(1)WORKED SOLUTIONS

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(1)WORKED SOLUTIONS. 4. Modeling the real world. Answers. 4. Skills check. 5. 1. lim f ( x ) 8 ∴ f (3) = 8. x o3. ∴ f (3) = 4. lim f ( x ) 4. x o 3. y. 3. 2. –4 0 –4. 2. ∑5. n=0. 4. 8. 6. x. x=3. ( ) = 1 − 1 = 10 1 2. n. 5. 2. Exercise 4A 1 2 3. x2 −1 x +1. lim. x →−1. lim x →1. x3 −1 x −1. lim f x →2 +. 5 6. a. discontinuous at x = ±1. b. discontinuous at x = ±2. c. continuous. d. discontinuous at x = −4 and x = 1. e. discontinuous at x = 1. f. continuous. Exercise 4C. = −2. 1. =3. b. § lim ¨ x o2 ©. x2 x 2 ·. c. § lim ¨ x o2 ©. x 6 64 ·. does not exist. 7. x 3. 3. x 8. x. x o0. lim ( x 6) x o6. x o0. 2 3. lim [ x ] 2. x o 3. x. d. lim [ x ] 3. x →1. ∴ lim f ( x ) x o1. ∴ lim f ( x ) x o1. f. lim. x o1. x o1. lim. x o2. 2. x x. x 1 x. x o0. 1. 1 x. 1. 1. h. lim x oa. = lim x →1. 1 =2 x. 1− x 2−x. =0. 6x. 4 + 12 x + 9 x 2 − 4 − 8 x − 4 x 2 6x 5x 2 4 x. x o0. lim f ( x ) 1. 1+. 1 x (2 3 x ) 2 4 (1 x ) 2. = lim. x o1. ( ). = lim x →1. 1. = lim x →0. ∴ f is continuous at x = −2. does not exist. x2 1. x o0. lim f ( x ) 1 ∴ lim f ( x ) 1. Also, f (−2) = 1 lim f ( x ) 1. lim. g. ∴ f is not continuous at x = 1. x o1. e. does not exist. x o2.

(2) = lim

(3) ,. which does not exist. x →1. x o2. § ( x 3 8) ( x 3 8) · ¨ ¸ ¸ = xlim o2 x3 8 ¹ © ¹ 3 = lim ( x + 8) = 0. = lim x o0. lim+ f ( x ) = 2. lim f ( x ) 1.

(4). 2 ( x 1) ( x 1) · lim §¨ x2 1 ·¸ = lim §¨ ¸ x o0 © x o0 © x x ¹ x ( x 1) ¹. x o 3. lim− f ( x ) = 0. ( x 2) ( x 1). x →−2. x. 0. ¸ = xlim o2 ( x 2) ¹ ( x − 1) = − 3 = xlim →−2. x 2. |x| 1 ∴ lim | x | does not exist lim | x | 1 lim x o0. Exercise 4B. 3. x o4. x → 2−. x o3. 2. x 3. lim. ∴ lim [ x ] does not exist.. 1.

(5). a. lim f ( x ) = 5. ( x ) = 31. ⎧⎪3 x − 1 x < 2 lim ∴ x →2 ⎨ 1 x ≥2 ⎪⎩ x 2 − 1 4. 3. a=1. 4. –8 ∞. 4. ∴ 9a − a = 4 8a = 4. 8 (0, –1 ). ∴k=. ∴6=8. 2. x a x a. ∴ f is not continuous at x = 1 © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. 6x 2. = lim x o0. = lim x oa. 5x 4. 2. 6. 3. (x a) (x a) x a. = lim (x + a) = 2a x →a. Worked solutions: Chapter 4. 1.

(6) WORKED SOLUTIONS 2. a. lim. b. lim. c. lim. d. 3. 2x x +2. x →∞. 3x 2. x →∞. x of. lim x of. e. lim. f. lim. x of. b. =2. x2 − 1. 2x 2 x 1. 2 3. 3x 2 5 x 1. gradient = lim h →0. 0. 4x 3 2. c. x 1. y = x3. 4x 3 2 1 x. 2. b. y=1. d. y = −1. e. no horizontal asymptote. y=0. c. 2. d. y = −x 2. converges. b. converges. d. diverges. e. converges. ∞. ∑. n =1. c. (−1). n =0. 2. converges, ∑. ( ) S 3.14. 1. =. 1 1+ 2. diverges since ∞. converges, ∑ 5 ∞. converges, ∑. n =1. e. n. n. n =1. d. n. () 1 3. 3 10. n. =. n. ∞. 2 −3. n =1. 7. converges, ∑. =. 1−. 2. f ′(x). lim. (x 2xh h x ) h. 1 3. =. h o0. ∞. =∑. n. n =1. = −2x f ′(1) = −2(1) = −2 e. !1. y=. x x +1. f ′( x ) = lim h→0. 5 2. () 2 7. = lim h→0 = lim h→0. n. ∞. −∑. n =1. () 3 7. =. n. a. f. h. ( x + h + 1) ( x + 1) h 2. y. 1. (x = 2). x2. 1. lim. 2. ( x h) h. h o0. b 4. h o0. −1 < 2x < 1 2x must be positive ∴ 0 < 2x < 1 ∴ x < 0 1− 2. 1 . = 40. x. 3x x 1. ∴ 1 2. ∴2. x. = 1 8. ∴ x = −3. 1 ∴ −0.25 < x < 0.5. f ′(2) = 2. 1. a. = lim h →0. −2x − h x 2 (x + h)s. ,. x3. =−. 1 4. -2 x3. =2. ⇒ x = − 1 ⇒ x = −1 ∴ point is (−1, 1). y = 2x − 1. (x = 1). 2. 2. [2(1 h) 1] [2(1) 1] (1 h) 1 4h 2h h. 2. hx 2 (x h). 2. 2. 3. 2. 'y 'x. h o0. x 2 (x h). 2. f ′(x) = - x 3. f ′(x) = 2 ⇒. Exercise 4E. 2. 3. x. lim. 2. 2. 4. −2. 1. hx 2 (x h). 2x x. . x 2 x 2 2xh h 2. lim. 7 8. 1. ( x + h + 1) ( x + 1). 1. r = 2x. x. = lim h→0. ( x + 1). f ′(x). ∞. 35. x +h x − x + h +1 x +1 h ( x + h ) ( x + 1) − x ( x + h + 1) ( x + h + 1) ( x + 1) h. f ′(0) = 1. 2 3 7 − 7 = 2 − 3 = −7 5 4 5 4 20 7 7. converges, ∑ 4(−0.6)n −1 = 4 = 2.5 1 + 0 .6. u1 = 35. 2. h o0. 3 10 = 1 1 3 1− 10 n. 2. lim (2x h). n =1. 3. 2. lim h o0. converges. c. 2 3. 3.14. =. f. =. S. 5 3. n. 3 3h h 2. (x = 1) (x h) (x ) h 2. a. b. 1 3h 3h h 3 1 h. h o0. y=3. ∞. 1 2. ∴ gradient = 3. a. a. 1 −2 + h. 2. 3. (1 h) 1 (1 h) 1. Exercise 4D 1. ( )= −. 2−2+h 1 = h(−2 + h) −2 + h. 2 gradient = lim (3 3h h ) 3. =4. x. =. (x = 1) 3. 'y 'x. 0. x 2 3x 5. x of. (x = −2). 2 2 2 +1 Δy −2 + h − −2 −2 + h = Δx (− 2 + h) − (−2) h. =3. 5x 2. 2 x. y. 2. 3 2. (1 4h 2h ) 1 h. f ′(x) = 4x −. f ′(x) = 3 ⇒ 4x − ⇒ 4x. 3. 4 2h. gradient = lim (4 + 2h) = 4 h →0. 1 x2. ⇒ (x ⇒x. 1 x2. =3. − 1 = 3x ⇒ 4x3 − 3x2 − 1 = 0 2. − 1) (4x2 + x + 1) = 0. = 1 ∴ point is (1, 3). © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 2.

(7) WORKED SOLUTIONS. Exercise 4F 1. 2. Exercise 4H. a. y = x 2 + 2x + 1 f ′(x) = 2x + 2. b. y = x3 − 1. c. y. 2 x. f ′(x) = 3x 2 f ′(1) = 3 2. f ′(x) =. d. y = x −1. e. y=. x. f ′(x) = −. ∴ f ′(2) =. 1 2 x 1. f ′(x) =. x +3 ,. −2 9. ∴ f ′(3) =. 2. f ′(x) =. 1 , x. 1. f ′(0) = 2. 1 , 2 x +3. 1. f. y=. a. Average velocity =. 3. 2 x. y = 4 − x − 3x 2. b. y = 2x 4 − 3x + 1. c. y = 4x 3 −. 1. dy dx. 2. f ′(1) =. , f ′(4) =. a. 1 4. 1 − 16. dy dx. 2. = −10a − 5h b. 1. y=9−x b. When x = −1, gradient = 8,. y. 3 dy dx. =2 1. 1. Normal is y − 8 = − 2 (x + 1) i.e. y = − 2 x + dy dx. 1 (x 1). ∴. 2. 1 (x 1). 2. a. y = 4 − 3x − 3x 2. dy dx. d. y=. dy dx. 1 1. =. y = x 2 − 3x. x2 x2. dy dx. 4. y. y= x x. 1 x. At (1, 2). dy dx. =8. dy dx dy dx. =. ≠0. 3 2. 1. 2 x 2 x. = 1−. 3x 2. 1 3x. 1. 3 x2 dy dx. 3 ∴ gradient of normal =. Equation of normal: y − 4 =. 1. ( ). y = (2x + 3)5. b. y = 2 − 3x = (2 − 3x) 2. Equation of tangent is y = 2 ∴ normal is x = 1. = 5(2x + 3)4 (2) = 10(2x + 3)4. 1. 1 (2 3x) 2 (3) 2. 2 2 3x. c. y=. 2 x. dy dx. d. y=. − 3x + 5x3, so −3 2. dy dx. = −3 ( 5x 2 + 1). 3 (5x 2 2. =. 1 2. 2. = − x 2 − 3 + 15x2. −1 2. 3. 15x. 1) 2 (10x). (5x 2 1). -. dy dx. 3. 1 2. 1. (3x 2 - 2 x ) 2 (6x − 2) =. At x = 1,. ∴ no points. 3. y = 3 x 2 - 2 x = (3x 2 - 2x ) dy dx. ( ). = 1−1 = 0. (x + 1) 3. dy dx. 5x + 1. 3 −9 , 2 4. x2. 1 3. 1 3. 1. 2. 1. dy dx. a. ∴ no points. ≠0. . . y = − 1 x + 11. −1 19 , 2 4. = 2x − 3. 1. dy dx. (0, 1). ∴x=. 2x − 3 = 0 e. 15 2. = 3x 2. ∴x=0. 3x 2 = 0 c. dy dx. y = x3 + 1. y. x 3 x. = −3 − 6x 1. 1 x. dy dx. At (−1, 4). y − 1 = −1(x − 2) y = −x + 3. −3 −6x = 0 ∴ x = x − 2 b. − 3 + 15x 2. = 12x − 4. Exercise 4I. At (0, −1) y + 1 = −1(x − 0) y = −x − 1 3. 2 x2. 3. (0, −1), (2, 1). At (2, 1). 3. 4. = −1. x−1=±1. ∴ (x − 1)2 = 1 x = 0 or 2. 3x. y = 8x − 6. When x = −1, gradient = 2. 1 x 1. 4. 4x . = −2x. ∴ tangent is y − 8 = 2(x + 1) i.e. y = 2x + 10. 2. 3 x4. Equation of tangent: y − 2 = 8(x − 1) dy dx. 2. a. c. 4 −3 x 3. = −2x −2 − 3 + 15x 2 = −. At (1, −2). Exercise 4G. = 4x 3 − x −3 + 2x 2 + 32 x −2. 2 3x 2. y = 2(3x 2 − 2x) = 6x 2 − 4x dy dx. velocity = lim (average velocity) = −10a hl 0. = 8x 2 − 3. y = 2 − 3x x + 5x = 2x−1 − 3x + 5x 3. d. 2. dy dx. + 2x 2 +. 2. 12 - 5(a + h ) -12 + 5a 2 h. = − 1 − 6x. = 12x 2 + 3x−4 + 4x − 12x 2 . x (a + h ) - x (a ) h. =. 1 x3. dy dx. 3 x -1 3x 2 - 2 x. = 2 and y = 1, so tangent is. y − 1 = 2(x − 1) i.e. y = 2x − 1 3. x -3 x dy = 32 dx x. y=. =1−. At (1, −2),. dy dx. 3 x. 1. = 3 so gradient of normal = − 3 1. ∴ equation of normal is y + 2 = − 3 (x − 1) 5 1 i.e. y = − 3 x − 3. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 3.

(8) WORKED SOLUTIONS 4. 1. y. 3x 6x 1. dy dx. 2. 6. −2. = −(3x − 6x + 1) (6x − 6). 6x − 6 = 0 ∴ x = 1 5. 2. = (3x 2 − 6x + 1)−1. 2. 1 § 2· ¨1 x ¸ © ¹. y. 1 x. dy dx. =. ⎞ 1⎛ 1− x 2 ⎟ 2⎜ ⎝ ⎠. dy dx. =. 1. −1 2. y. 1 3 6 1. dy dx. (6x 6). . 2. (3x 6x 1). y = (x − 1)4 (3x - 2) 3 ,. 2. 1− x x. 1. a. (3 x - 2). 2( x -1) (7 x - 5) 1. (3 x - 2)3. 2. x 7. y. x3. u(x) = x2 − 7. x −x. x (2x) (x 7)3 x. dy dx. x. v(x) = (x + 3)3. v′(x) = 3(x + 3)2. x. =. = (x − 1)3 (x + 3)2 + (x + 3)3 (1). b. 21 − x. = 4x (x + 3)2. y. 2. x 1. dy dx. u(x) = (2x − 3). u′(x) = 4(2x − 3). v(x) = (4x + 1)3. v′(x) = 12(4x + 1)2. c. = 8(2x − 3) (4x + 1)2 (5x − 4). u′(x) = 1. v(x) = (x − 1)−1. v′(x) = −(x − 1)−2. dy dx. = −(x + 1) (x − 1)−2 + (x − 1)−1. dy dx. = (x − 1) [−(x + 1) + (x − 1)] =. d. −2 (x − 1). 2. −1. = (1 2x) 5. 1 2. + (1 2x). dy dx. =. -1. (x. 4. - 3 x + 1). 2. 1. =. 1−2x. 3- 4x3. (x. 4. 1. u′(x) = x. x. 1 2. 2. 1 2. 1. v′(x) = − 2 x. 1. 1. 1. 1. 1 2. 1 2.

(9). −1 2. ( ). 1 2 2. =. 1 2 2. 1. x (1 − x ). 2. 1. e. (4x3 − 3) =. 3. ( x 2 + 1) 2. = (x 4 − 3x + 1)−1. 1 x. 1 − 3x. y = (x4 − 3x + 1)−1 ⇒. 1. (1 x 2 ) 2 x 2 (1 x 2 ) 2 x. 1− x. [−x + (1 − 2x)] =. + 1). x x. 1 2. 1 2. 2. (x 4 − 3x + 1) 2. 1 1. dy dx. v(x) = (1 2x) 2 v′(x) = 12 (1 − 2x) 2 (−2) = −(1 2x).

(10) ÷ (x. 3 − 4 x3. v(x) = 1 − x. u′(x) = 1. = −x (1 2x). =. 1. y = x 1 2x 1. 3x 1. 3 2. u(x) = 1 + x 2. −2. 1 2. y. 2. + 1) − x. 1 2. 1 2. = −(x 4 − 3x + 1)−2 (4x 3 − 3). y=. = (x + 1) (x − 1)−1. u(x) = x. x4. dy dx. u(x) = x + 1. 2. 1. y. v′(x) = (x 2 1). 1. ( x 2 + 1). = 4(2x − 3) (4x + 1)2 (10x − 8). dy dx. = (x. = 4(2x − 3) (4x + 1)2 [3(2x − 3) + (4x + 1)]. 4. u′(x) = 1 1 2. = ( x 2 1) 2 x 2 ( x 2 1). = (2x − 3)2 12(4x + 1)2 + (4x + 1)3 4(2x − 3). x 1 x 1. 4. v(x) = (x 2 1). y = (2x − 3)2 (4x + 1)3. y. x x. 4. 2. u(x) = x. = (x + 3)2 (4x). 3. 6. 2x 3x 21. u′(x) = 1. 2. 2. 2. 2. u(x) = x − 1. dy dx. v′(x) = 3x 2 2. 3. = (x + 3)2 (3(x − 1) + (x + 3)). 2. u′(x) = 2x. v(x) = x 3. y = (x − 1) (x + 3)3. dy dx. 2. Exercise 4K. 1. Exercise 4J 1. 2. 3. −1 4. 1. + 4(x − 1)3 (3x - 2) 3. 1 3. =. 1 2. ⎛ 1 − 12 ⎞ ⎜− 2 x ⎟ = − 4 ⎝ ⎠. -. 4. 1 2. 2. 2( x -1). =. 1,

(11). 1. 2. = (x −1)4 3 (3x - 2) 3 × 3 + 4(x − 1)3 (3x - 2) 3. - 3 x + 1). 2. y = ( x − x) 2 dy dx. −1 −1 ⎛ ⎞ = 1 ( x − x ) 2 ⎜ 1 x 2 − 1⎟ 2 2 ⎝ ⎠. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 4.

(12) WORKED SOLUTIONS 1 4. x x

(13). . x. 4. 1 2. 2. x x. y = ⎛⎜ x ⎝1−. f.

(14). 1 2. x. 4 xx x. x. 3. 2. dy dx. 3x. 2. 3. v′(x) = 3(1 − x ). 3. 3x. 3. u′(x) = 3x 2.

(15). 1 x. 2. (1 − x ). u(x) = x 3 v(x). x x

(16). 5. 3. 1 x

(17) 32 x 2 1 x

(18) 6 1 x

(19). 2. (. −1 x 2. −1 2. ). 2. 6x . 4. x.

(20). f ′(0) = −2. f ″(−1) = 12. f (x) = x − 4x + 16x − 16 4. f (x) = x4 + rx2 + sx + t. f ″(−1) = 16 ⇒ 12 + 2r = 16. 5. (x = −1) u′(x) = 4. v(x) = x 2 + 1. v′(x) = 2x. 2. (x 1). 2. 4 4x. = (t − 4)2 (3 − 2t) (25 − 10t). 2. 2. 2. (x 1). 8. y dy dx. 2. 4x. s ′(4) = 0 ms−1. 2. b. = 8(4 + x 2)−1. ( ) 5. dy dx. 16 25. 4x. 8. 4. f (x). 3. 1

(21). f ′(x) = 2 (1 − (2 + x). = (t − 4) [−80t + 40t 2 + 320 − 160t + 150 − 160t + 40t 2].

(22). 2. = (t − 4) (80t 2 − 400t + 470). 25 (x 16. − 1), y. acceleration = s″ (4) = 0 ms−2. 25 16 25 16. x. c 3 80. 1 (2 x)

(23) 1 −1 − 3. 2 2 3 (2 + x) 3 1− 1 2+ x. ). (2 + x)−2. s″(t) = 80t 3 − 720t 2 + 2070t − 1880 jerk = s′ ″(t) = 240t 2 − 1440t + 2070 s′ ″(1) = 240 − 1440 + 2070 = 870 ms−1. 2 1 3. 3. =. 2. gradient of normal =. 2. s ″(t) = (t − 4)2 (−80 + 40t) + 2(t − 4) (70 − 80t + 20t 2). (x = 1). Equation of normal: y − 5 = 1 2x. s′(t) = (t − 4)2 (75 − 80t + 20t 2). 16x. = −8(4 + x 2)−2 (2x). At 1, 8 ,. Velocity = s′(t) = (t − 4)3 2(3 − 2t) (−2) s ′(t) = (t − 4)2 (3 − 2t)[−4(t − 4) + 3(3 − 2t)]. At x = −1, gradient = 0 3. (3). = 3(t − 4)2 (3 − 2t)2. u(x) = 4x 4(x 1) 8x. (2). s (t) = (t − 4)3 (3 − 2t)2 a. 2. (1). Solve equations (1), (2), (3) to find r = 2, s = –8, t = 5.. 2. 2. 3. f ″(x) = 12x2 + 2r. 4. x 1. dy dx. f ″(x) = 12x 2. f ′(−1) = −16 ⇒ s − 2r = −12. 4. 4x. y. f ′(x) = 4x 3 − 2. f ′(x) = 4x3 + 2rx + s. (2 − x ) 4 2 (1 − x ). 3x. =. f (x) = x − 2x − 1. f (−1) = 16 ⇒ r − s + t = 15. 5 3x 2. 2 1. 2.

(24). x3. 4. f (−2) ≠ f ′(−2) ≠ f ″(−2), but f (2) = f ′(2) = f ″(2) = 0 so x = 2. 5. x. 2. f (x) = (x + 2) (x − 2)3 ⇒ x = −2 or x = 2. 2. 6x 6x 2 3x 2 2 1. f ″(x) = 2x −3 =. f ″(x) = 12x2 − 24x. 5. 5. 1 x. f ′(x) = 4x3 − 12x2 + 16. 1 x

(25) 32 x 2 4 1 x

(26). 2. f (x) = 4x + 1 + f ′(x) = 4 − x −2. 1 2 x. 3. =. 1. 1 2 x 4 x. ⎞ ⎟ ⎠. Exercise 4L. § 12 · ¨ x 2¸ © ¹. 1 2. 6. f (x) =. 1 x. = x −1. f ′(x) = −x −2. f ″(x) = 2x −3. f (4) (x) = 24x−5. f (5) (x) = −120x−6. f (n) (x) =. f ′ ″(x) = −6x −4. n. (1) n !. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. x n 1. Worked solutions: Chapter 4. 5.

(27) WORKED SOLUTIONS. Exercise 4M 1. a. b. y = x − 3x + 1 dy dx. = 2x − 3 3 2. 2x − 3 = 0 ∴ x x. 3 2. x<. dy dx. −. y=. ( ) − 3( ) + 1 = − 3 2. 2. 3 2. 5 4. 2. a. 3 2. x> +. ∴minimum value = b. 3. ( when x = ). −5 4. a. 3 2. x = −1, 2 , 2. iii. ] − ∞, −1 [. i. x=. iii. ] − ∞, − 12 [ = − 6x + 6 −6x + 6 = 0 ⇒ x = 1 (1, 2). x. x<1. x=1. x>1. dy dx. +. 0. −. = − 6x 2 + 12x or 2. (0, −3). maximum at (1, 2). (2, 5). f is increasing for −∞, 1[. x<0 0<x<2 x>2. dy dx. −. +. f is decreasing for 1, ∞[. −. ∴ minimum value = −3. (at x = 0). maximum value = 5. (at x = 2). b. y dy dx. 12x 3 − 6x 2 − 6x = 0 6x (2x 2 − x − 1) = 0. x. 1 2. x<. dy dx. or 1 1 2. ( 1 , 2. (0, 4) 1 2. −. 1 2. (2 − x 2). 2x. 59 ) 16. 0<x<1. x>1. −. +. +. ? minimum values are. 59 at 16. =. 1 2 2.

(28). 2 2x 2. 2. 2 (at x =1) maximum value = 4 (at x = 0). 2 x2. = 0 ⇒ x = ± 1 (1, 1). (−1, −1). 2x. 2. x. x < −1 x = −1 −1 < x < 1 x = 1 x > 1. dy dx. x = 1 and.

(29) 1.

(30). 2dxd 2. (−2x) + (2 − x 2) 2. 2 x2. (1, 2) 2 2x 2. <x<0. 1 2. 2x. 1 2 2. x 2 (2 x 2). 6x (2x + 1) (x − 1) = 0. −. 0. +. 0. increasing for ]−1, 1[. dy dx. decreasing for ⎡⎣ − 2, − 1 [ ∪ ] 1, 2 ⎤⎦. = 4x 3 − 12x 2. x = 0 or 3. 4x 2 (x − 3) = 0 (0, 0). (3, −27). x. x<0. 0<x<3. x>3. dy dx. −. −. +. ∴ horizontal point of inflexion at (0, 0) minimum value = −27 (at x = 3). x = −1 or 1 iii ]−1, 1[. c. y= dy dx. 2. x. (x 1) x (2x) 2. 2. (x 1). x 1. =0. 1 x 2. x. x < −1. x = −1. dy dx. −. 0. 2. (x 1). 2. 1, 12

(31) 1, 21

(32). x=±1. when. (. 2. −1 < x < 1 x = 1 x > 1 +. ). 0. −. ( ). minimum at −1, − 12 , maximum at 1, 12. Exercise 4N i. −. minimum at (−1, −1), maximum at (1, 1). y = x 4 − 4x 3. 4x 3 − 12x 2 = 0. a. =x. x 2. = 12x 3 − 6x 2 − 6x. x = 0,. . x 2 x2. y = 3x 4 − 2x 3 − 3x 2 + 4 dy dx. ] 32 , ∞ [. ] − 12 , ∞ [. ii. dy dx. x. 1. 1 2. ∪. ] 12 , 32 [. ∪. y = −2x3 + 6x 2 − 3). x=0. d. −. ] − 1, 12 [. ii. y = −3x 2 + 6x − 1 dy dx. −6x 2 + 12x = 0 ∴ 6x(−x + 2) = 0. c. 1 3. i. 2. ii. ]−∞, −1 [ ∪ ] 1, ∞[. increasing for ]−1, 1[ decreasing for ]−∞, −1 [ ∪ ] 1, ∞[. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 6.

(33) WORKED SOLUTIONS. dy dx. =. 4 3. 1 3. 2 3. x −. 2 3. 1 3. 4 3. x ,. x. x<. dy dx. −. 1 2. 2 3. x. dy dx. =0. 0. 2. d y dx. 1 2. x>. 2x. 2. dx. =x. 1 2. 2. 1 2 2. y=. d.

(34). (−2x) + 2x(2 − x ). ). 4x 3x. 3. 2x. 2. 2

(35) 2 3 2. x<− 3. ⎛ (0, 0), ⎜ ⎝. 2. 4 3 3. 2. −. ⎞ ⎛ ⎟, ⎜− ⎠ ⎝. 2 3. ,. x=− 3. +. = 20x 3. 2 3. 0. 2. 4 3 3. 2 3. ,. ⎞ ⎟ ⎠. −. = 12(x 2 + 2x − 3)−1. = −12(x 2 + 2x − 3)−2 (2x + 2). dy dx. = 0 ⇒ x = −1 x. x < −1. x = −1. x > −1. dy dx. +. 0. −. 24 (x 1). x. 2. 2x 3.

(36). 2. maximum at (−1, −3) 3x 3 x (3 x). y. e. x=0. <x<0. 12 x 2 2x 3. dy dx. = 0 ⇒ 4x − 3x 3 = 0, x (4 − 3x 2) = 0. x = 0, ±. 2. f ″(1) > 0 ∴ minimum at (1, −4). 1 2 2. (2 − x ) 2. 2dxd 2. 1 2 x 2. dy dx. x = ±1. f ″(−1) < 0 ∴ maximum at (−1, 4). 2. 2. x 3 2x (2 x. x. 5x 4 − 5 = 0, x4 = 1,. 2. x. dy dx. = −12x 2 < 0 ∴ maximum at (0.794, 0.191). = 5x 4 − 5. d y. 2. dy dx. 2. dy dx. +. increasing for ] 1 , ∞ [ , decreasing for ] − ∞, 1 [ y. 1 32. x. y = x 5 − 5x. c. ⎞ ⎛ minimum at ⎜ 1 , 34 ⎟ ⎜2 3 ⎟ ⎝ 2 ⎠. e. 1 2. −4x 3 + 2 = 0 ∴ x3 =. = −4x 3 + 2. = 0.794. § · ¨ 1 , 3 1 ¸ ¨ 2 2(2) 3 ¸ © ¹. 1 2. 1 2. x=. 2 3. x −. ∴ 4x − 2 = 0 ∴ x =. y = −x 4 + 2x − 1. b. 1. 4. 1. y = x 3 (x − 2) = x 3 − 2 x 3. d. 3x 3 3x x. 2. (3x x 2)3 (3x 3) (3 2x). dy dx. 0. 3x x. 2.

(37). 2. 9x 3x 2 9x 6x 2 9 6x. 0<x<. x dy dx. 2 3. 2 3. x=. +. x>. 0. ⎛ maxima at ⎜ −2 , 4 ⎝ 3 3 minimum at (0, 0), increasing for decreasing for. @ @. 2 3. 2 3. 3x 2 6x 9. −. ⎞ ⎛ ⎟ and ⎜ ⎠ ⎝. 2. x (3 x). 2. 4 3 3. 2 3. ,. ⎞ ⎟, ⎠. 1. a. dy dx. 2, . 2 3. >. . @ 0,. 2 3. 2 ,0 3. >. . @. 2 , 3. 2. dx. 2. b. = 6x + 6x − 12. 2. = 12x + 6. f ″(−2) = −18 < 0 ∴ f has a maximum at (−2, 17) f ″(1) = 18 > 0 ∴ f has a minimum at (1, −10). 3(x 2 2x 3). 2. 2. x (3 x). 3(x 3) (x 1). 2. 2. x (3 x). 2. x = −3 or 1. +. 0. −. x>1. 0. +. 3. y. 1 x x2 8. i. dy dx 2. 2. 2. 2. maximum at −3, 1 , minimum at (1, 3). 2. d y.

(38). ( ). @. 6(x 2 + x − 2) = 0 6(x + 2) (x − 1) = 0 x = −2 or 1. 2. x < −3 x = −3 −3 < x < 1 x = 1. dy dx. y = 2x + 3x − 12x − 3 dy dx. =0 ⇒. x. Exercise 4O 3. 3x x. dy dx. = (−x. 2. + 8) − 2x (1 − x) x 2 − 2x − 8 = 2 2 (x 2 + 8) 2 (x + 8). = 0 ⇒ (x − 4) (x + 2) = 0 x = −2 or 4. x. x < −2. x = −2. −2 < x < 4. x=4. x>4. dy dx. +. 0. −. 0. +. (. ). (. maximum at −2, 41 minimum at 4, −81. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. ). Worked solutions: Chapter 4. 7.

(39) WORKED SOLUTIONS ii. f is increasing for ]−∞, −2 [ ∪ ] 4, ∞[ f is decreasing for ]−2, 4[. iii. y 1 ) 4 1 4. (–2,. –4. y. x 1

(40) 3. i. dy dx. dx. 2. 6 x. 4. ii. 2. dx. d y. x. 9(x. e. x=0 x=0. x>0. −. 0. +. 2. x. 3x x 1. i. dy dx. (x 1). 2. dx. = (x − 1). 2. dx. = 4x − 3 2. = 6x. dx. = 12x. 2. 2. d y dx. x>0. +. 0. +. ii. 2. 1 2. d y dx. 2. (x − 1). =. 4x x

(41) −. (. 4x − x. ). 1 2 2. 2x. (4 2x). − (2 − x). (. (. 4x x. 1 4x − x 2. 4x − x. 2. ). ). −1 2 2. 1 2 2. 1. 2. − 12x + 6 − 6x + 12x (x − 1). 3. 3. x<1. x>1. −. +.

(42). ( 4 − 2x ). a. s (0) = 10 m. b. −5t 2 + 5t + 10 = 0 t2 − t − 2 = 0 (t − 2) (t + 1) = 0.

(43). 4x x. ∴ t = 2 seconds c. 3 2 2.

(44). 4x x 2. 4x. 4x x. 2. <0.

(45). 2. s (t) = −5t 2 + 5t + 10. v (t) = −10t + 5. a(t) = −10. v (2) = −15 ms. a(2) = −10 ms–2. –1. 4x x 2 4 4x x 2. 2. 3. Exercise 4Q. 4x x 2 (2 x) 2. dx. 4. concave up for ]1, ∞[, concave down for ]−∞, 1[. ]−∞, 0 [ ∪ ] 0, ∞[. 1 2 2. . d y. x. dx. 4x x 2 where x ∈ [0, 4]. 2. 2. ( 6 x − 6 ) − ( 3 x − 6 x) 2 (x − 1 ). 2. concave up for. dy dx. 2. ≠ 0 ∴ no points of inflexion. 2. d y. no point of inflexion. i. 2. (x 1). 2. x=0. 2. dx. (x 1). 2. x<0. d y. y. 2. (x 1) (6x 6) (3x 6x)2. ⇒ x=0. =0. 2. 3 x 6x. 6. 2. x. ii. 2. 2. (x 1). d y. 3. 2. c. 0. 6x (x 1) 3x. d y. 2. d y. inflexion. −. y = x 4 − 3x + 2 i. ≠0. x>1. concave up for ]0, ∞[. dy dx. 3. 2. y. concave down for ]−∞, 0[ b. 2. x<1 2. point of inflexion at (0, 0) ii. dx. 4. 9. concave down for ]− ∞, 1[∪]1, ∞[. = 6x. 2. x<0. d y dx. dx. =0 ⇒. 2. 2. dx. = 3x 2 − 1. d y. 2. d y. 4 1) 3. 2 x 1

(46). 2. 2. 2. dy dx. dx. 2. d y. y = x3 − x i. 2. d y. 1. x 1

(47) 3. ∴ no points of. 1 (4, – ) 8. Exercise 4P a. 2 3. 2. –1 8. 1. 2. d y. (1, 0). 0. –2. concave down for [0, 4]. 1 ) 8. (0,. 1 8. –6. d. ii. The diver is moving downwards and speeding up when he hits the water.. 3 2. 3 2 2.

(48). ∴ no points of inflexion. 2. s = 50t − 15t 2 a. v = 50 − 30t = 0 when t maximum height = 5. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. 5 3. ( ) = 41 m 5 3. 2 3. Worked solutions: Chapter 4. 8.

(49) WORKED SOLUTIONS b. 20 = 50t − 15t 2. d. t = 2s and t = 4s.. 3t 2 − 10t + 4 = 0. e. 0 − 2s distance = s (2) − s (0) = 6 32 − 0 = 6 32. t = 0.4648 or 2.8685. 1. v = 50 − 30t v (0.4648) = 36.1 ms. 1. v (2.8685) = −36.1 ms–1. a = −30 ms –2. d. 50t − 15t 2 = 0. s = 7t + 5t − 2t a. 10 s 3. 3. v = 7 + 10t − 6t 2. a = 10 − 12t. v (0) = 7 ms–1. a(0) = −10 ms–2. v (2) = 3 ms–1. c(x) = 20 000 + 180 x − 0.1x2 a. c′(x) = 180 − 0.2 x. b. c′(100) = 180 − 0.2 × 100 = 160 euros / tank. Initially the particle is moving in a positive direction and is slowing down. b. a(2) = −14 ms–2. 2. c. c(101) − c(100) = 159.9 ⇒ cost of producing 1 extra tank is nearly the same as the marginal cost function.. a. i. The particle is moving in a positive direction and slowing down. 4. s = 10t 2 − t 3 a s (3) = 63m ∴ average velocity = b. v = 20t − 3t 2. a = 20 − 6t. v (3) = 33 ms–1. a(3) = 2 ms–2. c. Speeding up.. d. 20t − 3t 2 = 0. 63 3. 5. s (t) =. 1 3 t 3. 20 3. = 21 ms–1 b. − 3t + 8t. i. a (t) = 2t − 6. 3. t 2 − 6t + 8 = 0 t = 2 s or 4 s. c. t v t a. 0<t<2 + 0<t<3 −. r(x) = x(7 − 0.002 x). Break-even points: r(x) = c(x) when 7x = 0.002 x2 = 500 +3x i.e. 0.002 x2 − 4x + 500 = 0 4 o 16 - 4 0.004. Average cost per 100 units = To minimize cost, need. (t − 2) (t − 4) = 0 ii. iii. For profit, need to make x memory sticks where 134 < x < 1870.. 2. b. c ′(x) = 3 euros / unit ⇒ it will always cost 3 euros to make an extra memory stick. = 134 or 1870 (3 sf).. 3. v (t) = t 2 − 6t + 8. ii. ⇒x=. direction changes when t = 20 s. a. p(x) must be > 0 so 0.002 x < 7 i.e. x < 3500 ∴ domain is 0 < x < 3500. t (20 − 3t) = 0 t = 0 or. 1. 3. 1. ∴ rock hits the ground again when t = 3. 1. total distance = 6 32 + 1 3 + 1 3 = 9 1 m. Exercise 4R. 5t (10 − 3t) = 0 2. 1. 4 − 5s distance = s (5) − s (4) = 6 32 − 5 3 = 1 3. speed = 36.1 ms–1 upwards (when t = 0.4648) and downwards (when t = 2.8685) c. 1. 2 − 4s distance = s (4) − s (2) = 5 3 − 6 32 = −1 3. –1. ⇒ 500 − 2<t<4 − t>3 +. t>4 +. v and a have the same sign for 2 < t < 3 and t > 4 ∴ the particle is speeding up at these times. iii the particle is slowing down for 0 < t < 2 and 3 < t < 4 a (2) = −2 ms–1 a (4) = 2 ms–2 the particle changes direction from positive to negative when t = 2s and from negative to positive when t = 4s.. 1000 x2. d dx. c (x ) x. æç c ( x )ö÷ ÷ çç çè x ø÷÷. = 500 +. 1000 x. =0. =0. ⇒ x2 = 2 ⇒ x = 1.41 (3 sf) ∴ costs are minimised by making 141 units. 4. b. r(x) = 35x − 3 and p(x) = r(x) − c(x) so p(x) = 35x − 3 − 400 − 20x + 0.2x2 − 0.0004x3 = 15x − 403 + 0.2x2 − 0.0004x3 p′(x) = 15 + 0.4x − 0.0012x2 = 0 ⇒ x = 367, so 367 jackets must be made to maximise profit.. c. Minimising costs will not necessarily maximise profits.. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 9.

(50) WORKED SOLUTIONS. Exercise 4S. x 2 + y 2 − 3x + 4y = 2. c. dy dx 3 − 2x dy = dx 2y + 4. 2x + 2y. y. 1. (x, 10 – x2). 4x − 3(x 2 2y dy + 2xy 2) + 2y dy = 0. A = 2x (10 − x 2) = 20x − 2x 3. 4x − 6x 2y dy − 6xy 2 + 2y dy = 0 dx. dy 3xy − 2x = 2 dx y − 3x y. 10 3. A″ = −12x. 2. (x + y)2 = 5 − 2x. e. A″ < 0. 3. x. dy dx. 800 – 2x. A = x (800 − 2x) = 800x − 2x 2 A′ = 800 − 4x A′ = 0 ⇒ x = 200 A″ = −4 < 0 ∴ maximum maximum area = 200 × 400 = 80 000 m 2 3. =. =. 1 xy. 1 xy. xy xy. x2. f. dx. dy dx. 1+. 3. x. −1 = − (1 + x + y) x+y. x 3 + x 2y = x − y. dx. dy 2 (x dx dy dx. 2. + 1) = 1 – 3x2 – 2xy. = 1 − 3x2. 2. − 2xy. x +1. x2 − y2 = 9. (5, 4). dx. y−4=. y. 2. =4. 3. 2x + 4y + πx = 8 8 2x S x 4 A = xy 1 (8x − 2x 2 − 4 1 A′ (8 − 4x − 2πx) 4. 4. 8 = 0.778 y = 1 4 2S A″ 1 (−4 − 2π) < 0 ∴ maximum 4 8 m by 1m dimensions: 4 2S. 3y + x = 4. b. + 2x = 0. dy dx. = 3x 2 ?. − 10 3. x 2 − 3 xy + 2y 2 = 5.

(51). ∴. dy −x = dx 3y. dy dx. x. 2. 4y. 3. ( 3, 2). dx. dx. − 3 dy + 4y dy = 3 y − 2x dx. dy dx. 5. y4 = x3 + 1 4y 3. (x − 1). dx. 3 y 2x 4y 3 x. 0. Tangent: y = 2. 2. dy dx. 4 3. 3 4. 3 2y. =. 2x − 3 x dy y + 4y dy = 0. Exercise 4T 6y. (1, −2) dy dx. x. a. 4 x 3. 5 4. 4. y 2 = 3x + 1. y=. x y. =. −9. y+2=. πx 2). dy dx. (x − 5). dx. A = 0 ⇒ x (4 + 2π) 8. 2. 5 4. 2y dy = 3. y. 1. 5 x 4. y=. x. Sx. dy dx. 3x 2 + x 2 dy + 2xy = 1 −. 2x − 2y dy = 0 ∴. x + 2y +.

(52). 2(x + y) 1 dy = −2. ∴ max.. height = 10 − 10 = 20 (3.65 by 6.67). 10 3. base = 2. dx. 2. 20 − 6x 2 = 0 ⇒ x = 10 , 3. dx. dx. A′ = 20 − 6x 2. If x =. =0. 2x 2 − 3x 2y 2 + y 2 = 9. d. x. 0. dy dx. −3+4. Normal: x = 3. x + y − 6x − 8y = 0 2. 2. 2x + 2y dy − 6 − 8 dy = 0 dy dx. dx 3x y4. dx. dy dx. =0 ⇒. x=3. 9 + y 2 − 18 − 8y = 0. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 10.

(53) WORKED SOLUTIONS y 2 − 8y − 9 = 0 (y − 9) (y + 1) = 0 y = 9 or −1 stationary points (3, 9), (3, −1) 6. 3x 2 + 2xy + y 2 = 3 6x +. 2x dy dx. dy dx. (3x y) xy. (2x 2y 1). 2. dx 2. 3+. d y dx. dy dx. +. 2. +. dx. 7. 2. =1. 1. A = πr 2. 2. A = 2πr 2 + 2πrh. dy dx. +y. 2. d y dx. 2. +.

(54) dy dx. 2. =0. 2. dx. +1=0. 2. 3. =6. x2 = 3. 2x + x. dy dx. dy dx. (2x y) x 2y. =. . +y+. =0 4. dy dx. At ( 3 , 0). 2y dy dx. = −2 At (− 3 , 0). dy dx. b. dy dx. = 2x − 2 x dy y + 2y dy. 1+. dy dx. = 2x − 2x. dy dx. + 2x. dy dx. =. 1−. dy dx. dy dx. dx. = 2 cm s−1. b. dt. dt. dx. (l 2 + w 2 + h ) dw dt. = −2 cm s−1 l = 12 cm, w = 5 cm. = l dw + w dl. dt. dt. = 2 dl + 2 dw dt. dt. = 2(2) + 2(−2) = 0 cms−1 Let x = diagonal of the rectangle. x2 = l2 + w2. (2x − 2y − 1) (2x − 2y + 1). 2x dx = 2l dt. dl dt. + 2w. dw dt. (l = 12, w = 5, x = 13). 13 dx = 12(2) + 5(−2) dt. dx 14 = cms −1 13 dt. 2 2x − 2y + 1 dy dx.

(55) . (2x 2y 1) 2 2. (2x 2y 1). 2. dy dx.

(56). 5. dv dt. = 1.5 m3s−1 v = 81 m3 x = 3 81 m. Let x = side length of cube V = x3. dy dx 2 (2x 2y 1) 44. dA dt. (2x − 2y − 1) 2x − 2y + 1. =. A = 6x 2. dv dt. = 1.5 × =. 2. 4(2x 2y 1) 4(2x 2y 1) (2x 2y 1). dt. 1 2 2. dp dt. 2x − 2y + 1 − 2x + 2y + 1 2x − 2y + 1. (2x − 2y + 1). dh dt. = 12(−2) + 5(2) = −14 cm2 s−1 p = 2l + 2w. c. 2. 4−4. + 2h. A = lw. dx. d y. =. dx. dt. l dl + w dw + h dh. dl dt. − 2y + 2y dy. dy dx. (2x 2y 1) 2 2. 2. dx. dx. dr dt. + 2πr. dw dt. + 2w. =. − 2y dy = 2x − 2y − 1. =1−. 1 − dy =. dl dt. dx dt. dA dt. 2x − 2y − 1 2x − 2y + 1. =. c.

(57). 1+. dh dt. + 2πr. Let x = diagonal of the box. x2 = l2 + w2 + h2. x + y = x 2 − 2xy + y 2 a. dr dt. = 2S r. dt. ∴ tangents are parallel 9. (from b). = (4S r 2S h) dr 2S r dh. dr dt. ). 3. 3. dA dt. a. = −2. ). = 4πr. dt. x=± 3. 2 2x − 2y + 1. dA dt. 2. y=0. dA dt. 2x dx = 2l. x + xy + y = 3 2. dx. Exercise 4U. 2. d y. (. = 1 − dy. (3 2) 1. d y. +2−2. 2. 2. =0. dx. d y. 3+x. dx. (1, −2). 3x + x dy + y + y dy = 0 dx. 2. d y. ∴. (. =. 3. dy dx. + 2y + 2y =−. 8. 3. dA dt. ×. dx dv. 1 3x 2. × 12x. ×. dA dx. 6 x. =. 6 3. 81. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. = 1.39 m2 s−1. Worked solutions: Chapter 4. 11.

(58) WORKED SOLUTIONS y. 6. 2. f (2) = 8. 5m. 0. When x = 3 m,. dx dt. −1. = 0.5 ms. 2x dx + 2y dy = 0 dt. dt. dy dt. dy dt. 4. ∑ 3⎜⎜ ( −1n). 1 2. n =0. x dy +. =. 1 2. =. 1 2. dt. 1 2. 5. y dx dt. (3)(−0.375) +. 1 1 2. 1 a. (4)0.5. sum =. 6. dA dt. (r = 5). 2. dr dt. = 2πr dr dt dr = 10 π dt. dr1 dt. 1 5S. a 1−. dr2 dt. = 1.2 ms−1. ( ). 3. 2. =. 1. a. y=−1. = 1.5 ms−1. 2S r1. dr. 1. dt. 3 x x. 1. a. lim. b. lim. x →1. 7. =−1. x. lim 3. d. lim 3 x. e. f. x o0. 1 x 2. x →0. lim x →∞. lim. x →−∞. +x. x. 5x. 2. 2. 3. 2x + 1 7 3. x +1. 2x 1. (0, 1). 2. x 1 2. dy. 2( x 1) 2 x (2 x 1). dx. ( x 1). 2. dy dx. 2. =2 y = 2x + 1. 1. 1.10 (3sf ) 2. y. ( − 5 , −1). Tangent: y − 1 = 2x. does not exist since the domain. =4. 9. x + y = −3 y dy dx. 2. gradient = −1. x x 1. = x 1 ( x + 1) 2. −1 2. 1. + ( x + 1) 2. x 2( x 1). =0 =0. y = − 1 x +1. Normal: y − 1 = − 2 x. is ]−∞, −1] ∪ [1, ∞[ c. r 5. If x = 0,. 2. x −1 x. x →0. x2. ( 5 , −1). Review exercise x3 − 3 x +1. 1+ a −1. =1 + a 2. 1. 2. x. 2. dt. = 2π (9 × 1.2 − 1 × 1.5). ✗. 2. x 3 2x 2 5 x 2 x 3. dr. = 18.6π = 58.4 m2s−1. 2. 2. x 2x 5. 5. 2S r2. 2. a (1 + a ). 2 3 x x. A = S r12 S r22 dA dt. 2. 2. x 2x 5. b. 1 1 a. 2. 1+ a. y. 3. = 0.0637 cms−1. = 2 .5. 3. 1− − 1 5. < 1 provided a ≠ 0. 2. = 2 cm s 2. =. Geometric series with, r =. 2 −1. A = πr. 10. n⎞ ⎟ ⎟ ⎝ 5 ⎠. ⎛. ∞. = 0.4375 m s 7. is a geometric series with r = − 1 5 hence it converges.. ∑ 3⎜⎜ ( −1n). 2 −1. dA dt. n⎞ ⎟ ⎟ ⎝ 5 ⎠. ⎛. ∞. =0. dA dt. xy. sequence converges since an → 0 as n → ∞.. 3 n 2. n =0. = −0.375. A=. 2. an. ∴ 0.375 ms−1 down the wall b. lim f ( x ) 4. x o 2. 2n 3. 3. x 2 + y 2 = 25. 3(0.5) + 4. x d2 x !2. f (x) is not continuous at x = 2. x. 3m. a. x 2 2x ® 3 ¯ x 6x. y. 1. 2( x 1) 2. =. 3x 2 1. 2( x 1) 2. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 12.

(59) WORKED SOLUTIONS 1. 3x 2 2( x 1). 1 2. 3 x 2 2( x 1) 2. = −1. c. y. 2. x 3. x ≠ −1. x 1 1. (3x + 2)2 = 4(x + 1). dy. 9x 2 + 12x + 4 = 4x + 4. ( x 1) 2 2 x . dx. 1. If x = 0,. at 8 , . 10. dy dx. 9. =. 1 2. 8 27. 1 2. 7. 9. ,. = 3x. = −1. x. x x 1. dy dx. 1§. 2. 2. 1 x ( x 1). 14 1 7. x 37. x = 3.0782 y = −2.2031. 2 x 1.

(60)

(61) 2. y+ 12 a. 1 8. 8 3.

(62). 3. . 1 8. ª 2 «( x ¬. 0,

(63) 1. 8. e. 1. 1 2. 1. § 2 ¨ x (x ©. 1). 1 2. ·2 ¸ ¹. 1 2. 2. º2 x » ¼.

(64). 1 2. 1. 2. 2. x 1.

(65). = (1 − 3x)7 3(3x + 5)2 3 + (3x + 5)3. 6. y = (x + 2 + (x − 3)8)3 dy dx. 13. 2. = 3(1 − 3x) (3x + 5) (−30x − 32) = −6(1 − 3x)6 (3x + 5)2 (15x + 16). = 3(x + 2 + (x − 3)8)2 (1 + 8 (x − 3)7). f (x) = ax 3 + 6x 2 − bx f ′(x) = 3ax 2 + 12x − b f ″(x) = 6ax + 12 f ″(1) = 0 ∴ 6a + 12 = 0 ∴ a = −2 f ′(−1) = 0 ∴ −6 − 12 − b = 0 ∴ b = −18. = 3(1 − 3x)6 (3x + 5)2 [3(1 − 3x) − 7 (3x + 5)] 14. y. x 33 x. a. (0, 0). 1. x 3x 3 1. y = ( 4 x 2 − 3 x + 1)5 dy. ·2 ¸ ¹. 1. = 9(1 − 3x)7 (3x + 5)2 − 21 (1 − 3x)6 (3x + 5)3. dx. 1 2. ( x 1) 2 x. 1 2. 7(1 − 3x)6 (−3). b. 1). 2 x 1. y = (1 − 3x)7 (3x + 5)3 dy dx.

(66). 1). 8. 2. = 2x or y = 2 x −. · 2x¸ ¹. 1. (3.08, −2.20) 12. 2. 2. f c(0) 3 1. 1 2. 1 2. 2. f c( x ) 3 > g ( x )@ g c( x ) 2. 2. ( x 2 1). ( x 1) 2 x. 14. 14x 4 − 35x 3 − 35x 2 + 19x +37 = 0. ? f (0). 1. 1. § 2 2 ¨ x (x ©. 37. > g ( x )@. 1. 1 2. · 2§ x ( x 1) ¸ ¨ 1 ¨ 2© ¹ ©. ( x − 1) 7. 3. 1 2 § 2 2 · ¨ x ( x 1) ¸ © ¹. 2. 1. (2x 4 − 5x 3 − 5x 2 +3). f (x ). 3. y. 14x 4 − 35x 3 − 35x 2 + 21x = 2x − 37. 11. + 4x + 3. 1. (8x 3 − 15x 2 − 10x +3) = −7 =. 2. 2 ( x + 1) 2.

(67) 2. y. . 9. dy dx. 3 2. 2( x 1). 8. d. 5 normal: y + 1. If x =. =1. 8. 2. ( x 3). 2. x x 1 is parallel to the line x + y = −3. ? y. dy dx. . 2. 4 x ( x 1) ( x 3 ). x (9x + 8) = 0 x = 0 or. 1. 2 ( x 1). 2. 9x + 8x = 0. ( x 1). x 3 2. = 5 ( 4 x 2 − 3 x + 1) (8 x − 3) 2. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. 2. x 3 ( x 3 3) 0 2. 0 or x 3. 3. x = ± 27 ( 27 , 0 ) ( − 27 , 0 ). Worked solutions: Chapter 4. 13.

(68) WORKED SOLUTIONS b. (1 x 3 ). dx. 1 x. 2 3. d y. 2. =. 2. 2. c. dx. d. 2. d y. d y dx. 2. 2. 2. ( x 3). −1 < x < 1. If x = 1, f ″(x) < 0 ∴ maximum at (1, −2). x=1 x>1. −. 0. c. f ″(x) = 0 x. +. b. f (−x). 2 x. 2 x. 2. ( x ) 1. i. increasing for ]−∞, − 3 [ ∪ ] − 3 , 1[ ∪ ]3, ∞[. ii. decreasing for ]1, 3 [ ∪ ] 3 , 3[. e. x 1. 8. ∴ function is odd. ∴. 4. ( x 1)2 2 x (2 x ). dx. ( x − 1). 2. 2. <0. x = –√3. 17. 2 4 x. 2. –2. f (x ) a b. ( x 3). dy dx. dy. At (0, 1). dx. dy dx. ∴. 1 4. 5y 1. = −1 =1 4. x=1. Review exercise. –4. 16. = 5y 4 − 1. At (0, 0). 0. –2. x = y 5 − y 0 = y (y 4 − 1) dx dy. 4. –4. x = √3. –8. y = 0, ±1 (0, 0) (0, 1) (0, −1). y x = –1. 8 x. 0 (3, 0) 4 –4 (0, –3) (1, –2) –4. –8. <0. d. (4.20, 0.0979). y=1. 2. ( x 1). x > 4.1958 −. y. = −f (x). 2. x = 4.1958. ∴ point of inflexion at (4.20, 0.0979) d. x = 1, x = −1, y = 0. ⇒. x < 4.1958 +. f ″(x). 2x. a. 3. If x = 3, f ″(x) > 0 ∴ minimum at (3, 0). 2. 2. 2. < 0 ∴ maximum at (−1, 2). function decreases for ]−1, 1[. −2 x − 2. 3. 12 x 72 x 108 x 72. ii. 2. 2. 4. 2. ( x 3). function increases for ]−∞, −1 [ ∪ ] 1, ∞[. dy. 2. 2. ∴ minimum at (1, −2). 0. 2. 2. ( x 3)(12 x 24) 4 x (6 x 24 x 18). i. =. 2. 3. 2. +. dy dx. 2. ( x − 3). 2. x 1. c. − 24 x + 18. 2. 5. x x < −1 x = −1. y. x = 3 or 1. ( x 3). x. >0. 2. 2. ( x 3) (12 x 24) 2( x 3)2 x (6 x 24 x 18). f ″(x). ≠ 0 ∴ no points of inflexion. dy dx. 15. 3. 2. If x = −1,. 2. 1 x=±1. 2. = 3. dx. ? x. ⇒. f ′(x) = 6 x. 2 3. x3. −5 3. x. 3. If x = 1,. d y. 1. =0 ∴1=. 2. dx. f ′(x) = 0. 2. dy. 1. 2. (1.5, 0). (x,. Let l = distance. x). l 2 = (x − 1.5)2 + x. 2. x 3. (0, − 3). (3, 0) x = ± 3. y=1. f ′( x ) =. ( x − 3) 2( x − 3) − 2 x ( x − 3). 2. 2. 2. ( x − 3). l. 2. ( x 3). 2( x 3)(3 x 3) 2. ( x 3). 2. 1. ( x 2 2 x 2.25) 2. 2. 2( x 3)[ x 3 x ( x 3)] 2. = x 2 − 2x + 2.25. dl. 1. dx. 2. 1. ( x 2 2 x 2.25) 2 (2 x 2) x 1. 2 2. 2. ( x 3). 2. 1. ( x 2 x 2.25) 2. 6( x 3)( x 1) dl dx. =0⇒x=1. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 14.

(69) WORKED SOLUTIONS x. x<1. x=1. x>1. dy dx. −. 0. +. 4 y. ∴ minimum distance when x = 1. 2. x 180. xy = 180. Let r = radius, x = side of square. printing area, A = (x − 2)(y − 3). 4πr + 4x = 80. A = xy − 3x − 2y + 6. x = 20 − πr A = 2πr + x 2. x. = 2πr + 400 − 40πr + π r 2. dA dr dA dr. 2 2. d A. r dr dt. = 0 ⇒ 3x 2 = 360. x d A dx. 2. 720 x. 3. dt. x. 2. y = 16.43. 0 ∴ maximum. ∴ dimensions are 11.0 cm by 16.4 cm. = 4π + 2π2 > 0 ∴ minimum. 5. dx dt. 1 1 2x. = (1 + 2x)−1. 20. acceleration. 2S dh dt. = 3 cm min−1. 2. d x dt. = −4 cm min−1. 2. = −(1 + 2x)−2 2 dx dt. 2 (1 + 2 x ). v = πr 2h dv. 360. 120 = 10.95 2. 20. 2. dA dx. = 0 ⇒ 4πr + 2π2r = 40π. 2S. dr. = −3 +. x 2 = 120. 2r + πr = 20. 2. dA dx. = 4πr − 40π + 2π2r. r. x. 180 3 x 360 6. 2. = 2πr 2+ (20 − πr)2. 3. y. minimum distance = 1.25. S r 2 dh 2S rh dr dt. dt. 3. At x = 2, acceleration =. −2 125. = π(81) (−4) + 2π (9)(12)(3) = 324π cm3 min−1 increasing at a rate of 324π cm3 m−1. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 4. 15.

(70) WORKED SOLUTIONS. 7. The evolution of calculus. Answers. Exercise 7A. Skills check 1. a. y = x lnx. b. y= dy dx. dy dx. x. ⎛ 1 ⎞ 1 ⎜ (2 − x ) 2 2e2 x − 3 +e2 x − 3 1 (2 − x )− 2 ⎟ 2 ⎝ ⎠ (2 − x ) 4 (2 − x ) e. 2 x −3. 2( 2 − x ). =. dy dx. 2. a. 2. 3x8 dx =. 3. − 5x4 dx = − x5 + c. 2 x −3. =. y=. − 2x dx = − x2 + c 9. + lnx = 1 + lnx. e 2 −x. =. c.

(71) 1 x. 1. e. 2 x −3. +e. 2 x −3. x. x. +c. dx = x−5 dx =. 5. 3. x 3 dx = x 2 dx. 5. 3 2. 1. 6. (9 − 4 x ). 2( 2 − x ) 4 x − 1. 1. 4. x 3. x. 1. dx =. 3. x. 3 2. 2 5. −4. 4x. +c. 4. 5. x2 c. dx = x. 3 2. −1. +c=. −3 2. 1. 2 x. dx = 2 x 2 c. c. 3. 2x x. 7. 4. = 4x3 + 4x−5 = 4 x 3 +. 1 2. 2x 3 2. dx = 2x dx =. 4 x. 4. − x. 5. y = 3x − 2 y = x − 2x + 4 2. 2. x − 2x + 4 = 3x − 2 x2 − 5x + 6 = 0 (x − 2)(x − 3) = 0 x = 2 or 3 (2, 4) or (3, 7). 8. 7x. b. y=1−x. c. (1 − x)2 = 2x + 1 1 − 2x + x2 = 2x + 1 x2 − 4x = 0 x(x − 4) = 0 x = 0 or 4 Check in (1) if x = 0, LHS = 1 RHS = 1 ᅚ if x = 4, LHS = −3 RHS = 3 ᅜ ∴ x = 0 (0, 1) y = 6 + 3x y = x3 − 5x. y 2x 1 1 − x = 2 x + 1 (1). x. +c = 4 x2 +c 3. 1 dx = − x 4 dx = − 1 x 7. 3. 4 x 21. 3 4. 1. a. ( )+c −4 3. c. (5x − x ) dx. ⎛ 5x 2 − 1 ⎞ dx = ⎜ 2 ⎟ 5x ⎠ ⎝. 1 5. 2. 3. = 5x + 3. b. −3 4. 7. Exercise 7B. 1 5x. −2. +c. (x + 3)(2x − 1)dx = (2x2 + 5x − 3)dx 3. 2. = 2 x + 5 x − 3x + c 3. 2. c. x −1 x. −3. d. + 3x = x − 5x. 2.

(72) dx = x. 2. dx = (x−2 − x−4)dx. 4. x 3. = −x −1 +. 3. 6 + 3x2 = x4 − 5x2 x4 − 8x2 − 6 = 0 x = −2.948 or 2.948 (−2.95, −10.9) or (2.95, 10.9) 3 s(t) = 3t4 − t3 + t v(t) = s′(t) = 12t3 − 3t2 + 1 a(t) = v ′(t) = 36t2 − 6t. 3. 2. −7. 5. 2. 6 x. −x 4. 1 x. +c=. e. ( x + 3)( x − 4 ). © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. x. 5. 1 x. +. 1 3x. 3. +c. (x2 + 2 + x−2)dx 3. =. . x 3. 1 x. + 2x −. +c. dx = (x−3 − x−4 − 12x−5)dx =. −x 2. =−. −2. −3. −4. + x + 12 x + c 3. 1 2x. 2. +. 1 3x. 3. +. 4 3. x. 4. +c. Worked solutions: Chapter 7. 1.

(73) WORKED SOLUTIONS ⎛ x− ⎜ ⎝. f. 1 1 ⎞ ⎟ dx = ( x 2 5x 3 ) dx ⎠. 5 3. x. =. 2 3. 2. dy dx. = (3x2 − 4). 3. 2 3. 2. x 2 − 5x 3. x. 3 2. 15 2.

(74) 3 2. 4. (1, −21 ). 1 2. t. 2. + 3t +. 1 2. =. 1 2. +3+1+c ∴c=−5. f (t) =. t 2. 2. + 3t +. 1 t. 1 t. +c. 3. 3. −2 (2 x + 1) 2. −2 (2 x + 1) 2. 2. (). ( 4 x − 1). =. +c =. 3 2. 1. 3. −5. dx = (4x − 1)−5 dx. 5. ( 4 x − 1) 4 ( −4 ) 2 4. −4. +c =. y = 2x4 + 12x3 + 27x2 + 27x + 12 =. = (2 x 3)4 15 8. = (2x + 1)(x − 1) = 2x + x − 2x − 1 2. 0= A=. x 2. 3. −1 16 ( 4 x − 1). dx = 2(3 x ). 3−x. 2. +c. 3. . 2 (3 − x ). +c. 4. 1 4. dx 3. 3. z = 2 − 12 + 27 − 27 + c ∴ c = 12. 1 2. +c. − 2 2 x 1 dx = − 2(2 x 1) 2 dx. y = 2x4 + 12x3 + 27x2 + 27x + c. 4. 8. 2. 4. x 2. (3 x − 1) 24. (3x − 1)7 dx =. = (2x + 3)3 = 8x3 + 3(2x)2(3) + 3(2x)32 + 33. A=. + 2t + 1. 1. =. t 2. dA dx. 2. t 2. 1 = c ∴ s(t) = t3 +. = 8x3 + 36x2 + 54x + 27. 5. + 2t + c. Exercise 7C. f (t) =. dy dx. 2. t 2. 1. y = x3 − 4x − 1. −. s(t) = t3 +. x c. −1 = 8 − 8 + c ∴ c = −1 f ′(t) = t + 3 −. 2 = c ∴ v(t) = 3t2 + t + 2. 2 3. (2, −1). a(t) = 6t + 1 v(t) = 3t2 + t + c. c. y = x3 − 4x + c. 3. 8. 4. +c =. −3 4. −8 (3 − x ). 4. +c. 3. 1 1 § · § 2 3 1 x · dx = ¨ 2(2 5x ) 3 (1 x ) 3 ¸ dx 1 ¨¨ ¸ © ¹ 3 ¸ © (2 5x ) ¹. 5. 3. + x − x2 − x + c. 2. 3. −1−1+c ∴c=. 1 3. +. 7 6. =. 2 (2 − 5 x ) −5 2 3. =. −3 (2 − 5 x ). 3. (). 4. +. (1 − x ) −4 3. 3. +c. 2. 6. ds dt. 4. 3. x 3. +. = 3t −. −x −x+ 2. 7 6. 8 t. 4. −. 3 (1 − x ) 4. 3. +c. 2 § 3 · ¨ 4 2 3x 6(3x 2) ¸ dx © ¹. 6. 2. 5. 3. 2. s = 3t + 8 + c. 2 1 § · = ¨ 4(2 3x ) 2 6(3x 2) 3 ¸ dx © ¹. t. 2. 1.5 = 1.5 + 8 + c. ∴ c = −8. s = 3t + 8 − 8 t. 2. 2. 7. d y dx. 2. = 6x − 1. dy dx. = 3x − x + c. = 3x − x − 6 2. y = x3 −. =. −8 (2 − 3 x ). 2. −. (). 6 (3 x + 2) 3 5 3. ∴c=−6. 9. 2. − 6x + c. +c 5. −. 6 (3 x + 2). 3. 5. +c. Exercise 7D 1. − 5e −2x dx 1. 2 e. 3x +2. 5e 2. c. dx = e −3x −2 dx. 2. x 2. 3. (). 2 x. 2. x 2. 0 = 8 − 2 − 12 + c ∴ c = 6 y = x3 −. =. 4 (2 − 3 x ) −3 3 2. 3. 2. 4 = 12 − 2 + c dy dx. 5. 3. 2. − 6x + 6. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. =. 1 −3x −2 e 3. +c. Worked solutions: Chapter 7. 2.

(75) WORKED SOLUTIONS 3. ⎛3 x ⎜ e − e ⎝. 2 e. x. 4. x. 3 dx =. 3. 6 7. 2x. 3. 4. =. 3 ln 2. (16 − 4) =. +c. 41−x dx = − 4. 1− x. −2 −2 x. m. . 4. =a. 1 a. dx =. x. 1. ª¬1 7 6 º¼. u. m ln (m ). a. 1 a ln (m ). Exercise 7G. +c. 0. max+b + c. 1. Exercise 7E 1 3x. 2. − dx = −6ln|x| + c. 1 3. dx = ln|x| + c. 6 x. 0. ⎡ (2r − 1) 5 ⎤ (2r − 1) dx = ⎢ −1 ⎣ 10 ⎥⎦ −1 4. dx = − 1 ln|2 − 3x| + c. 4. 5 3 5x. dx = −ln|3 − 5x| + c. 5. −2(4 + 3x) −1 dx =. 3. Not possible, x ≠ ± 1, 1 ∈ [0, 2] 0 (2 x. + 1). (2x + 1) −3dx. =. 3. 0 1. 1

(76). 1. 1 4 9. 5 6 3. 2 ⎛ 3x + 1 ⎞ dx = ⎡ 3x − 1 ⎤ = 27 − 1 − 3 + 1 ⎜ ⎢ 2 2 ⎟ x⎥ 1 ⎝ x ⎠ ⎣ ⎦1 2 3 2. 3 3 x 4 0. . 2 x 1.

(77) dx = [ln|3x + 4| − 2ln|x + 1|]. 1. = ln7 − ln4 − ln4. 3. ln. = 38 2. 1. e +4. −1. e. x. 7. 2. 3 ⎤ ⎡ ( 4 x + 1) 2 ⎥ 3 4 x 1dx = 3 = ⎢ 0 ⎢⎣ 4 32 ⎥⎦ 0. x. dx =. (). −2e1−3x dx = 2 ⎡⎣e1−3 x ⎤⎦. −1. 3. =. 2 3. −1. = 2 − 4 + 4e e. 2. 10xdx. 8 0. 2. 1 ln 10. 2. ª¬10 x º¼ 0. 1 (100 1) ln 10. −1 9. (e −5 − e4) =. 1. = (1 − 4e ) − (−1 − 4e). 3 ª 2º (4 x 1) «¬ »¼ 0. 2. 2. (1 + 4e −x)dx −1. −1. = 1 (27 − 1) = 13 3. 7 16. 1. = ⎡⎣ x − 4e − x ⎤⎦. 2. 1 2. 0. = (ln7 − 2 ln2) − (ln4 − 2 ln1). = 12 − 1 + 1 3. 2 9. Not possible, x ≠ − 1 1. Exercise 7F. 2. 121 5. 1 ⎡ ⎤ = − 1 ⎡⎣(2 x + 1)−2 ⎤⎦ = − 1 ⎢ 1 2 ⎥ 0 4 4 (2 x + 1) ⎣ ⎦0. ln|4 + 3x| + c. 3. 242 10. 1. dx. 3. 2 3. (( −1) − ( −3)5) =. Not possible, s ≠ 0 1. 1 2 3x. 1 10. 2. 4. 3. . 1)dx. = (4 − 4) − (2 − 1) = −1. =. 1. 1 2. 4. mu du. =1. 1. (x. ª 12 º «¬2x x »¼ 1. du a. ∴ dx =. du a. 4. x. 1. 6. dx Let u = ax + b. max+b dx = mu. 1 9. ª¬ (1 3x )6 º¼ 2. = 13072. +c. ln 4. 0. 2 3 (6). 2(1 − 3x)5dx. +c dx = 32 x dx = − 2 ln 3. ax+b. 36 ln 2. 0. du dx. 1. 1. 3. ª¬ 2 x 1 º¼ 1. 3 ln 2. 3.2x+1dx. 5 3. 1. 5. 3 ln 3. 2x. x ⎞ − x −1 3 ⎟ dx = (e − 2e ) dx ⎠ x = 3 e 3 + 2e −x −1 + c. 2 (1 − e ) 3e. 5. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. 99 ln 10. Worked solutions: Chapter 7. 3.

(78) WORKED SOLUTIONS. Exercise 7H 1. 5. y = x4 + 3x3 − 3x2 − 7x + 6 y. y = x4 − x = x(x3 − 1) y. O. x. 1. 0. x. 1. 0. 2 ⎡ 5 ⎤ (x − x)dx = ⎢ x − x ⎥ −1 ⎣ 5 2 ⎦ −1. 4. 0. 1 1 5 2. −2. (x4 + 3x3 − 3x2 − 7x + 6)dx.

(79). −3. 7 10. −2. 4 2 ⎡ 5 ⎤ = ⎢ x + 3x − x 3 − 7 x + 6x ⎥ 5 4 2 ⎣ ⎦ −3. 1. 1. 2 ⎡ 5 ⎤ (x4 − x) dx = ⎢ x − x ⎥ 5 2 0 ⎣ ⎦0. 1 5. ∴ area 2. O. –3 –2. 7 10. . 3 10. 1. . 2. . 3 10. = 1 sq. unit. 32 5. 12 8 14 12. 243 5.

(80). 243 27 63 18 4. 2.

(81). = −12.4 − ( −10.35) = −2.05. y = x2 − 2x − 3 = (x − 3)(x + 1). 1. (x4 + 3x3 − 3x2 − 7x + 6)dx. y. −2 1. –1 O. 4 2 ⎡ 5 ⎤ = ⎢ x + 3x − x 3 − 7 x + 6x ⎥ 4 2 ⎣5 ⎦ −2. x. 3. 1 5.

(82). 3 1 7 6 − ( −12.4) = 14.85 4. 2. area = 2.05 + 14.85 = 16.9 sq. units 6. -1. A=. (x − 2x − 3)dx −3. 1 3. 32 3. −1. 3.

(83). 1 3 − ( − 9 − 9 + 9).

(84) . 1 3. ∴ required area = 3. 4 ln3. x. 0. 2. 2. 32 3.

(85). + 16 = 16 sq. units. 8. 2. 2x dx =. 1. ln3. 1 (4 ln 2. 9. 1 ln 2. 2). y = 2e −x+1 − 1. [2 x ]. 2 1. 2 ln 2. sq. units. x = 0, x = 3. 3. ( e x − 3) dx = ⎡⎣e x − 3x ⎤⎦ ln3. A=. |2e−x+1 − 1|dx = 3.32 sq. units 0. 0. ∴ area = e + 3ln3 − 12 + 3ln3 − 2 = e3 + 6ln3 − 14.

(86) 2

(87) 1 2. y = 2x x = 1, x = 2 A=. 3. = 3 − 3 ln 3 − 1 = − (3 ln 3 − 2). 1 5 5. = 6.3 sq. units. 3. = e − 9 − 3 + 3 ln 3 = e3 + 3 ln 3 − 12. 5. (x−2 + 1) dx = ⎡⎢ −1 + x ⎤⎥ 1 ⎣x ⎦1 2. 16 3. ( e x − 3) dx = ⎡⎣e x − 3x ⎤⎦. 3. sq. units. 2. 3. ln3. 16 3. 1 x = 1, x = 5. 1. y. A=. ª x3 º 2 « 3 x 3x » ¬ ¼ 1. −1. 1 3 . 2 (0 4 2 ) 3. 5. (x2 − 2x − 3) dx. 7. sq. units. 1 3. 4. 3 ª 2 º (4 x ) 2 «¬ 3 »¼ 0. (4 − x) dx . 1. 1 3. x=4. 1 2. 0. 1. 3. x = 0,. 4. A=. ⎡ 3 ⎤ = ⎢ x − x 2 − 3x ⎥ 3 ⎣ ⎦ −3. =. 4x. y=. 2. 10. 1 x2. y. A=. 2. x = −1, x = 2. 1 x +2 −1. 2. dx = ⎡⎣In| x + 2|⎤⎦ −1. = ln4 − ln1 = ln4 = 2ln2 sq. units. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 7. 4.

(88) WORKED SOLUTIONS 11. 2 3 4x. y 3. 2 dx 3 − 4x. 1. −. =. x = 1, x = 3. 1 2. =. 3. ⎡⎣In|3 4 x |⎤⎦. 2 −4. −. (ln9 − ln1) =. 1 2. y= 4−x. 3. y =4−x. 1. x = 4 − y2 ⎡ (4 − y2) dy = ⎢ 4 y − 0 ⎣ 2. A=. ln9 = −ln3. ∴ area = ln3 sq. units 12. 8 8. y = −x3 + 6x2 + x − 30 x-intercepts: −2, 3, 5 (−x3 + 6x2 + x − 30)dx. 4. 3.

(89). 54 9 90 4 16 2 60

(90) 2. x2 = 4 − y. 5. ⎡ ⎤ (− x3 + 6x2 + x − 30)dx = ⎢ − x + 2 x 3 + x − 30 x ⎥ 4 2 3 ⎣ ⎦3 4. 625 4. 250 . 2. 1. x = (4 − y ) 2 4.

(91). 25 2. A=2. 150 51.75

(92). 3. 5. x 0 d x d1 ® ¯2 x 1 d x d 2. Area =. x2dx +. A=. 1. y.

(93). (4 2) 2 1. 2. ° x ® 2 °¯ x. 5 6. sq. units. 3. 2. y = 1,. y 1 A =. y = 10 1 2. ( y 1) dy 1. 2 3 10 2. ª2 º «¬ 3 ( y 1) »¼ 1. y = 0, y = 4. 2 (9) 3. 3 2. −1 2. 3 2 ⎡ ⎤ = ⎢2x − x + x ⎥ 3 2 ⎣ ⎦ −1. 4 2

(94) 2

(95). x=y. 8 3. 4. ⎡ 3⎤ y2 dy = ⎢ y ⎥ = 64 sq. units 0 ⎣ 3 ⎦0 3. (2 − x2 + x) dx. Area =. 2 4. y=–x. 2 − x2 = −x x2 − x − 2 = 0 (x − 2) (x + 1) = 0 x = 2 or − 1. = 18 sq. units. A=. x. –2. 3. 10. y= x. 2. · dy = ª 4 y 1 º 2 ¸ «¬ »1 y¼ y ¹ 2 1. y = 2 – x2. O. Exercise 7I. 2. 2. 2. = 3 sq. units. x. 1. y. y. 1. 81. y = x2 + 1. §4 ¨ ©. 1 1. x=4−. 2. Exercise 7J. ⎡ 3⎤ ⎡ 3⎤ = ⎢2 x 2 ⎥ + ⎢x ⎥ ⎣ 3 ⎦ 0 ⎣ 3 ⎦1. 1. y. y=2. 2. 2. 0. 2 3. 1. 1 2. = 4 1 sq. units. x dx. x dx +. Area =. y=. sq.units. 1 2. 2. 1 2. 4 3. 3 ª º 2 « 2 (4 y ) 2 » 3 ¬ ¼3. 8

(96) (2 2). 0 d x d1 1d x d 1. 1 2. 2. 2 ⎤ ⎡ ⎤ ⎡ = ⎢ x ⎥ + ⎢2x − x ⎥ 3 2 ⎣ ⎦0 ⎣ ⎦1. 14. 1 x 4. 2. (2 − x)dx. 1. 3. (4 − y ) dy. −x+4=. 2. 0. 1 3. y=. 4. 1 2. 4 (0 1) 3. = − 43.75 + 51.75 = 8 ∴ area = 93.75 + 8 = 101.75 sq. units. 1. x. O. 5. y. y=4. 4. = − 51.75 − 42 = − 93.75. 13. 2. ⎤ ⎥ ⎦0. y. 3. 2 ⎡ −x 4 ⎤ + 2 x 3 + x − 30 x ⎥ =⎢ 4 2 ⎣ ⎦ −2. 3. y 3. sq.units. y = 4 − x2 y = 3,. −2. 81 4. 16 3. 3. 3. y = 0, y = 2. 2. =. 9 2. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. 1 3. 1 2. sq. units. Worked solutions: Chapter 7. 5.

(97) WORKED SOLUTIONS 2. 4. y. A=. (16 − x2 − (x2 − 4x)) dx −2 4. y = x3. y = x2. (16 − 2x2 + 4x) dx. = −2. 4. O –1. ª16 x 2 x 3 2 x 2 º «¬ »¼ 2 3. x. 1. 64 . x3 = x2 x2 (x − 1) = 0 x = 0 or 1. 4. 3.

(98). 4. y. 6. 1. ⎡ ⎤ (x2 − x3)dx = ⎢ x − x ⎥ 3 4 0 ⎣ ⎦0 1 1 1 sq. units 3. 3. 32 32 16 8. = 72 sq. units. 1. A=.

(99) . 128 3. y = 2x2. 12 y. 3. y = x4 – 2x2. 4. y=2–x. O. –2. y = 4 – x2. x. 2. x4 − 2x2 = 2x2 x4 − 4x2 = 0 O. –2. x2 (x2 − 4) = 0. x. 2. x2 (x − 2) (x + 2) = 0. –2. x = 0, ± 2. 4 − x2 = 2 − x x2 − x − 2 = 0 (x − 2) (x + 1) = 0 x = −1 or 2. 2. 9 2. y = –x. −1. (2 – x2 + x)dx −1. sq. units. 2 3. −1 1.

(100) 1 2. y. 5 16 y = 16 –. 1 5. 5. ). y = x2– 4x. 1 2. −2 1 2. (−10x3 − 25x2 − 5x + 10)dx −1 −1 3 2 ⎤ 25 x 5x + + − 10 x ⎥ 3 2 ⎦ −2 +. ⎡ 4 = ⎢ 5x ⎣ 2. 1. 3 2 ⎡ 4 ⎤2 + ⎢ −5x − 25x − 5x + 10 x ⎥ 3 2 ⎣ 2 ⎦ −1. x2. –4 –2 O –5. 4. 16 − x = x − 4x 2x2 − 4x − 16 = 0 x2 − 2x − 8 = 0 (x − 4) (x + 2) = 0 x = 4 or − 2 2. + 32. (10x3 + 25x2 + 5x − 10)dx. Area =. 2 2 ⎡ 5 ⎤ Area = 2 ( x 3 x )dx = 2 ⎢ 3 x 3 − x ⎥ 5 2 0 ⎣ ⎦0. 3 5. −32 3. 2x3 + 5x2 + x − 2 = 8 − 4x − 20x2 − 8x3. x = − 2, −1,. 1. 2. 5. 2x3 + 5x2 + x − 2 = 0. x. 1. )(. − 32 −. 15. (1, 1). –1 O –1. 32 3. 10x3 + 25x2 + 5x − 10 = 0 y=x. 1. (. = 128 sq. units. (see qn. 1) 7. y=x. 2. 5 ⎡ 3 ⎤ = ⎢ 4x − x ⎥ = 3 5 ⎣ ⎦ −2. 2. y. 4. −2 2. (4 − x2 − (2 − x)) dx =. =. (4x 2 – x 4)dx. −2. 2. Area =. 2. (2x2 − (x4 − 2x2)) dx =. A=. x. 5 2. 2. + 20 3.

(101) 5

(102) − (. 25 5 10 − 40 200 10 20 3. . 5 32. 2. 25 5 24. 8. 10 305 20 3. 96. 3. 3. −. 5 2.

(103). + 25 − 5 − 10 3. 2. ). 1265 96. = 13.2 sq. units (3 sf ). © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 7. 6.

(104) WORKED SOLUTIONS 8. 1 1 x. x4 − 4. 1. (x > 0). 0. (x4 − 4) (1 + x) = 1. = 2 [ −e −x ]0 = 2 (−e −1 + 1) = 2 1 − 1 or 1.26 sq. units 1. ( ). x5 + x4 − 4x − 5 = 0 x = 1.449 1.449. A= 0. 1 1 x.

(105). x 4 4 dx = 5.41 sq. units. e. y. 13. y=x. y. 9. O. y = √x. 1. A= O. 0.4755. ( x − e1−x + 1) dx = 7.00 sq. units. y dy =. = 1 2. y dy. a. 3. 1. a. 4. ª 2 32 º «¬ 3 y »¼ 0. 3. 1. 1 x. dx. 3 5 3 5. + ln3 sq. units. Exercise 7K. 0. ª 2 32 º «¬ 3 y »¼ a. 0. = (1 − 0) + ln3 − ln1. y = x2 x = y a. 3. 1. 0.4755. 1 2. 1 2 3. 3 ⎡ 5⎤ = ⎢ 3 x 3 ⎥ + [ ln| x |]1 5 ⎣ ⎦0. 4. A=. x. 3. x dx +. x x=4. 4. 2 3. 1 y= x. y = e1–x – 1. 10. e −x dx. Area = 2. v(t) = t (t − 4) v(t). 3. 42 − a2 = a2 3. 2a 2 = 8 3. a2 = 4 2. O. a = 43. t. 4. y. 11. y = x2 4. distance = | O. –2. x. 1 y=2–x. 4. ⎡ 3 ⎤ = ⎢ t − 2t 2 ⎥ ⎣3 ⎦0. x2 = 2 − x. =. x2 + x − 2 = 0 (x + 2) (x − 1) = 0. 2. x = −2 or 1 x2 dx +. = 32 m 3. v(t) = 5 + 4t − t2 = (1 + t) (5 − t). (2 − x) dx. 1. 12. −32 3. 1 2. 2 ⎡ 3⎤ ⎡ ⎤ A = ⎢ x ⎥ + ⎢2 x − x ⎥ = 1 + (4 − 2) − 2 − 12 2 ⎣ 3 ⎦0 ⎣ ⎦1 3. =. − 32 =. 2. 0. 5 6. 64 3. v(t). 1. A=. 0. (t2 − 4t) dt|. (. ) –1. O. 5. t. sq. units 1. y y = e–x. a. y = ex. (5 + 4t − t2)dt. distance = 0. 1. 3 ⎡ ⎤ = ⎢5t + 2t 2 − t ⎥ 3 ⎣ ⎦0. –1 O. 1. x. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. = 5 + 2 − 1 = 20 m 3. 3. Worked solutions: Chapter 7. 7.

(106) WORKED SOLUTIONS 5. b. 4. (5 + 4t − t2) dt. distance =. 5.

(107).

(108)

(109). 6. 3 ⎡ t ⎤ 2 ⎢5t + 2t − 3 ⎥ ⎣ ⎦5. 3. 5. = 25 50 125 20 + (30 + 72 − 72) 3. (. − 25 + 50 − 125 80 3. 3. 3. 10 3. . v=π 3S 7. = 30 m. 6. a(t) = 1 − e –2t 0 ≤ t ≤ 3 v (t) = t + v(0) = 0 v (t) = t +. 1 −2t e +c 2 1 ∴0= 2 + 1 –2t e − 1 2 2. (t +. 3. distance = 0. 1 2. 2. ). 7. (3 3 1) = 16.1 cu. units 36 x 2 y 2 =. 2. x (36 − x 2 ) 144. =. 2. x 4. −. 4. x 144. 6. Exercise 7M. 3. 1. x 2.

(110). 5 2. 3x 4. 117S 4. 2. dx cu. units. y. 2. 0. y=. 2 § · v = π ¨ x 2 x ¸ dx = π 4 2© ¹ 5 3 ª º S « x » S 125 2 4 ¬ 4 ¼2. 2. (10 + 5e −0.5t ) dt. y=x. 5. a(t) = −2.5e −0.5t < 0 ∴ always negative distance =. y dy. = 7.2π cu. units. = ª« t 1 e 2t 1 t º» = 3.25 m 2 ¬2 4 ¼0 −0.5t 4 v(t) = 10 + 5 e b. 3. 5 ⎛ x2 x4 ⎞ ⎡ 3 ⎤ dx = p ⎢ x − x ⎥ v=π ⎜ − ⎟ 4 144 12 720 0⎝ ⎠ ⎣ ⎦0 = π (18 − 10.8). − 1 dt. 2. a. x2 = y 3 ª 7º S «3 y3 » ¬ 7 ¼1. 4 3. 1. x 12. y. 4. x = y3. 6. c ∴c=−. 1 −2 t e 2. 2. y = x2 3. 3. ). 2. 3 (2x − x2) dx = p ⎡ x 2 − x ⎤ ⎢⎣ 1 3 ⎥ ⎦1 8 1 º 2S ª S « 4 1 » cu. units 3 3 ¼ 3 ¬. (5 + 4t − t2) dt|. 3 ⎡ ⎤ = ⎢5t + 2t 2 − t ⎥ + 3 ⎣ ⎦1. y2 = 2x − x2. v=π. 6. 5. 2x x 2 2. 1. +|. y. = 26.3 m. Exercise 7L 1. O. y = (x − 1) − 1 = x − 2x 2. 1. v=π. 0. 1. (x2 − 2x)2 dx = π. (x4 − 4x3 + 4x2) dx 0. –4. 1. ⎡ ⎤ = p ⎢ x − x 4 + 4x ⎥ 5 3 ⎣ ⎦0 5. = 2. 8 S 15. 3. x − 4 = x2 − 4x x2 − 5x + 4 = 0. cu. units or 1.68 cu. units. (x − 1)(x − 4) = 0 x = 1 or 4. x. y=1+ 2. v=π. (1 + 0. 4. x )2 dx = π 2 (1 + 2 x + x) dx. (. 3. y. ) (. 3. ). (3 + 2 2 ) cu. units or 24.4 cu. units 2. x 2. v=π. 2y dy 0. 2. 1 4. (x4 − 8x3 + 15 x 2 + 8x − 16) dx 1 4. ª º S « x 2 x 4 5x 3 4 x 2 16 x » ¬5 ¼1 5. ¬. S ª¬ y 2 º¼ 0 = 4π cu. units. (x4 − 8x3 + 16x2 − x2 + 8x − 16) dx. =π. S ª«. x2 = 2y 2. ((x2 − 4x)2 − (x − 4)2)dx 1. =π. = p 2 + 4 (2 2 ) + 2 = p 4 + 4 (2 2 ) 3. v=π. 4. 2. 3 2 ⎡ ⎤ = p ⎢x + 4 x 2 + x ⎥ 3 2 ⎦0 ⎣. 4S 3. 4 x. 1. 2. 108S 5. 1024 5.

(111) . 512 320 64 64 . 1 5.

(112). 2 5 4 16 º» ¼. cu. units. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 7. 8.

(113) WORKED SOLUTIONS y. 3 2. O. 4 5. y. 2 5. 2a b. c. 24 5. 8 16. 2. 5. 5. x 2 32 24 5x. 5. y. 2. x. 2. y=x. 32 5. b. a. y = x2. y2 = 2x. y = √x. x2 = 2x. x(x − 2) = 0. O. x = 0 or 2 y = 0 or 2 ⎛ 2 ⎛ y 2 ⎞2 ⎞ v = π ⎜ y − ⎜ ⎟ ⎟ dy ⎝ 2⎠ ⎠ 0⎝. x. 1. 2. 1. Area = 0. 4 ⎛ ⎞ v = π ⎜ y 2 − y ⎟ dy 16 ⎠ 0⎝. 2. (. 2. 5 ⎡ 3 ⎤ v =S ⎢y − y ⎥ =S ⎣ 3 20 ⎦ 0. 8 3. 1. ( x 2 x 2 )dx 1. ). 3 ⎡ 3 ⎤ = ⎢ 2 x 2 − x ⎥ = 2 − 1 = 1 sq. units 3 3 ⎣ ⎦0 3 3 3. − 32 = 16 S cu. units 20. 15. y. 3. y. 4. y = 2x – 1 O. 1. 1. y = x2. O –1. 0.7459. x. 3.1149. x. 1. ⎛ 2 2 ⎜ (1 + 3x − x ) − ⎝ 0.7459 3.1149. y. y = 2x − 1 y 1 2. x. x = y2. x 2 = ( y + 1). 2. 1. −1. ( y + 1) 4. 1. 2. dy − π. 0. 1. a. y4 dy. S. ✗.

(114) S

(115). 7S 15. 1 5. 1. = ax +. b x. 2. 2 . cu. units. 4. b. when x = −2, ∴ b = 8a y=. ax 2. 2. dy dx. b 4. 1 24.

(116) . 4. (x + x−2 − x−4) dx 1 2. x. =0. 1 2. 1 1. 4. 3. 3.

(117). 41 24 1. (5x 2 4 x 2 )dx. dx = 0. 1 ª 52 º 2 «¬2 x 8x »¼ 1. 5. 1. (2 (4) 2 8(4) 2 ) − (2 − 8). = (64 − 16) − ( − 6) = 54. (1). 2. b x. − +c 0 = 2a +. (−1, 2). 2=. (2) − (3) − 2 3a 2. 1 dx x 3 1. c. (−2, 0). ∴ −2. ∴ − 2a +. . 4. ( −2, 0) = stationary point. =0. 1 2. 5x − 4. 1. (−1, 2). 2. 2. Review exercise dy dx. 2. −3 ⎡ 2 ⎤ ⎡ 2 ⎤ = ⎢ x − x −1 + x ⎥ = ⎢ x − 1 + 1 3 ⎥ 3 ⎣2 ⎦1 ⎣ 2 x 3 x ⎦1. 1. 5. 8 12. ⎛ 1 1 ⎞ ⎜ x + 2 − 4 ⎟ dx = x x ⎠ 1⎝ 2. 4. ⎡ ⎤ ⎡ ⎤ = p ⎢ ( y + 1) ⎥ −p ⎢ y ⎥ 12 ⎣ ⎦ −1 ⎣ 5 ⎦0 3. ( ) ⎞⎟⎠ dx 2 x. = 41.3 cu. units. x2 = y4. 4. v=π. x. v=π. 1 2. a 2. b 2. +c. (2). +b+c. (3). 3a 2. b = 8a. b. 2. 4a ⇒ − 4 = 3a − 8a ⇒ −4 = −5a. e. d 1. = =. 1 1 4x 1 4 1 4 . = [ ln| x 3|]12 = ln1− ln2 = −ln2. dx =. . 1 4. [ln|1 − 4x|]1e. (ln |1 − 4e| − ln3) (ln(4e − 1) − ln3) = − 1 ln. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. 4. ( 4e3−1). Worked solutions: Chapter 7. 9.

(118) WORKED SOLUTIONS. Review exercise 1. (x − 4 + ). 3. 3. 2. v(t) = t3 − 4t = t(t2 − 4) = t(t − 2) (t + 2). 2. x. 1. ⎡ 3 ⎤ dx = ⎢ x − 4 x − 3 ⎥ x 3 ⎣ ⎦1. v(t). = ( 3 4 3 3) . O. (. 1 3. ). 3. 4. 4 2. ⎡ ⎤ (t3 − 4t)dt = ⎢ t + 2t 2 ⎥ 4 0 ⎣ ⎦0 2. 4. = 4 − 8 = −4. 3x + 6. a. x. b.

(119)

(120). c. 1 2 3x. d. 2 1 4 x. 1 x. x. 6. dx = 3x2 +. 2. x. 3. 81 4.

(121). x. dx =. 1 3. 4. dx = x3 −. 6 x. +c 3. x 3. + 1 +c x. ln |2 − 3x| + c 1. dx = 2(1 4 x ) dx =. 2 (1 − 4 x ) 2 1 −4 2. +c. = − 1 − 4x + c. 2e3x 3 e x

(122) dx =. e. 41 m 4. v(t) = t3 − 3t2 + 2. =. v(t) (0, 2). 5. −2 −3 x e 3. x. 2e. + 3e 3 + c =. 3 x. x.

(123). e 3 dx. −2 −3 x e 3. + 33 ex + c. x 2 2 x 1 2 x 3x 2. 2x2 − x. t. 1. 2. 3. 1 2. 25 4. 18 (4). ∴ total distance = 4 + 25 =. O. 3. ⎛ 2 1⎞ ⎜ x − 2 ⎟ dx = x ⎠ ⎝. x 1 dx =. 3. ⎡ 4 ⎤ (t − 4t)dt = ⎢ t − 2t 2 ⎥ 2 ⎣4 ⎦2 3. 2.

(124). 4 3 = −4 3 + 20. ∴ total area = 2 4 3 − 20 = 8 3 − 40 sq. units. t. 2. 3. 4x (2, 2). 4x − 2. 1. ⎡ ⎤ (t3 − 3t2 + 2) dt = ⎢ t − t 3 + 2t ⎥ 4 ⎣ ⎦0 0 1. 4. 1 4. =. 2 ∴. 5 4. −1+2=. 2. 4. (4 8 4) .

(125) 5 4. = x +2+ 2. 2. 2 x + 3x 2x − 1 1. 2. ⎡ ⎤ (t3 − 3t2 + 2) dt = ⎢ t − t 3 + 2t ⎥ 4 ⎣ ⎦1 1 2. 2. 2 x + 3x 2x − 1. dx = 1. 3. 4. x 2

(126) dx 2 2x 1. 2. ⎡ 2 ⎤ = ⎢ x + 2 x + ln 2 x − 1 ⎥ ⎣2 ⎦1. 5 4. 1. ∴ total distance = 5 + 5 = 5 m 4. 2 2x − 1. = (2 + 4 + ln3) − ( 2 + 2 + ln1). 2. =. y. 7 2. + ln3. y. 6. y=5 –√3 –1 O. y=. x2 − 4 +. x −4+ 2. 3 x. 2. 1 √3. x. 3 x. 2. x. –1 O. =0. x − 4x + 3 = 0 4. 2. (x2 − 1)(x2 − 3) = 0 x = ± 1, ± 3. y=. 1 x +1. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. ∴x+1=. 1 y. Worked solutions: Chapter 7. 10.

(127) WORKED SOLUTIONS 1 y. x= 5 1. 9. −1. 1 y. 5. 2. 1. 1. = ln5 − 4. 1. S. (32 1). A=. 0. 1 2. (x + 1) dx −1. 2 3. 3 ª2 2º ( x 1) «¬ 3 »¼ 1. sq. units. ln 3. S ln 3. 3−2x dx 0. 1. ª¬32 x º¼ 0. x 0. 0. 1. (3−x)2 dx = 2π. 2S 2 ln 3. . –1 O. x. 1 1. y. =. –1 O. v = 2π. ∴ area = 4 − ln5 sq. units 7. y = 3x. 3. 1

(128) dy = >ln y y @. = (ln5 − 5) − (ln1 − 1). 8. y. y = 3–x.

(129). 8S 9 ln 3. 8 9. cu. units. y. O. a. x. a. (3ax − 3x2 ) dx = 4 0 a. ⎡ 3ax 2 3⎤ ⎢ 2 −x ⎥ =4 ⎣ ⎦0 3a 2. 3. − a3 = 4. 3. a 2. 4. a3 = 8. ∴a=2. © Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute. Worked solutions: Chapter 7. 11.

(130) WORKED SOLUTIONS. 9. The power of calculus. Answers. 2. Skills check. a. y = sin (x 5 − 3). dy dx. = 5x 4 cos( x 5 − 3). b. y = cos (e x). dy dx. = − e x sin(e x ). c. y = csc (x 2 + 11). sin q 2. 1−. 1 − tan T 2. 1. a. =. 1 + tan T 2. 1+. tan 2T. =. sin q 2. ∴ cos 2T = b. cos 2 q. cos 2 T − sin 2 T. = cos2θ − sin2θ = cos2θ. cos 2 T + sin 2 T. cos 2 q. d. 2. 1 − tan T 2. tan T tan T 1 tan T tan T. tan(T T ). 2 tan T 2. 1 tan T. ∴ tan 2T = 2. y = tan (ln(2x + 1)),. f. y = sec dy dx. 2. 1 − tan T. a. f (x) = 3e 2x − 2x 2,. b. g (x) = (x + 1) ln (x 2 + 2x + 1). f (x) = 6e 2x − 4x. = (x + 1) ln (x + 1) 2 = 2(x + 1) ln (x + 1) 2( x 1) ( x 1). g(x). ex x 1. + 2 ln (x + 1) = 2 + 2 ln (x + 1) ( x 1)2 xe x e 2. 2. c. h(x ). hc( x ). ( x 1) e. x2. g. x2. (2 x 2 2 x 1). a. Exercise 9A b. a. y = cot x. dy dx. b. y = csc x. dy dx. y = sin 3x. dy dx. =3 cos 3x. d. y = tan (5x − 3). dy dx. = 5 sec (5x − 3). e. y = cos (8 − 3x). dy dx. = 3 sin(8 − 3x ). f. y = csc. dy dx. = − 1 csc. g. y = cot. ( ). (. 7 − 2x 13. ). dy dx. csc 2 x. c. 2. =. ex + 1). = 1 e x (e x + 1) 2 sec ( e x + 1 ) tan ( e x + 1 ) −1. 2. x. ) ( e + 1) 2( e + 1) sin ( e + 1 ) sec ( e + 1 ) 2( e + 1). =. e sec. =. e. (. x. x. e + 1 tan x. x. x. x. 2. x. = −sec 2 x sin (tan x) cos (cos(tan x)). (. 2 csc 2 7 − 2 x 13 13. cot. ). x −3 4. = (3 − 2x) sin 2x + 2 (3x − x 2) cos 2x. = e 1−x sec2 x − e 1−x tan x = e 1−x (sec2 x − tan x). d. y. sinx dy , dx x. e. y. 2 x 3 dy , sin 2 x dx. f. ( ) ( ) x −3 4. = 2cos x − (2x − 1) sin x. y = e 1−x tan x dy dx. = − csc x cot x. 4. + 1)). y = (3x − x 2) sin 2x dy dx. Exercise 9B. x −3 4. 2 sec 2 ( ln(2 x 2x + 1. y = (2x − 1) cos x dy dx. Proofs using differentiation from first principles. c. =. Exercise 9C 1. 1. (. dy dx. y = sin (cos (tan x)) dy dx. 2. ( x 1)2. = −(12x 2 − 4x + 7) csc 2(4x 3 − 2x 2 + 7x + 17). e. 2 tan T. 2 x csc( x 2 11)cot( x 2 11). y = cot (4x 3 − 2x 2 + 7x + 17) dy dx. 1 + tan T. dy dx. x cos x - sin x. =. x2. =. 2 sin2 x − 2(2 x + 3) cos2 x 2. sin 2x. tan x 2 x. y dy dx. −1 ⎛ ⎞ = ⎜ 2 − x sec 2 x + 1 (2 − x ) 2 tan x × 1 ⎟ (2 − x ) ⎠ 2 ⎝. dy dx. =. 2. 2(2 − x ) sec x + tan x. © Oxf ord University Press 2012: this may be reproduced f or class use solely f or the purchaser’s institute. 3. 2( 2 − x ) 2. Worked solutions: Chapter 9. 1.

(131) WORKED SOLUTIONS 2. a. b. x. dy dx. 2cos S = 1. Exercise 9D. 6. 1. 3. x. 1 x. b. f (x) = arcsin 3x, f ' ( x ). 1 (2 x 1). 2. 2. 3S 2. 2. b. = 2x tan x + x 2 sec 2 x. ( ) 3S 4. 16. dx. dy dx. 2 tan 2T 2. 1 − tan T. tan 3U 2. c. dy dD. =. 1− x. 2. − arccos x x. 2. d 2. dy dx. 3U U cos 2 2 U 3U 2 cos cos 2 2. dy dx. 3 sec 2 3U 2 2. e. sin M cos M 2. 2 sin M − 1 sin M − cos M. =. 2. sin M − cos M. (sin M cos M )(sin M cos M ) sin M cos M dy dM. 1 x arccos x x. 2. 1 x. 2. 2 arctan x . 2x 1 1 x. 2. 1 x 2 arcsin x 1 x. 2. 1 x. 2. 1. 1. 2 x 1 (1 x 2 ) 2 arcsin x 2. x arcsin x 1 x. 2. y = (4x 2 + 1) arctan 2x 2. 2(4 x 1). dy dx. 8x arctan 2 x . dy dx. 8x arctan 2 x 2. 2. sin M − cos M. y = sin φ + cos φ. y. 4 sec 4T. 2 sin. dy dx. 2. x . y = (2x + 1) arctan x dy dx. (sin M sin 2M cos M ) sec M. =. arc cos x x. 0. sec E tan E dy dT. = tan 4T. dy dx. = e x sec x tan x + e x sec x = sec 0 tan 0 + sec 0 = 1. dy dE. sec E. sin U sin 2 U cos U cos 2 U. y. 2arcsin x 2. 1 x. 8. tan E sin E. y. 2x. y dy. 2. y = sin2 α + cos2 α = 1. y. y = 2x arcsin x. −x. 4. a. y=. 2. 2x 2x 1. 2. + 9S sec 2 3S. y = e x sec x x = 0. y. a. dy dx. 3x 4. x. y = x tan x. 1. f ' (x ). 2. g. e. 2. f (x) = arctan (2x + 1), f ' ( x ) 1 4x 4x 1. S. x. 2. 2. 2. d. 2. 2. = −3 cos x + 3x sin x. 3S 9S. c. 3. 2. 2. b. 2. 2. = sin x + (x − 2) cos x. = 3S tan. 3. 1 x. 1 x. c. 2. dy dx. 1. . 1. ? f c( x ) . x=0. 3cos S 3 S sin S f. 1 1 cos y. 2. 4. 1 siny. 2. 5S 4. x. sec 2 5S. y = −3x cos x dy dx. . = sin 0 + (−2) cos 0 = −2 e. dx dy. sin y ∴. 3 2. 4. = −sec 2 x. y = arccos x ∴ x = cos y dx dy. 3sin 7S. = −3sin 3x. y = (x − 2) sin x dy dx. a. 7S 12. y = tan (−x) = −tan x dy dx. d. = 2cos 2x. y = cos 3x dy dx. c. S. y = sin 2x. cos M sin M. 3. a b. 1 4x. 2. d dx. (arcsin x + arccos x). d dx. (arctan x + arctan (−x)). 1 1 x. © Oxf ord University Press 2012: this may be reproduced f or class use solely f or the purchaser’s institute. 1. 2. 1 x. 1 1 x. 2. . 0 2. 1 1 x. 2. Worked solutions: Chapter 9. 0. 2.

(132) WORKED SOLUTIONS c. ⎛ 2 arctan x − arcsin 2 x ⎞ ⎜ ⎟ 2 ⎝ x +1⎠. d dx. § 2 2 2 x (2 x ) · ¨ ( x 1)2 2 ¸ ( x 1) © ¹. 2 1+ x. 2. 2. 2. 1 4 x ( x 2 1)2. 2 1+ x. 4. a. x = siny 1 cos y dy. dy dx. tan 2y = 1 ⇒. dy dx. = cot 2 y. 2. x y. cosy = dy dx. dy dx. ?. (1 − siny). = 2x. dy dx. ?. dy dx. dy dx. 1 dy y dx. 2x e. sin y. cos y. dy dx. 1 y. =. c. d. P (0, 0). f (x) = 3sec 3x. f (0) = 3. 2. f (x) = sin (2x) − 1 y. sin.

(133) 1 2S 3. y. 3 2. 1. f'.

(134). 1 1 x S. 3. x S 3. 3 2. 1. , y