartic 64–65 reciproca l 66–67 reflections 77 regions 53–54 simultaneous equations 43–45 sine 192–194, 195–196, 203 stretching 75–77 tangent 193, 195–197 translating 71–73 identities 146, 209 identity symbol 146 images 79 ‘implies that’ symbol 19 indefinite integrals 290–291 indices 2–3, 312 fractional 9–11 laws of 2–3, 9 negativ e 9–11 inequalities 46–54 on graphs 51–52 linear 46–48 quadratic 49–50 regions satisfying 53–54 inflection, point of 273–274 initial va lue 317 integration 287–304 areas between curv es and lines 302–304 areas under curves 297–298 areas under x -axis 300–301 definite 295–296 finding functions 90–91 indefinite 290–291 polynomials 289, 290 symbol 290 xn 288–289 intersection points 42–43, 68–69 intervals 270 inv erse trigonometric functions 213 irra tional numbers 12 isosceles right-angled triangles 208acute angles 206–207 algebr a 1–17 alge braic fractions, simplifying 138–139 alge braic methods 137–153 angles acute 206–207 in all quadrants 203–207 negativ e 204 obtuse 203 positive 203 reflex 211 appr oximations 167 area trapezium 303 triangle 100–101, 185–186, 303 under curve 297–298 assumptions 106 asymptotes 66 horizontal 277, 278 translation 73 vertical 277 bases 2, 319, 325 binomial estimation 167–168 binomial expansion 158–168 in ascending powers of x 164 combinations 161 factorial notation 161 general ter m 165 ignoring large pow ers of x 167–168 Pascal’ s triangle 159–160, 161 solving binomial prob lems 165–166 brackets expanding 4–6, 163–164 minus sign outside 2 CAST diagram 205 chords 123 perpendicular bisectors 123–125, 129–130 circles 113–130 equation 117–119 finding centre 129 intersections of straight lines and 121 triangles and 128–130 unit 203, 209 circumcentre 117 circumcir cle 128–129 column vectors 235–237 combinations 161 common factors 7 completed square for m 22 completing the square 22–23 solving quadratic equa tions by 23–24 conjectures 146 constant of integr ation 288, 293 coordinates 90 cos θ of any angle 207 graph 192–193, 195–197, 203 quadrants wher e positive or negative 205 quadratic equa tions in 219–221 ratio 174 value f	[ -0.008804206736385822, 0.1193762943148613, 0.0967804566025734, 0.006190256681293249, -0.06818682700395584, -0.07585109025239944, -0.052523281425237656, 0.0053118327632546425, -0.10357135534286499, 0.04361988231539726, 0.011120540089905262, -0.06547959893941879, 0.03712446242570877, 0.03339...
or 30°, 45° and 60° 208 value f or multiples of 90° 192, 204 cos θ = k 213–215 cos (θ + α) = k 218 cos nθ = k 217 cos−1 θ 213 cosine rule 174–176, 187–189 counter-e xamples 150–151 critical v alues 49 cube roots 10 cubic functions 60–62 cubic graphs 60–62 compared with quadra tic graphs 68 curves areas between lines and 302–304 areas under 297–298 gradients 256–257 definite integra ls 295–296 delta x (δx) 259 demonstra tion 146 denominator , rationalising 13–14 deriva tive finding 259–261 first order 271 second order 271–272 straight line 264 difference of tw o squares 7 differentiation 255–280 decreasing functions 270–271 finding deriva tive 259–261 from first principles 260 functions with two or more ter ms 266–267 increasing functions 270–271 modelling with 279–280 quadratic functions 264–265 second order deriv atives 271–272 sketching gradient functions 277–278 stationary points 273–276 xn 262–263 directed line segments 231 direction of v ector 231, 239–240 discriminant 30–31, 43–44 displacement 235, 248 distance 248 between points 100–101 division law of logarithms 321–323 domain 25Answers Index	[ 0.022473951801657677, 0.015355157665908337, 0.00029580481350421906, -0.0230964794754982, -0.04679643735289574, 0.0022889147512614727, -0.11842477321624756, 0.037406258285045624, -0.08646530658006668, -0.0455658994615078, 0.02838910184800625, -0.06168826296925545, 0.00373494578525424, 0.014...
400 Index 400tan θ = p 213–215 tan (θ + α) = p 218 tan nθ = p 217 tan−1 θ 213 tangents 43, 123–125, 256 equations 268–269 gradients 256, 268–269 theorems 146 transf ormations 71–80 to functions 79–80, 194–197 translations 71–73 tra peziums, areas 303 triangle law 231 triangles areas 100–101, 185–186, 303 circles and 128–130 equilatera l 208 isosceles right-angled 208 lengths of sides 174 –176, 179–180 right-angled 187–188 sizes of angles 174 –175, 180–181, 183–184 solving problems 187–188 trigonometric equations 213–221 trigonometric functions, inv erse 213 trigonometric gra phs 192–194 transfor ming 194 –197 trigonometric identities 209–211, 221 trigonometric ratios 173–197 signs in four quadrants 203–207 turning points 27–29, 264 unit circle 203, 209 unit vectors 236, 239 value initial 317 produced by definite integr al 295 vectors 231–250 addition 231–233 column 235–237 direction 231, 239–240 magnitude 231, 239–240, 248 modelling with 248–250 multiplication b y scalars 232 paralle l 232, 233 position 242–243 representing 235–237 solving geometric problems with 244 –246 subtraction 232 two-dimensional 236 unit 236, 239 zero 232 ve locity 248 vertices 127 y = ex 314 –316 y = mx + c 90–92 y-inter cept 91 zer o gradient 264, 273 zero v ector 232reciproca l functions 66–67 reciproca l graphs 66–67 reflections 77 refle x angles 211 regions 53–54 resultant 231 right-angled triangles 187–188 isosceles 208 roots 19 of function 25, 30–31 number of 30–31 repeated 19 scalars 232, 248 scale factors 75–77 semicircle, angle in 129 set notation 46 simplifying expressions 2–3 simultaneous equations 39–45 on graphs 43–45 linear 39–40 quadratic 41–42 sin θ of any angle 207 graph 192–194, 195–196, 203 quadrants wher e positive or negative 205 quadratic equa tions in 219–221 ratio 174 value f or 30°, 45° and 60° 208 value f or multiples of 90° 192, 204 sin θ = k 213–215 sin (θ + α) = k 218 sin nθ = k 217 sin−1 θ 213 sin2 θ + cos2 θ ≡ 1 209, 221 sine rule 179–181, 187–188 small change 259 speed 248 square r oots 10 statements 146 stationary points 273–276 straight lines 89–106 equation 90, 93–95 gradient 91–92 intersections of circles and 121 modelling with 103–106 stretches 75–77 substitution 39–40 surds 12–13 symmetry, lines of 27–29 tan θ of any angle 207 graph 193, 195–197 quadrants wher e positive or negative 205 quadratic equa tions in 219 ratio 174 value f or 30°, 45° and 60° 208 value f or multiples of 90° 193, 204 tan θ = sin θ _____ cos θ 29non-linear data, lo garithms and 328–330 norma ls 268 equations 268–269 number lines 46–48 obtuse angles 203 parabola 27 para llel lines 97–98 paralle logram law 233 Pascal’ s triangle 159–160, 161 periodic gra phs 192 perpendicular bisectors chords 123–125, 129–130 line segments 116 perpendicular lines 98–99 product of gr adients 98 ‘plus or minus’ symbol 20 point of inflection 273–274 points of intersection 42–43, 68–69 polynomials 139–144 dividing 139–141 factorising 143–144 integrating 289, 290 position vectors 242–243 positive integers 163 power la w of logarithms 321–323 powers 2–3, 312 principal va lue 213–214 products 4–6 proof 146–148 methods of 150–152 Pythagoras’ theorem 117, 187 quadratic equa tions form 19 signs of solutions 19 solving by completing the square 23–24 solving by factorisation 19–21 in trigonometric ratios 219–221 quadratic e xpressions 7 quadratic f ormula 21 quadratic functions 25–26 differentiating 264–265 graphs 27–29 modelling with 32–33 quadratic gr aphs, compared with cubic graphs 68 quadratic inequa lities 49–50 quadratic sim ultaneous equations 41–42 quartic functions 64–65 quartic graphs 64–65 quotient 140 range 25 ra te of change 105, 279 ratio theor em 244 rationa l numbers 9 rationa	[ -0.015750648453831673, 0.044856373220682144, 0.056311119347810745, -0.023283885791897774, -0.061915747821331024, 0.003554436843842268, -0.0269561018794775, 0.019527800381183624, -0.07021992653608322, -0.0016957970801740885, 0.05383745953440666, -0.056922752410173416, 0.003681753994897008, ...
lising denominators 13–14 real n umbers 7, 232key points summaries algebr aic expressions 17 algebr aic methods 156 binomial expansion 172 circles 135–136 differentiation 286 equations 58 exponentials and lo garithms 336–337 graphs 84 inequalities 58 integration 310 quadratics 37 straight line gra phs 111–112 trigonometric identities and equations 224–225 trigonometric ratios 201 vectors 254 known facts , starting proof from 151–152 laws of indices 2–3, 9 length 100–101 limits 260 integral betw een 295–296 line segments 114 directed 231 midpoints 114 perpendicular bisectors 116 linear inequalities 46–48 local maximum 273–276 local minimum 273–276 logarithms 319–330 base 319, 325 division law 321–323 graphs of 326 multiplication la w 321–323 natura l 320, 326–327 and non-linear data 328–330 power la w 321–323 solving equations using 324 –325 to base 10 320 magnitude-direction f orm 240 magnitude of v ector 231, 239–240, 248 mathematica l models 32, 104 mathematica l proof 146–148 methods of 150–152 maximum point 27–29 midpoints, line segments 114 minimum point 27–29 modelling with differentiation 279–280 exponential 317 linear 103–106 with quadratic functions 32–33 with vectors 248–250 modulus of vector 239 multiplication la w of logarithms 321–323 natur al logarithms 320, 326–327 natura l numbers 163	[ -0.07146811485290527, 0.022757621482014656, 0.019757868722081184, 0.04973829165101051, 0.026328211650252342, -0.058873217552900314, -0.09019046276807785, 0.001942924689501524, -0.09273918718099594, -0.022780228406190872, 0.008641071617603302, -0.10636290907859802, -0.025523578748106956, 0....

389 Answers 389Exercise 12K 1 2t – 3 2 2π 3 – 4 _ 3 4 48π 5 18 6 a Let x = width of garden. x + 2y = 80 A = xy = 2y b 20 m × 40 m, 800 m2 7 a 2π r2 + 2πrh = 600π ⇒ h = 300 – r2 ________ r V = π r2h = πr (300 − r2) = 300πr − πr3 b 2000π cm3 8 a Let θ = angle of sector . πr2 × θ ____ 360 = 100 ⇒ θ = 36 000 _______ πr2 P = 2r + 2πr × θ ____ 360 = 2r + 200πr ______ πr2 = 2r + 100 ____ r θ < 2π ⇒ Area < πr2, so πr2 > 100 ∴ r > √ ____ 100 ____ π b 40 cm 9 a Let h = height of rectangle. P = πr + 2r + 2h = 40 ⇒ 2h = 40 – 2r – πr A = π __ 2 r2 + 2rh = π __ 2 r2 + r (40 – 2r – πr) = 40r – 2r2 – π __ 2 r2 b 800 _____ 4 + π cm2 10 a 18x + 14y = 1512 ⇒ y = 1512 – 18x ___________ 14 A = 12xy = 12x ( 1512 – 18x ___________ 14 ) = 1296x − 108x2 ______ 7 b 27 216 mm2 Mixed exercise 1 f9(x) = lim h→0 10(x + h)2 − 10x2 ________________ h = lim h→0 20xh + 10h2 ____________ h = lim h→0 (20x + 10 h) = 20x 2 a y-coordinate of B = (δx)3 + 3(δx)2 + 6δx + 4 Gradient = ((δx )3 + 3(δx)2 + 6δx + 4) − 4 __________________________ (1 + δx) − 1 = (δx )3 + 3(δx)2 + 6δx __________________ (δx) = (δx)2 + 3δx + 6 b 6 3 4, 11 3 _ 4 , 17 25 __ 27 4 2, 2 2 _ 3 5 (2, –13) and (–2, 15) 6 a 1 – 9 __ x2 b x = ±3 7 3 _ 2 x − 1 _ 2 + 2 x − 3 _ 2 8 a dy ___ dx = 6 x − 1 _ 2 − 3 _ 2 x 1 _ 2 = 3 _ 2 x − 1 _ 2 (4 − x ) b (4, 16) 9 a x + x 3 _ 2 − x − 1 _ 2 − 1 b 1 + 3 _ 2 x 1 _ 2 + 1 _ 2 x − 3 _ 2 c 4 1 __ 16 10 6x2 + 1 _ 2 x − 1 _ 2 – 2x–211 a = 1, b = –4, c = 5 12 a 3x2 – 10x + 5 b i 1 _ 3 ii y = 2x – 7 iii 7 _ 2 √ __ 5 13 y = 9x – 4 and 9y + x = 128 14 a ( 4 _ 5 , − 2 _ 5 ) b 1 _ 5 15 P is (0, −1), dy ___ dx = 3x2 − 4x – 4 Gradient at P = −4, so L is y = −4x − 1. −4x −1 = x3 − 2x2 − 4x − 1 ⇒ x2(x − 2) = 0 x = 2 ⇒ y = −9, so Q is (2, −9) Distance PQ = √ ___________________ (2 − 0 ) 2 + (−9 − (−1) ) 2 = √ ___ 68 = 2 √ ___ 17 16 a x = 4, y = 20 b d2y ____ dx2 = 3 __ 4 x − 1 __ 2 + 96x3 At x = 4, d2y ____ dx2 = 15 ___ 8 > 0 (4, 20) is a local minimum. 17 (1, −11) and ( 7 _ 3 , − 329 ___ 27 ) 18 a 7 31 __ 32 b f9( x) = (x − 1 __ x ) 2 > 0 for all values of x 19 (1, 4) 20 a (1, 33) maximum, (2, 28) and (−1, 1) minimum b 20y x O(2, 28)(1, 33) (–1, 1)y = f(x) 21 a 250 ____ x 2 – 2x b (5, 125) 22 a P (x, 5 − 1 __ 2 x2) OP2 = (x – 0)2 + (5 – 1 __ 2 x2 − 0 ) 2 = 1 __ 4 x4 – 4x2 + 25 b x = ±2 √ __ 2 or x = 0 c OP = 3; f9 (x) > 0 so minimum when x = ±2 √ __ 2 , maximum when x = 0 23 a 3 + 5(3) + 32 − 33 = 0 therefore C on curve b A is (–1, 0); B is ( 5 _ 3 , 9 13 __ 27 ) 24 1.2 2.1 0, 5Velocity (cm/s)OTime (s) 25 10 ___ 3 , 2300π _______ 27 26 dA ___ dx = 4πx − 2000 _____ x 2 dA ___ dx = 0: 4πx = 2000 _____ x 2 → x 3 = 2000 _____ 4π = 500 ____ π 27 a y = 1 – x __ 2 − πx ___ 4

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390 Answers 390 Full worked solutions are available in SolutionBank. Online b R = xy + π __ 2 ( x __ 2 ) 2 = x (1 – x __ 2 – πx ___ 4 ) + πx2 ____ 8 = x – x2 __ 2 – πx2 ____ 4 + πx2 ____ 8 = x __ 8 (8 – 4x – πx) c 2 _____ 4 + π m2 (0.280 m2) 28 a πx2 + 2πx + πx2 + 2πxh = 80π h = 40 – x – x2 __________ x V = π x2h = πx2 ( 40 – x – x2 __________ x ) = π(40 x − x2 − x3) b 10 __ 3 c d2V ____ dx2 < 0 ∴ maximum d 2300π _______ 27 e 22 2 _ 9 % 29 a Length of short sides = x ___ √ __ 2 Area = 1 __ 2 × base × height = 1 __ 2 ( x2 __ 2 ) = 1 __ 4 x2 m2 b Let l be length of EF. 1 __ 4 x2 l = 4000 ⇒ l = 16 000 _______ x2 S = 2 ( 1 __ 4 x2 ) + 2x l ____ √ __ 2 = 1 __ 2 x2 + 32 000x ________ √ __ 2 x2 = x2 __ 2 + 16 000 √ __ 2 _________ x c x = 20 √ __ 2 , S = 1200 m2 d d2S ____ dx2 > 0 Challenge a x7 + 7x6h + 21x5h2 + 35x4h3 b d ___ dx (x7) = lim h→0 (x + h)7 − x7 ____________ h = lim h→0 7x6h + 21x5h2 + 35x4h3 ______________________ h = lim h→0 (7x6 + 21x5h + 35x4h2) = 7x6 CHAPTER 13 Prior knowledge check 1 a 5 x 5 _ 2 b 2 x 3 _ 2 c x 5 _ 2 − √ __ x d x – 3 _ 2 + 4x 2 a 6x2 + 3 b x − 1 c 3x2 + 2x d − 1 ___ x 2 − 3x2 3 a O xy –1 3 –3 b Oy x –15 –5 Exercise 13A 1 a y = 1 _ 6 x6 + c b y = 2x5 + c c y = x−1 + c d y = 2x−2 + c e y = 3 _ 5 x 5 _ 3 + c f y = 8 _ 3 x 3 _ 2 + c g y = − 2 _ 7 x7 + c h y = 2 x 1 _ 2 + c i y = −10 x − 1 _ 2 + c j y = 9 _ 2 x 4 _ 3 + c k y = 3x12 + c l y = 2x−7 + c m y = −9 x 1 _ 3 + c n y = −5x + c o y = 3x2 + c p y = 10 __ 3 x 0.6 + c 2 a y = 1 _ 4 x4 − 3 x 1 _ 2 + 6x−1 + c b y = x4 + 3 x 1 _ 3 + x−1 + c c y = 4x + 4x−3 + 4 x 1 _ 2 + c d y = 3 x 5 _ 3 − 2x5 − 1 _ 2 x−2 + c e y = 4 x − 1 _ 3 − 3x + 4x2 + c f y = x5 + 2 x − 1 _ 2 + 3x−4 + c 3 a f(x) = 6 x2 − 3 x − 1 _ 2 + 5x + c b f(x) = x6 − x−6 + x – 1 _ 6 + c c f(x) = x 1 _ 2 + x − 1 _ 2 + c d f(x) = 2 x5 − 4x−2 + c e f(x) = 3 x 2 _ 3 − 6 x − 2 _ 3 + c f f(x) = 3 x3 − 2x−2 + 1 _ 2 x 1 _ 2 + c 4 y = 4 x 3 ____ 3 + 6x2 + 9x + c 5 f(x ) = −3x−1 + 4 x 3 _ 2 + x 2 ___ 2 − 4x + c Challenge y = − 12 ____ 7 x 7 _ 2 − 4 ____ 5 x 5 _ 2 + 3 ____ 2 x 2 + 1 __ x + c Exercise 13B 1 a x 4 ___ 4 + c b x 8 ___ 8 + c c −x−3 + c d 5 x 3 ____ 3 + c 2 a 1 __ 5 x5 + 1 __ 2 x4 + c b x 4 ___ 2 − x 3 ___ 3 + 5 x 2 ____ 2 + c c 2 x 5 _ 2 − x3 + c 3 a −4x−1 + 6 x 1 _ 2 + c b −6x−1 − 2 _ 5 x 5 _ 2 + c c −4 x − 1 _ 2 + x 3 ___ 3 − 2 x 1 _ 2 + c 4 a x4 + x−3 + rx + c b 1 _ 2 x 2 + 2 x 1 _ 2 − 2 x − 1 _ 2 + c c px5 ____ 5 + 2tx − 3x−1 + c 5 a t3 + t −1 + c b 2 _ 3 t3 + 6 t − 1 _ 2 + t + c c p __ 4 t4 + q2t + px3t + c 6 a 2x − 3 __ x + c b 4 _ 3 x3 + 6x2 + 9x + c c 4 _ 5 x 5 _ 2 + 2 x 3 _ 2 + c

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391 Answers 3917 a 1 _ 3 x3 + 2x − 1 __ x + c b 1 _ 2 x2 + 8 _ 3 x 3 _ 2 + 4x + c c 2 x 1 _ 2 + 4 _ 3 x 3 _ 2 + c 8 a 3 _ 5 x 5 _ 3 − 2 ___ x 2 + c b − 1 __ x 2 − 1 _ x + 3x + c c 1 __ x x4 − 1 _ 3 x3 + 3 _ 2 x2 − 3x + c d 8 _ 5 x 5 _ 2 + 8 _ 3 x 3 _ 2 + 2 x 1 _ 2 + c e 3x + 2 x 1 _ 2 + 2x3 + c f 2 _ 5 x 5 _ 2 + 3x2 + 6 x 3 _ 2 + c 9 a − A __ x − 3x + c b 2 _ 3 √ __ P x 3 _ 2 – 1 ___ x 2 + c c − p __ x + 2 qx 3 _ 2 _____ 3 + rx + c 10 − 6 __ x + 8 x 3 _ 2 ____ 3 − 3 x 2 ____ 2 + 2x + c 11 2x4 + 3x2 − 6 x 1 _ 2 + c 12 a (2 + 5 √ __ x )2 = 4 + 10 √ __ x + 10 √ __ x + 25x = 4 + 20 √ __ x + 25x b 4x + 40 x 3 _ 2 _____ 3 + 25 x 2 _____ 2 + c 13 x 6 ___ 2 − 8 x 1 _ 2 + c 14 p = −4, q = −2.5 15 a 1024 − 5120x + 11 520x2 b 1024x − 2560x2 + 3840x3 + c Exercise 13C 1 a y = x3 + x2 − 2 b y = x4 − 1 ___ x 2 + 3x + 1 c y = 2 _ 3 x 3 _ 2 + 1 __ 12 x 3 + 1 _ 3 d y = 6 √ __ x − 1 _ 2 x 2 − 4 e y = 1 _ 3 x3 + 2x2 + 4x + 2 _ 3 f y = 2 _ 5 x 5 _ 2 + 6 x 1 _ 2 + 1 2 f(x ) = 1 _ 2 x 4 + 1 __ x + 1 _ 2 3 y = 1 − 2 ___ √ __ x − 3 __ x 4 f(x ) = 3x3 + 2x2 − 3x − 2 5 y = 6 x 1 _ 2 − 4 x 5 _ 2 ____ 5 + 118 ___ 5 6 a p = 1 _ 2 , q = 1 b y = 4 x 3 _ 2 + 5 x 2 ____ 2 − 421 ___ 2 7 a f(t) = 10 t − 5 t 2 ___ 2 b 7 1 _ 2 8 a f(t) = −4.9 t2 + 35 b 23.975 m c 35 m d 2.67 seconds e e.g. the ground is flat Challenge 1 f2(x) = x 3 ___ 3 ; f4(x) = x 4 ___ 12 b x n+1 _____________________ 3 × 4 × 5 × … × (n + 1 ) 2 f2(x) = x + 1; f3(x) = 1 _ 2 x2 + x + 1; f4(x) = 1 _ 6 x 3 + 1 _ 2 x 2 + x + 1 Exercise 13D 1 a 152 1 _ 4 b 48 2 _ 5 c 5 1 _ 3 d 2 2 a 5 1 _ 4 b 10 c 11 5 _ 6 d 60 1 _ 2 3 a 16 2 _ 3 b 46 1 _ 2 c 11 __ 14 d 2 1 _ 2 4 A = −7 or 4 5 28 6 −8 + 8 √ __ 3 7 k = 25 __ 4 8 450 mChallenge k = 2 Exercise 13E 1 a 22 b 36 2 _ 3 c 48 8 __ 15 d 6 2 4 3 6 4 10 2 _ 3 5 21 1 _ 3 6 4 __ 81 7 k = 2 8 a (−1, 0) and (3, 0) b 10 2 _ 3 9 1 1 _ 3 Exercise 13F 1 a 1 1 _ 3 b 20 5 _ 6 y xO –2 y x O 4 –1 c 40 1 _ 2 d 1 1 _ 3 y x O–3 3 2y x O e 21 1 __ 12 y x O 5 2 2 a (−3, 0) and (2, 0) b 21 1 __ 12 3 a f(−3) = 0 b f(x ) = (x + 3)(−x2 + 7x − 10) c f(x ) = (x + 3)(x − 5)(2 − x) d (−3, 0), (2, 0) and (5, 0) e 143 5 _ 6 Challenge 1 a 4 1 _ 2 b 9 c 9a ___ 2 d 4 1 _ 2 e 9 ___ 2a 2 a B has x -coordinate 1 ∫ 1 0 (x3 + x2 − 2x)dx = [ 1 __ 4 x4 + 1 __ 3 x3 − x2] 1 0 = 1 __ 4 + 1 __ 3 − 1 = − 5 __ 12 So area under x-axis is 5 __ 12 Area above x-axis is ( 1 __ 4 04 + 1 __ 3 03 − 02) − ( 1 __ 4 x4 + 1 __ 3 x3 − x2) = 5 __ 12 So the x-coordinate of a satisfies 3x4 + 4x3 − 12x2 + 5 = 0 Then use the factor theorem twice to get (x − 1)2(3x2 + 10x + 5) = 0

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392 Answers 392 Full worked solutions are available in SolutionBank. Online b A has coordinates ( −5 + √ ___ 10 _________ 3 , −80 + 37 √ ___ 10 _____________ 27 ) The roots at 1 correspond to point B. The root −5 − √ ___ 10 _________ 3 gives a point on the curve to the left of −2 below the x-axis, so cannot be A. Exercise 13G 1 a A(−2, 6), B(2, 6) b 10 2 _ 3 2 a A(1, 3), B(3, 3) b 1 1 _ 3 3 6 2 _ 3 4 4.5 5 a (2, 12) b 13 1 _ 3 6 a 20 5 _ 6 b 17 1 _ 6 7 a, b Substitute into equation for y c y = x − 4 d 8 3 _ 5 8 3 3 _ 8 9 a Substitute x = 4 into both equations b 7.2 10 a 21 1 _ 3 b 2 5 _ 9 11 a (−1, 11) and (3, 7) b 21 1 _ 3 Mixed exercise 1 a 2 _ 3 x 3 − 3 _ 2 x 2 − 5x + c b 3 _ 4 x 4 _ 3 + 3 _ 2 x 2 _ 3 + c 2 1 _ 3 x 3 − 3 _ 2 x 2 + 2 _ x + 1 _ 6 3 a 2x4 − 2x3 + 5x + c b 2 x 5 _ 2 + 4 _ 3 x 3 _ 2 + c 4 4 _ 5 x 5 _ 2 − 2 _ 3 x 3 _ 2 – 6 x 1 _ 2 + c 5 x = 1 _ 3 t3 + t2 + t − 8 2 _ 3 ; x = 12 1 _ 3 6 a A = 6, B = 9 b 3 _ 5 x 5 _ 3 + 9 _ 2 x 4 _ 3 + 9x + c 7 a 9 _ 2 x − 1 _ 2 − 8 x − 3 _ 2 b 6 x 3 _ 2 + 32 x 1 _ 2 − 24x + c 8 a = 4, b = −3.5 9 25.9 m 10 a f(t) = 5 t + t2 b 7.8 seconds 11 a −1,3 b 10 2 _ 3 12 a − 2 x 3 _ 2 ____ 3 + 5x − 8 √ __ x + c b 7 __ 3 13 a (3, 0) b (1, 4) c 6 3 _ 4 14 a 3 _ 2 x − 1 _ 2 + 2 x − 3 _ 2 b 2 x 3 _ 2 − 8 x 1 _ 2 + c c A = 6, B = −2 15 a dy ___ dx = 6 x − 1 __ 2 − 3 __ 2 x 1 __ 2 = 3 __ 2 x − 1 __ 2 (4 − x) b (4,16) c 133 (3 sf ) 16 a (6,12) b 13 1 _ 3 17 a A(1, 0), B(5, 0), C (6, 5) b 10 1 _ 6 18 a q = − 2 b C (6,17) c 1 1 _ 3 19 – 9 __ x − 16 x 3 _ 2 _____ 3 + 2 x 2 − 5x + c 20 A = −6 or 1 21 a f9( x) = (2 − x2)(4 − 4x2 + x4) ___________________ x2 = 8x−2 − 12 + 6x2 − x4 b f 0( x) = −16x−3 + 12x − 4x3 c f(x) = − 8 __ x − 12x + 2 x 3 − x 5 ___ 5 − 47 ___ 5 22 a (−3, 0) and ( 1 _ 2 , 0) b 14 7 __ 24 23 a ( − 3 _ 2 , 0) and (4, 0) b 55 11 __ 24 24 a −2 and 3 b 21 1 __ 12 Challenge 10 5 __ 12 CHAPTER 14 Prior knowledge check 1 a 125 b 1 __ 3 c 32 d 49 e 1 2 a 66 b y21 c 26 d x4 3 gradient 1.5, intercept 4.1 Exercise 14A 1 a 2 O13y 4 x –1–2–3–4 2 2.634 1y = (1.7)x y = 4 b x ≈ 2.6 2 a 2 O13y 4 x–1–2–3–4 2 –1.434 1y = (0.6)x y = 2 b x ≈ −1.4 3 2 O13y 4 x–1–2–3–4 234 1y = 1x 4 a True, because a0 = 1 whenever a is positive b False, for example when a = 1 __ 2 c True, because when a is positive, ax > 0 for all values of x 5 2ba cd O1y –3x

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393 Answers 3936 k = 3, a = 2 7 a As x increases, y decreases b p = 1.2, q = 0.2 Challenge Oy y = 2x – 2 + 5 x21 4 Exercise 14B 1 a 2.7183 b 54.5982 c 0.0000 d 1.2214 2 a Oy x1y = ex e = 2.71828… e3 = 20.08553… 3 a Oy x2y = ex + 1 y = 1 b y xy = 4e–2x O4 c y x y = –3–1y = 2ex – 3 O d y x4 y = 4 – ex3 e y xy = 6 + 10e 16 61 2 Ox f y xy = 100e–x + 10110 10 O 4 a A = 1, C = 5, b is positive b A = 4, C = 0, b is negative c A = 6, C = 2, b is positive 5 A = e2, b = 3 Oy = f(x) e2y x 6 a 6e6x b − 1 _ 3 e − 1 _ 3 x c 14e2x d 2e0.4x e 3e3x + 2ex f 2e2x + ex 7 a 3e6 b 3 c 3e−1.5 8 f9(x) = 0.2e0.2x The gradient of the tangent when x = 5 is f9(5) = 0.2e1 = 0.2e. The equation of the tangent is therefore e = 0.2e × 5 + c, so c = 0. Exercise 14C 1 a £20 000 b £14 331 c t20 000V (£) OV = 20 000e–t 12

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394 Answers 394 Full worked solutions are available in SolutionBank. Online 2 a 30 000 b 38 221 c tP (thousands) 30 10094 OP = 20 + et 50 d Model predicts population of the country to be over 200 million, this is highly unlikely and by 2500 new factors are likely to affect population growth. Model not valid for predictions that far into the future. 3 a 200 b Disease will infect up to 300 people. c N t200N = 300 – 100e–0.5t300 O 4 a i 15 rabbits ii 132 rabbits b The initial number of rabbits c dR ____ dm = 2.4 e 0.2m When m = 6, dR ____ dm = 7.97 ≈ 8 d The rabbits may begin to run out of food or space 5 a 0.565 bars b dp ___ dh = –0.13 e −0.13h = − 0.13p , k = −0.13 c The atmospheric pressure decreases exponentially as the altitude increases d 12% 6 a Model 1: £15 733 Model 2: £15 723 Similar results b Model 1: £1814 Model 2: £2484 Model 2 predicts a larger value c 20k OModel 1 Model 2T t d In Model 2 the tractor will always be worth at least £1000. This could be the value of the tractor as scrap metal. Exercise 14D 1 a log4 256 = 4 b log3 1 _ 9 = –2 c log10 1 000 000 = 6 d log11 11 = 1 e log0.2 0.008 = 3 2 a 24 = 16 b 52 = 25 c 9 1 _ 2 = 3 d 5–1 = 0.2 e 105 = 100 0003 a 3 b 2 c 7 d 1 e 6 f 1 _ 2 g –1 h –2 i 10 j –2 4 a 625 b 9 c 7 d 9 e 20 f 2 5 a 2.475 b 2.173 c 3.009 d 1.099 6 a 5 = log2 32 < log2 50 < log2 64 = 6 b 5.644 7 a i 1 ii 1 iii 1 b a1 ≡ a 8 a i 0 ii 0 iii 0 b a0 ≡ 1 Exercise 14E 1 a log2 21 b log2 9 c log5 80 d log6 ( 64 __ 81 ) e log10 120 2 a log2 8 = 3 b log6 36 = 2 c log12 144 = 2 d log8 2 = 1 _ 3 e log10 10 = 1 3 a 3 loga x + 4 loga y + loga z b 5loga x – 2loga y c 2 + 2loga x d logax – 1 _ 2 loga y – loga z e 1 _ 2 + 1 _ 2 loga x 4 a 4 _ 3 b 1 __ 18 c √ ___ 30 d 2 5 a log3 (x + 1) – 2 log3 (x – 1) = 1 log3 ( x + 1 _______ (x − 1 ) 2 ) = 1 x + 1 _______ (x − 1 ) 2 = 3 x + 1 = 3(x –1)2 x + 1 = 3(x2 – 2x + 1) 3x2 – 7x + 2 = 0 b x = 2 6 a = 9, b = 4 Challenge loga x = m and loga y = n x = am and y = an x ÷ y = am ÷ an = am − n loga ( x __ y ) = m − n = loga x − loga y Exercise 14F 1 a 6.23 b 2.10 c 0.431 d 1.66 e –3.22 f 1.31 g 1.25 h –1.73 2 a 0, 2.32 b 1.26, 2.18 c 1.21 d 0.631 e 0.565, 0.712 f 0 g 2 h –1 3 a 5.92 b 3.2 4 a (0, 1) 1y y = 4x x O b 1 _ 2 , 3 _ 2 5 a 0.7565 b 7.9248 c 0.2966

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395 Answers 395Exercise 14G 1 a ln 6 b 1 _ 2 ln 11 c 3 – ln 20 d 1 _ 4 ln ( 1 _ 3 ) e 1 _ 2 ln 3 – 3 f 5 – ln 19 2 a e2 b e _ 4 c 1 _ 2 e4 – 3 _ 2 d 1 _ 6 ( e 5 _ 2 + 2) e 18 – e 1 __ 2 f 2, 5 3 a ln 2, ln 6 b 1 _ 2 ln 2, 0 c e3, e−5 d ln 4, 0 e ln 5, ln ( 1 _ 3 ) f e6, e−2 4 ln 3, 2 ln 2 5 a 1 _ 8 (e2 + 3) b 1 _ 5 (ln 3 + 40) c 1 _ 5 ln 7, 0 d e3, e−1 6 1 + ln 5 ________ 4 + ln 3 7 a The initial concentration of the drug in mg/l b 4.91 mg/l c 3 = 6 e − t __ 10 1 _ 2 = e − t __ 10 ln ( 1 _ 2 ) = – t ___ 10 t = –10 ln ( 1 _ 2 ) = 6.931… = 6 hours 56 minutes 8 a (0, 3 + ln 4) b (4 – e–3) Challenge As y = 2 is an asymptote, C = 2.Substituting (0, 5) gives 5 = A e0 + 2, so A is 3. Substituting (6, 10) gives 10 = 3e6B + 2. Rearranging this gives B = 1 _ 6 ln ( 8 _ 3 ) . Exercise 14H 1 a log S = log (4 × 7x) log S = log 4 + log 7x log S = log 4 + x log 7 b gradient log 7, intercept log 4 2 a log A = log (6x4) log A = log 6 + log x4 log A = log 6 + 4 log x b gradient 4, intercept log 6 3 a Missing values 1.52, 1.81, 1.94 b log xlog y 0.20 0.40.60.811.21.40.5 011.522.5 c a = 3.5, n = 1.4 4 a Missing values 2.63, 3.61, 4.49, 5.82 b x 2 0 4 6 8 101 0234567log y c b = 3.4, a = 105 a Missing values −0.39, 0.62, 1.54, 2.81 b O 1 –1 –2 2 31234log R log m c a = 60, b = 0.75 d 1,600 kcal per day (2 s.f.) 6 a Missing values 2.94, 1.96, 0.95 b log Rlog f 0.50 11.522.533.50.5 011.522.533.54 c A = 5800, b = −0.9 d 694 times 7 a Missing values 0.98, 1.08, 1.13, 1.26, 1.37 b P = abt log P = log (abt) log P = log a + log bt log P = log a + t log b c tlog P 50 10152025303540450.2 00.40.60.811.21.4 d a = 7.6, b = 1.0 e The rate of growth is often proportional to the size of the population 8 a log N = 0.095t + 1.6 b a = 40, b = 1.2 c The initial number of sick people d 9500 people. After 30 days people may start to recover , or the disease may stop spreading as quickly. 9 a log A = 2 log w – 0.1049 b q = 2, p = 0.7854 c Circles: p is approximately one quarter π, and the width is twice the radius, so A = π __ 4 w 2 = π __ 4 (2r) 2 = π r 2 . Challenge y = 5.8 × 0.9x

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396 Answers 396 Full worked solutions are available in SolutionBank. Online Mixed exercise 1 a Oy x1y = 2–x y = 0 b Oy x4y = 5ex – 1 y = –1 c Oy x1y = ln x x = 0 2 a 2 loga p + loga q b loga p = 4, loga q = 1 3 a 1 _ 4 p b 3 _ 4 p + 1 4 a 2.26 b 1.27 c 7.02 5 a 4x – 2x + 1 – 15 = 0 22x – 2 × 2x – 15 = 0 (2x)2 – 2 × 2x – 15 = 0 u2 – 2u – 15 = 0 b 2.32 6 x = 6 7 a –e–x b 11e11x c 30e5x 8 a e 8 + 5 ______ 2 b ln 5 ____ 4 c – 1 _ 2 ln 14 d 3 + √ ___ 13 ________ 2 e 0 f e 4 ___ 2 9 a £950 b £290 c 4.28 years d £100 e OP t100950 P = 100 + 850e–t 2 f A good model. The computer will always be worth something10 a y = ( 2 ____ ln 4 ) x b (0, 0) satisfies the equation of the line. c 2.43 11 a We cannot go backwards in time b 75°C c 5 minutes d The exponential term will always be positive, so the overall temperature will be greater than 20°C. 12 a S = aV b log S = log (aVb) log S = log a + log (V b) log S = log a + b log V blog S 1.26 1.70 2.05 2.35 2.50 log V 0.86 1.53 2.05 2.49 2.72 c log Slog V 0.50 0.00 1.00 1.50 2.00 2.50 3.000.50 0.001.001.502.002.503.00 d The gradient is approximately 1.5; a = 0.09 13 a The model concerns decay, not growth b OR t140 R = 140ekt c 70 = 140e30k 1 _ 2 = e30k ln ( 1 _ 2 ) = 30k k = 1 __ 30 ln ( 1 _ 2 ) k = – 1 __ 30 ln (2), so c = – 1 __ 30 14 a 6.3 million views b dV ___ dx = 0.4e0.4x c 9.42 × 1016 new views per day d This is too big, so the model is not valid after 100 days 15 a 4.2 b i 1.12 × 1025 dyne cm ii 3.55 × 1026 dyne cm c divide b ii by b i 16 a They exponentiated the two terms on RHS separately rather than combining them first. b x = 2 ± √ __ 5

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397 Answers 397Challenge a y = 9x = 32x, log3(y) = 2x b y2 = (9x)2 = 92x, log9(y2) = 2x c x = − 1 __ 3 or x = −2 Review exercise 3 1 −4.5 2 √ __ 7 3 a All equal to √ ____ 145 b (x − 1)2 + (y + 3)2 = 145 4 a −2i − 8j b | ⟶ AB | = | ⟶ AC | = √ ___ 85 c | ⟶ BC | = √ ___ 68 cos ∠ABC = 85 + 68 − 85 _____________ 2 × √ ___ 85 × √ ___ 68 = 1 ___ √ __ 5 5 12 6 a 5N b 7 7 m = 50 √ __ 3 + 30, n = 50 8 a √ _____________ (−75) 2 + 180 2 = 195 > 150 = √ _________ 90 2 + 120 2 b Boat A: 6.5 m/s; Boat B: 5 m/s; Both boats arrive at the same time – it is a tie. 9 lim h→0 5 (x + h) 2 − 5 x 2 _____________ h = lim h→0 5x 2 + 10xh + 5 h 2 − 5 x 2 _____________________ h = lim h→0 10xh + 5 h 2 ___________ h = lim h→0 10x + 5h = 10x 10 dy ___ dx = 12x2 + x − 1 __ 2 11 a dy ___ dx = 4 + 9 __ 2 x 1 __ 2 − 4x b Substitute x = 4 into equation for C c Gradient of tangent = −3 so gradient of normal = 1 __ 3 Substitute (4, 8) into y = 1 __ 3 x + c Rearrange y = 1 __ 3 x + 20 __ 3 d PQ = 8 √ ___ 10 12 a dy ___ dx = 8x − 5x−2, at P this is 3 b y = 3x + 5 c k = − 5 __ 3 13 a P = 2, Q = 9, R = 4 b 3 x 1 __ 2 + 9 __ 2 x − 1 __ 2 − 2 x − 3 __ 2 c When x = 1, f9(x) = 5 1 __ 2 , gradient of 2y = 11x + 3 is 5 1 __ 2 , so it is parallel with tangent 14 f9 (x) = 3x2 – 24x + 48 = 3(x − 4)2 > 0 15 a A (1,0) and B (2,0) b ( √ __ 2 , 2 √ __ 2 – 3) 16 a V = πr2h = 128π, so h = 128 ____ r 2 S = 2π rh + 2πr2 = 256π _____ r + 2πr2 b 96π cm217 a dy ___ dx = 6x + 2 x − 1 __ 2 b d2y ____ dx2 = 6 − x − 3 __ 2 c x3 + 8 __ 3 x 3 __ 2 + c 18 a 2x3 − 5x2 − 12x b x(2x + 3)(x − 4) c xy O 4 –3 2 19 6 3 _ 4 20 4 21 a −x4 + 3x2 + 4 = (−x2 + 4)(x2 + 1); x2 + 1 = 0 has no real solutions; so solutions are A(−2, 0), B(2, 0) b 19.2 22 4 1 _ 2 square units 23 a P(−1, 4), Q(2, 1) b 4.5 24 a k = −1, A(0, 2) b ln 3 25 a 425 °C b 7.49 minutes c 1.64 °C/minute d The temperature can never go below 25 °C, so cannot reach 20 °C. 26 a −0.179 b x = 15 27 a x = 1.55 b x = 4 or x = 1 __ 2 28 a logp 2 b 0.125 29 a x = 2 b x = ln 3 or x = ln 1 = 0 30 a Missing values 0.88, 1.01, 1.14 and 1.29 b tlog P 01 0 20 30 400.2 00.40.60.811.21.4 c P = abt log P = log (abt) = log a + t log b This is a linear relationship. The gradient is log b and the intercept is log a. d a = 5.9, b = 1.0 Challenge 1 a 0 b 1 2 a f9 (−3) = f9(2) = 0, so f9(x) = k(x + 3)(x − 2) = k(x2 + x − 6); there are no other factors as f(x) is cubic b 2x3 + 3x2 − 36x – 5 3 51.2 4 a f(0) = 03 – k(0) + 1 = 1; g(0) = e2(0) = 1; P(0, 1) b 1 _ 2

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398 Answers 398 Full worked solutions are available in SolutionBank. Online Practice paper 1 a 1 __ 3 b 5 √ __ 2 2 y = 2 __ 3 x + 8 __ 3 3 a error 1: = − 3 ___ √ __ x = −3 x − 1 _ 2 , not −3 x 1 _ 2 error 2: [ x5 __ 5 − 2 x 3 __ 2 + 2x] 2 1 = ( 32 ___ 5 − 2 √ __ 8 + 4) − ( 1 __ 5 − 2 + 2) not ( 1 __ 5 − 2 + 2) − ( 32 ___ 5 − 2 √ __ 8 + 4) b 5.71 (3 s.f.) 4 x = 30°, 90°, 150° 5 a 2x2(x + 3) b 2x2(x + 3) = 980 ⇒ 2x3 + 6x2 − 980 = 0 ⇒ x3 + 3x2 − 490 = 0 c For f( x) = x3 + 3x2 − 490 = 0, f(7) = 0, so x − 7 is a factor of f(x) and x = 7 is a solution. d Equation becomes (x − 7)( x2 + 10x +70) = 0 Quadratic has discriminant 102 − 4 × 1 × 70 = −180 So the quadratic has no real roots, and the equation has no more real solutions. 6 x + 15y + 106 = 0 7 a log10 P = 0.01t + 2 b 100, initial population c 1.023 d Accept answers from 195 to 200 8 1 + cos4 x − sin4 x ≡ 1 + (cos2x + sin2x)(cos2 − sin2x) ≡ (1 − sin2x) + cos2x ≡ 2cos2x 9 Magnitude = √ ___ 29 , angle = 112° (3 s.f.) 10 a 56.5° (3 s.f.) b £49.63 11 a O 27–28–1y = g(x) b −4, −1, 4 12 x = 3 or − 5 __ 3 13 a 1 − 15x + 90 x2 b 0.859 c Greater: The next term will be subtracting from this, and future positive terms will be smaller . 14 a f(x ) = ∫ ( x − 3 _ 2 − 1 − x −2 ) dx = −2 x − 1 _ 2 − x + x −1 + c = − x 2 + 2 √ __ x − 1 ____________ x + c b c = 5 __ 3 + 2 __ 3 √ __ 3 15 a 52 + 12 − 4 × 5 + 6 × 1 = 12, so (5, 1) lies on C. Centre = (2, −3), radius = 5 b y = − 3 __ 4 x + 19 __ 4 c 12 15 __ 32

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399 Index 399elimination 39–40 equations circle 117–119 normal 268–269 solving using logarithms 324–325 straight line 90, 93–95 tangent 268–269 equilater al triangles 208 exhaustion, proof by 150 expanding brack ets 4–6, 163–164 exponential functions 312–318 gradient functions of 314–316 graphs 312–313, 326 inverses 326 modelling with 317 exponents 2–3, 312 expressions expanding 4–6 factorising 6–8 simplifying 2–3 factor theorem 143–144 factorial notation 161 factorising expressions 6–8 polynomials 143–144 quadratic equa tions 19–21 force 248 for mula completing the square 22–23 distance between points 100–101 gradient of str aight line 91–92 quadratic 21 fractions , simplifying algebraic 138–139 function notation first order deriv ative 260, 289 second order deriv ative 271 functions 25–26 cubic 60–62 decreasing 270–271 domain 25 finding 90–91 gradient see gradient functions increasing 270–271 quadra tic see quadratic functions quartic 64–65 range 25 reciproca l 66–67 roots 25, 30–31 transfor ming 79–80, 194–197 fundamental theorem of calculus 296 geometric prob lems, solving with vectors 244–246 gradient curve 256–257 normal 268–269 paralle l lines 97 perpendicular lines 98–99 straight line 91–92 tangent 256, 268–269 zero 264, 273 gradient functions 260–261 of exponentia l functions 314–316 rate of change of 271–272, 274 sketching 277–278 gra phs 59–80 cosine 192–193, 195–197, 203 cubic 60–62, 68 exponential functions 312–313, 326 inequalities on 51–52 intersection points 42–43, 68–69 logarithms 326 periodic 192 quadratic functions 27–29 quartic 64–65 reciproca l 66–67 reflections 77 regions 53–54 simultaneous equations 43–45 sine 192–194, 195–196, 203 stretching 75–77 tangent 193, 195–197 translating 71–73 identities 146, 209 identity symbol 146 images 79 ‘implies that’ symbol 19 indefinite integrals 290–291 indices 2–3, 312 fractional 9–11 laws of 2–3, 9 negativ e 9–11 inequalities 46–54 on graphs 51–52 linear 46–48 quadratic 49–50 regions satisfying 53–54 inflection, point of 273–274 initial va lue 317 integration 287–304 areas between curv es and lines 302–304 areas under curves 297–298 areas under x -axis 300–301 definite 295–296 finding functions 90–91 indefinite 290–291 polynomials 289, 290 symbol 290 xn 288–289 intersection points 42–43, 68–69 intervals 270 inv erse trigonometric functions 213 irra tional numbers 12 isosceles right-angled triangles 208acute angles 206–207 algebr a 1–17 alge braic fractions, simplifying 138–139 alge braic methods 137–153 angles acute 206–207 in all quadrants 203–207 negativ e 204 obtuse 203 positive 203 reflex 211 appr oximations 167 area trapezium 303 triangle 100–101, 185–186, 303 under curve 297–298 assumptions 106 asymptotes 66 horizontal 277, 278 translation 73 vertical 277 bases 2, 319, 325 binomial estimation 167–168 binomial expansion 158–168 in ascending powers of x 164 combinations 161 factorial notation 161 general ter m 165 ignoring large pow ers of x 167–168 Pascal’ s triangle 159–160, 161 solving binomial prob lems 165–166 brackets expanding 4–6, 163–164 minus sign outside 2 CAST diagram 205 chords 123 perpendicular bisectors 123–125, 129–130 circles 113–130 equation 117–119 finding centre 129 intersections of straight lines and 121 triangles and 128–130 unit 203, 209 circumcentre 117 circumcir cle 128–129 column vectors 235–237 combinations 161 common factors 7 completed square for m 22 completing the square 22–23 solving quadratic equa tions by 23–24 conjectures 146 constant of integr ation 288, 293 coordinates 90 cos θ of any angle 207 graph 192–193, 195–197, 203 quadrants wher e positive or negative 205 quadratic equa tions in 219–221 ratio 174 value f

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or 30°, 45° and 60° 208 value f or multiples of 90° 192, 204 cos θ = k 213–215 cos (θ + α) = k 218 cos nθ = k 217 cos−1 θ 213 cosine rule 174–176, 187–189 counter-e xamples 150–151 critical v alues 49 cube roots 10 cubic functions 60–62 cubic graphs 60–62 compared with quadra tic graphs 68 curves areas between lines and 302–304 areas under 297–298 gradients 256–257 definite integra ls 295–296 delta x (δx) 259 demonstra tion 146 denominator , rationalising 13–14 deriva tive finding 259–261 first order 271 second order 271–272 straight line 264 difference of tw o squares 7 differentiation 255–280 decreasing functions 270–271 finding deriva tive 259–261 from first principles 260 functions with two or more ter ms 266–267 increasing functions 270–271 modelling with 279–280 quadratic functions 264–265 second order deriv atives 271–272 sketching gradient functions 277–278 stationary points 273–276 xn 262–263 directed line segments 231 direction of v ector 231, 239–240 discriminant 30–31, 43–44 displacement 235, 248 distance 248 between points 100–101 division law of logarithms 321–323 domain 25Answers Index

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400 Index 400tan θ = p 213–215 tan (θ + α) = p 218 tan nθ = p 217 tan−1 θ 213 tangents 43, 123–125, 256 equations 268–269 gradients 256, 268–269 theorems 146 transf ormations 71–80 to functions 79–80, 194–197 translations 71–73 tra peziums, areas 303 triangle law 231 triangles areas 100–101, 185–186, 303 circles and 128–130 equilatera l 208 isosceles right-angled 208 lengths of sides 174 –176, 179–180 right-angled 187–188 sizes of angles 174 –175, 180–181, 183–184 solving problems 187–188 trigonometric equations 213–221 trigonometric functions, inv erse 213 trigonometric gra phs 192–194 transfor ming 194 –197 trigonometric identities 209–211, 221 trigonometric ratios 173–197 signs in four quadrants 203–207 turning points 27–29, 264 unit circle 203, 209 unit vectors 236, 239 value initial 317 produced by definite integr al 295 vectors 231–250 addition 231–233 column 235–237 direction 231, 239–240 magnitude 231, 239–240, 248 modelling with 248–250 multiplication b y scalars 232 paralle l 232, 233 position 242–243 representing 235–237 solving geometric problems with 244 –246 subtraction 232 two-dimensional 236 unit 236, 239 zero 232 ve locity 248 vertices 127 y = ex 314 –316 y = mx + c 90–92 y-inter cept 91 zer o gradient 264, 273 zero v ector 232reciproca l functions 66–67 reciproca l graphs 66–67 reflections 77 refle x angles 211 regions 53–54 resultant 231 right-angled triangles 187–188 isosceles 208 roots 19 of function 25, 30–31 number of 30–31 repeated 19 scalars 232, 248 scale factors 75–77 semicircle, angle in 129 set notation 46 simplifying expressions 2–3 simultaneous equations 39–45 on graphs 43–45 linear 39–40 quadratic 41–42 sin θ of any angle 207 graph 192–194, 195–196, 203 quadrants wher e positive or negative 205 quadratic equa tions in 219–221 ratio 174 value f or 30°, 45° and 60° 208 value f or multiples of 90° 192, 204 sin θ = k 213–215 sin (θ + α) = k 218 sin nθ = k 217 sin−1 θ 213 sin2 θ + cos2 θ ≡ 1 209, 221 sine rule 179–181, 187–188 small change 259 speed 248 square r oots 10 statements 146 stationary points 273–276 straight lines 89–106 equation 90, 93–95 gradient 91–92 intersections of circles and 121 modelling with 103–106 stretches 75–77 substitution 39–40 surds 12–13 symmetry, lines of 27–29 tan θ of any angle 207 graph 193, 195–197 quadrants wher e positive or negative 205 quadratic equa tions in 219 ratio 174 value f or 30°, 45° and 60° 208 value f or multiples of 90° 193, 204 tan θ = sin θ _____ cos θ 29non-linear data, lo garithms and 328–330 norma ls 268 equations 268–269 number lines 46–48 obtuse angles 203 parabola 27 para llel lines 97–98 paralle logram law 233 Pascal’ s triangle 159–160, 161 periodic gra phs 192 perpendicular bisectors chords 123–125, 129–130 line segments 116 perpendicular lines 98–99 product of gr adients 98 ‘plus or minus’ symbol 20 point of inflection 273–274 points of intersection 42–43, 68–69 polynomials 139–144 dividing 139–141 factorising 143–144 integrating 289, 290 position vectors 242–243 positive integers 163 power la w of logarithms 321–323 powers 2–3, 312 principal va lue 213–214 products 4–6 proof 146–148 methods of 150–152 Pythagoras’ theorem 117, 187 quadratic equa tions form 19 signs of solutions 19 solving by completing the square 23–24 solving by factorisation 19–21 in trigonometric ratios 219–221 quadratic e xpressions 7 quadratic f ormula 21 quadratic functions 25–26 differentiating 264–265 graphs 27–29 modelling with 32–33 quadratic gr aphs, compared with cubic graphs 68 quadratic inequa lities 49–50 quadratic sim ultaneous equations 41–42 quartic functions 64–65 quartic graphs 64–65 quotient 140 range 25 ra te of change 105, 279 ratio theor em 244 rationa l numbers 9 rationa

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lising denominators 13–14 real n umbers 7, 232key points summaries algebr aic expressions 17 algebr aic methods 156 binomial expansion 172 circles 135–136 differentiation 286 equations 58 exponentials and lo garithms 336–337 graphs 84 inequalities 58 integration 310 quadratics 37 straight line gra phs 111–112 trigonometric identities and equations 224–225 trigonometric ratios 201 vectors 254 known facts , starting proof from 151–152 laws of indices 2–3, 9 length 100–101 limits 260 integral betw een 295–296 line segments 114 directed 231 midpoints 114 perpendicular bisectors 116 linear inequalities 46–48 local maximum 273–276 local minimum 273–276 logarithms 319–330 base 319, 325 division law 321–323 graphs of 326 multiplication la w 321–323 natura l 320, 326–327 and non-linear data 328–330 power la w 321–323 solving equations using 324 –325 to base 10 320 magnitude-direction f orm 240 magnitude of v ector 231, 239–240, 248 mathematica l models 32, 104 mathematica l proof 146–148 methods of 150–152 maximum point 27–29 midpoints, line segments 114 minimum point 27–29 modelling with differentiation 279–280 exponential 317 linear 103–106 with quadratic functions 32–33 with vectors 248–250 modulus of vector 239 multiplication la w of logarithms 321–323 natur al logarithms 320, 326–327 natura l numbers 163

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