Pra-U STPM Maths(T) Semester 2 2022 CC039332b

196 Mathematics Semester 2 STPM Answers 7. (b) right x –2 2 –3 0 5 f(x ) not continuous 8. k = – 3 4 ; Continuous 9. c = 1 10. (a) 2 (b) x 0 542 f(x ) continuous STPM Practice 1 1. (a) 1 2 (b) 3 2. (a) (i) lim x → c – f(x) ≠ lim x → c + f(x) (ii) lim x → c – f(x) = lim x → c + f(x) ≠ f(c) (iii) lim x → c – f(x) = lim x → c + f(x) but f(c) is not defined (b) {k: –∞ < k < 0 or 0 < k < 3 2 or 3 2 < k < ∞} 3. a = –32, b = 3 4. (a) 1, 0, 2, 1 (b) Yes, No 5. (a) 8 27 (b) –2√7 6. (a) b = 3 (b) a = ±√6 7. (a) lim x → 0– f(x) = lim x → 0+ f(x) = 1, lim x → 0 f(x) exists. (b) lim x → 0 f(x) = 1 ≠ f(0) = 2 f(x) is discontinuous at x = 0 f(x) is continuous in the interval (–∞, 0) < (0, ∞) 8. a = 1, b = –1 9. lim x → 3– f(x) = –6 ≠ lim x → 3+ f(x) = 6 f(x) is discontinuous at x = 3 10. (a) 2 (b) – 1 2 13. (a) 3 8 (b) 2 14. No, not continuous at 1 15. (a) lim x → 0– f(x) = 8, lim x → 0+ f(x) = 2 + c, c = 6 (b) (i) f(x) continuous at x = 0 only when c = 6 (ii) f(x) continuous at x , 0. (quadratic) (iii) f(x) continuous at x . 0. (exponential function) 2 Differentiation Exercise 2.1 1. 3x 2 2. 4x3 3. 10x 4. – 2 x3 5. 2x + 5 6. 2x – 1 7. 12x 2 8. – 3 x4 9. 4x 10. 4x – 3 Exercise 2.2 1. 0 2. 3 3. –4 4. 5x 4 5. –4x–5 6. 1 3 x – 2 —3 7. – 1 x2 8. 3 2 x 9. – 1 2 x – 3 —2 10. 2 3 x – 1 —3 11. –15x–4 12. 70x 9 13. – 5 x5 14. 21 8 x 2 15. 16 x9 Exercise 2.3 1. 2x ln 2 2. 1 1 5 2 x ln 1 5 3. 10x + 3 4. 4x 3 – 18x2 5. 6x – 5 6. 2x – 4 x2 7. – 2 x3 – 1 x2 8. ex + cos x 9. 2 x 10. 8x + 2 x2 + 9 x4 11. – 2 3x – 2x 12. – 1 2x 13. 10x – 3 x2 + 8 x3 14. 1 x2 + 4 x3 15. – 3 2x 16. 2 cos x – 3 sin x 17. 13 3 4 18. 13 Exercise 2.4 1. 2x(2x 2 + 1) 2. 24x – 5 3. 8x 3 – 30x2 + 1 4. 12 x4 – 1 x2 – 2x 5. 3x 2 – 4x – 1 6. 5(cos x – x sin x) 7. 2x(2 sin x + x cos x) 8. (cos x) ln x + sin x x 9. 1 + ln x 10. ex (x 2 + 2x – 1) 11. ex (tan x – cos x + sec2 x + sin x) 12. sin x (1 + sec2 x) 13. x(2 cos x – x sin x) 14. 2x (sin x + cos x) + x2 (cos x – sin x) 15. cos 2x Exercise 2.5 1. –6x (2x 2 – 3)2 2. 2 (x + 1)2

Mathematics Semester 2 STPM Answers 197 3. 2(2 – x) (x + 2)3 4. –3x 4 + 6x2 + 8x (x 3 + 2x2 ) 2 5. 2x cos x – sin x 2x x 6. (ln x) – 1 (ln x) 2 7. 1 1 + cos x 8. ex (x – 2) (x – 1)2 9. sec2 x 10. –cosec2 x 11. sec x tan x 12. –cosec x cot x 13. 2 1 – sin 2x 14. ex (2 + ex ) 2 15. 1 x (1 – x ) 2 Exercise 2.6 1. 3e3x 2. 2x ex2 3. 2 x – 1 4. 6(x + 1)5 5. 36x 2 (2x3 + 1)5 6. 8x x2 + 1 7. 2 sin x cos x 8. –(x + 1)–2 9. 2x e(x2 + 3) 10. 3 cos (3x + π 4 ) 11. 4x sin (x2 + 1) cos (x2 + 1) 12. 2 cot 2x 13. –(ex + e–x )(ex – e–x ) –2 14. –cosec x 15. –4x sin(2 + 2x2 ) 16. (2x – 3)(43 – 6x) (3x + 4)4 17. –sin x 1 + cos x 18. x 1 + x2 19. 2 sec2 2x tan 2x 20. 2x + 5 2(x2 + 5x + 3) 21. tan4 x 24. – 1 21 1 2 π – 12 —3 2 25. –1 Exercise 2.7 1. (a) x + y dy dx = 0 (b) 8x + 18y dy dx = 13 (c) 2x + y + (x + 2y) dy dx = 0 (d) x 2 + (y2 – 3) dy dx = 0 (e) 3x 2 + y2 + (2xy – 5x) dy dx = 5y 2. (a) 2y 2 – 2xy3 3x 2 y2 – 4xy – 3 (b) cot x cot y (c) x y (d) 2x + 5y + 2 2y – 5x + 3 (e) sin 2x x3 e3x – cos 2x x2 e3x – 3y 6. 1 7. 3 8. – 1 3 Exercise 2.8 1. (a) 1 t (b) 3(t 2 + 1) 2t (c) –3t 2 (t – 1)2 (d) 2t 3t 2 + 1 2. (a) –cot q (b) –tan q 3. 1 4. cos t et 5. (0, 2), (0, –2) 9. (2, 2), (2, –2) Exercise 2.9 1. (a) 2 (b) 34 (c) 2 1 4 (d) –1 2. 1–2, 1 2 2 3. (a) – 10 13 (b) – 3 4 4. (a) (3, –5) (b) (2, –8) (c) (0, –8) 5. 12nπ + π 2 , 12 Exercise 2.10 1. Increase for x . 3, decrease for x , 3 2. Increase for x . 4, decrease for x , 4 3. Increase for x , –1, decrease for x . –1 4. Increase for x , –2 and x . 2, decrease for –2 , x , 0 and 0 , x , 2 Exercise 2.11 1. 1– 1 2 , – 25 4 2 2. 1 1 3 , – 14 27 2, (3, –10) 3. (1, 0) 4. (–2, 9) 5. (–1, –2), (1, 2) 6. (2, 108), (0, 0), (5, 0) 7. 1 1 4 ln 3 2 , 122 8. 1 π 6 , π 6 + 3 ), ( 5π 6 , 5π 6 – 3 2 9. 1e 1 —2 , 1 2e 2 10. 1 1 2 ln 2, 2 2 2 Exercise 2.12 1. (a) (0, 0) maximum, 1 2 3 , –4 27 2 minimum (b) (2, 10) point of inflexion (c) (2, 32) maximum, (6, 0) minimum (d) (–1, –1) minimum, (0, 0) point of inflexion 2. (3, 9) 3. (a) (0, 0) (b) (2, –11) (c) (–1, –25), (3, –173) (d) (2, 0), (1, –1)

198 Mathematics Semester 2 STPM Answers 4. 1e, 1 e 2 maximum 5. 1 1 2 , 1 2 e – 1 —2 2 maximum, 1– 1 2 , – 1 2 e – 1 —2 2 minimum 6. (a) (1, 3) 7. a = 2, b = 2, c = 3 2 8. max: 1 1 2 , 2 2 ; min: 1– 1 2 , 02 9. maximum 10. 1 – 3x (1 + x 2 )1 + x2 , x = 1 3 , maximum 11. dy dx = e2x (2ax + a + 2b) d2 y dx2 = 4e2x (ax + a + b) p = –4, q = 4 Exercise 2.13 1. (a) x = 1, y = 0 (b) x = 1, y = –1 (c) x = 0, y = 0 (d) x = 4, y = 1 (e) x = – 3 2 , y = 1 2 2. (a) x → ∞, y → 0; x → –∞, y → 0 (b) x → ∞, y → 0; x → –∞, y → ∞ (c) x → ∞, y → ∞; x → –∞ (d) x → ∞, y → ∞; x → –∞, y → –∞ (e) x → ∞, y → ±∞; x → –∞ 3. (a) y x y = 2 1–x _ x = 1 (0, 2) (b) y x y = x 1–x _ x = 1 y = –1 (0, 0) (c) ( ) 3 0, 4 – y x y = 1 x = 4 y = x – 4 x – 3 (3,0) – (d) y = x 2 ex y x (0, 0) max. point = (–2, 0.54) (e) y = x 3 +2x 2 –x –2 y x (1, 0) (0, –2) (f) y = x – 4 _3x y x y = 3 x = 4 (0, 0) (g) y = x e_ x y x 4. 1 2 5 ln 3 4 , 12.62 y x ( ) 5 2 – 4 3 ln , 12.6 – 5. x = e 1 —2 maximum y x 0 e2 2e 1 1 ( –, – )

Mathematics Semester 2 STPM Answers 199 6. (a) x-axis (c) 1 2 3 , 4 3 2 3 2, 1 2 3 , – 4 3 2 3 2 (d) y 2 =x ( 2 – x ) 2 (0, 0) ( ) 2 3 – 2 3 – 4 3, – ( ) 2 3 – 2 3 – 4 3 , – y x – 7. (a) y = x 2 + 2 (0, 2) y x (b) y = x 2 + 2 (0, ) y x – 1 2 – 1 8. (a) y = 1, x = 3 (b) (–2, 0), 10, – 2 3 2 ; No stationary point (c) (0, ) y x – 2 3 y = x = 3 x – 3 – x + 2 (–2, 0) – 9. – 3(x + 2)(x – 6) x 2 (x + 6)2 ; x = –2, 6 1–2, 3 2 2 minimum, 16, 1 6 2 maximum y x 3 (x – 2) x (x + 6) –2, 3 2 y = 1 x = –6 2 ( ) – y =– 1 (6, —) 6 10. 1 1 2 ln 2, 2 2 2 y x y = ex + 2e–x (0, 3) ( ) 2 1 2 – ln 2, 2 11. – 289 24 , – 3 2 , 4 3 y = – 1 f(x) y = f(x) y x ( – –, – –) 1 289 12 24 12. (a) – 1 (x – 4)2 (b) 4y + x = 12 (c) y = x – 3 (d) 0 1 4 y x 13. (a) x = 0, x = 2p (b) x = 0 and 2(p ± p) y x y = x 2 e– x (0, 0) (4, 2.17) 1 2 – y x y = x 3 e– x (0, 0) (6,10.75) 1 2 – Exercise 2.14 1. (a) y = 2x – 3 (b) x – y – 7 = 0 x + 2y + 1 = 0 x + y + 1 = 0 (c) y = 5x – 4 (d) y = 8x + 2 5y + x = 6 x + 8y + 49 = 0 (e) x + y + 2 = 0 x – y = 0 2. y = – 121 8 3. – 87 32 4. x + 3y + 6 = 0 5. 2y = 6x – 3, 3x + y + 3 = 0, 1– 1 4 , – 9 4 2 6. x + y – 3 = 0, x – y – 2 = 0

200 Mathematics Semester 2 STPM Answers 7. 1 3 2 , – 11 4 2, – 29 4 8. 2y = x + 1 9. 2y – x + 1 = 0 10. 3 4 t, 4y = 3qx – 2q3 11. (1 + 3t 2 )y – 2tx = (t 2 + 1)2 12. x = y; 1– 2 2 , – 2 2 2 Exercise 2.15 1. 12 m3 s –1 2. 0.26 m s–1 3. 1.6 cm2 s –1 4. 860 m2 s –1 5. 2 cm s–1 6. 4 45 cm s–1 7. 0.3 m s–1 8. 0.04 cm s–1 9. (a) 1 3 cm s–1 (b) 4 9 cm s–1 10. 2 9 cm s–1 Exercise 2.16 1. 1 3 2. 4 minimum 3. 0 minimum, 32 maximum 4. p2 2(4 + π) 5. (200π)– 1 —2 6. (b) x = 2 3 , Vmax = 7 11 27 cm3 7. 32x–2, width 3.42 cm, length 13.68 cm, depth 2.74 cm 8. y = 4x + 36 x ; 6 10. 6 m × 12 m STPM Practice 2 1. (a) 4x 3 – 2 – 2 x3 (b) 2(x2 + 4) (x2 – 4)2 (c) 6x + 3 4x 5 —4 (d) (x + 1) 1 —2 (4x – 5) 2(x – 2) 1 —2 (e) 6 cot 3x (f) etan x sec2 x (g) (13 – 5x)(x – 2)2 e–5x (h) x 2 + 2x + 2 (x 2 – 2)(1 + x 2 ) (i) cos2 x (4 cos2 x – 3) (j) 13e–2x cos 3x (k) 6(2x + 3) (4x2 + 9) 3 —2 (l) 3 – x (x + 2)2 (2x – 1) 1 —2 (m) 3 tan2 x cos 3x + 2 tan x sec2 x sin 3x (n) 7x4 + 8x3 – 1 2(x + 1) 3 —2 (o) 2e2x (1 – e3x ) (1 + 2e3x ) –2 2. – (x2 + 2) sin x x3 4. –1 5. –5 6. x2 1 + x 7. – 3 4 11. 81 64 12. y = x – π 2 + 2; y = –x + π 2 13. k = 2 15. (a) 1– 1 4 , –62, local maximum point. (b) 132– 1 3 , 02 (c) 32 0 – –1 – –1 3 4 –6 y x 16. y = x + 3 17. 2y + x = 3 18. 2t – 1 2t + 1 , x + 3y – 6 = 0 19. (a) a = –2, b = 4 (b) 8 20. 1–1, 1 3 2 21. (a) 3 (b) 1 3 (2 ± 13 ) 22. (a) x = –1, 2 24. (a) 5π (b) 5 12 25. 0.0224 mm2 s –1 26. 25 π cm s–1 30. x = 18 , v = 36 m3 31. 1–1, 1 2 e 2 y = – e-x 1 + x 2 y x (0, 1) (–1, e – ) 1 2 32. (a) x = –1, 1 3 , x = – 1 3 33. ± 3x – 4 2x – 2 y 2 = x 2 ( x – 2 ) y x (2, 0) y = 1 x – 32x2

Mathematics Semester 2 STPM Answers 201 34. (a) 60 πr 2 cms –1 (b) 3 πh cms –1 (c) 1 000π 3 s 35. (a) (0, 0), 1 4 3 , 32 27 2; 1 4 3 , 27 32 2 (b) asymptote : x = 0, x = 2, y = 0 4 3, _ y x (0, 0) – 32 27 ( ) x = 2 y x 4 3, – _ 32 27 ( ) (c) k , 0; k . 32 27 36. (a) x = 2, y = –1 (b) y = 1 2 x 0 –1 2 y = – x 2 – x y x 0 1 2 x y y = x |– | 2 – x 37. y x 0 1 – 2 1 – – 2 1 (– –, 2) 2 –1 1 y = (2x – 1)2 (x + 1) y x 0 1 – 2 1 – – 2 –1 1 y = – (2x – 1)2 (x + 1) 1 38. – 1 2x 3 —2 39. (b) (0, 4), (0, –4) 42. (a) x = 0, y = 0 (b) 1e, 1 e 2; maximum point (c) (i) (4.48, ∞) (ii) (4.48, 0.3348) (d) 1 y x x Inx 0 e, (4.48, 0.3348) y = 1 ( e ( 43. y = –x + 1 π 2 + 22 44. (a) dy dx = – 2x + y x + 2y (b) 2; – 1 2 (c) (–3, 6) is a minimum; (3, –6) is a maximum. 45. At (1, 6), gradient = – 2 9 At (1, –3), gradient = 29 9 46. (a) (0, 0) an inflexion point (b) (3, 27e–3) a maximum point (c) (3, 27e–3) (0, 0) y x f (x) 47. (a) A = 8xr 2 – 4x2 (b) x = r 8 3 Integration Exercise 3.1 1. 1 x2 + 1 , ln (x + x2 + 1 ) + c 2. 3(x + 3)2x – 3 + c 3. cos x – x sin x ; x cos x + c 4. 2x(x – 1) (2x – 1)2 ; 1 2 1 x2 2x – 1 2 + c 5. –tan x, –ln |cos x| + c Exercise 3.2 1. (a) x6 6 + c (b) 3ex + c (c) –5 cos x + c (d) 2 3 x 3 —2 + c (e) – x–2 2 + c (f) 2 sin x (g) 6 ln x (h) – x–3 3 2. (a) 4 3 x 3 – 5 4 x 4 + c

202 Mathematics Semester 2 STPM Answers (b) x4 4 – 2 3 x 3 + c (c) – 5 x – 2 x2 + c (d) 2 3 x 3 —2 + 2x + c (e) 16 3 x 3 – 12x 2 + 9x + c (f) – 9 x – 12x + x2 2 + c (g) 7x – 3e–x + 2ex + c (h) –9x–1 – 12x 1 —2 + x2 2 + c 3. (a) (x + 4)6 6 + c (b) (x 3 – 1)3 3 + c (c) 1 3 (x 2 – 1) 3 —2 + c (d) 2 3 x3 + 1 + c (e) 1 2 (ln x) 2 + c (f) 2 3 (1 + ex ) 3 + c (g) – 1 5 cos5 x + c (h) 1 4 tan4 x + c (i) 1 5 sin5 x + c (j) – 2 3 (cos x) 3 —2 + c Exercise 3.3 1. x – 2 ln|x + 2| + c 2. ln 3x – 2 1 – x  + c 3. 1 2 ln|x – 4| – 1 10 ln|5x + 2| + c 4. 1 2 ln x + 1 x + 3  + c 5. 25 49 ln x – 2 2x + 3  – 12 7 1 x – 2 + c 6. ln x 1 – x  – 1 x + c 7. ln x2 + 4 x + 2  + c 8. ln|2x – 1| – 1 2x – 1 + c 9. x2 2 – x + ln|x + 1| + c 10. x + 5 ln x x + 1  + c 11. 1 2 ln (x – 1)7 (x + 1) (2x – 1)7  + c 12. 12 ln (3 – x) 3 (2 – x)(4 – x) 2  + c 13. A = –1, B = –1, C = 1; ln x – 1 x  + 1 x + c 14. –x + 2 ln|ex + 1| + c Exercise 3.4 1. (a) (x 2 + 2)4 8 + c (b) 2 3 (x 2 – 4) 3 —2 + c (c) 1 6 (2x 2 – 5) 3 —2 + c (d) 2 3 (x + 9) 1 —2 (x – 18) + c (e) 1 – 5x 10(x – 3)5 + c (f) –2(8 + x)4 – x + c (g) 1 2 (sin–1 x + x 1 – x2 ) + c (h) –cos x + 2 3 cos3 x – 1 5 cos5 x + c (i) 1 5 sin5 x + c (j) 1 2 tan–1 x 2 + c (l) – 1 4 cos4 x + c (m) 4 3 (1 + x ) 3 —2 + c (n) x 99 – x2 + c 2. (a) – 1 36 (8x + 1)(1 – x) 8 (b) – 1 4 1 – 4x2 + c (c) 5 + 4x 12(4 – x) 4 + c (d) sin–1 1 x 3 2 + c (e) 2 sin–11 x 2 2 – 1 2 x4 – x2 + c (f) –cos x + 1 3 cos3 x + c (g) sin4 x 4 + c Exercise 3.5 1. (a) 1 30 (3x – 9)10 + c (b) – 1 35 (2 – 5x) 7 + c (c) – 1 8(4x + 5)2 + c (d) 4 15 (3 + 5x) 3 —4 + c (e) – 2 3 (1 – x) 3 – 21 – x – 1 1 – x + c 2. (a) – 1 5 e2 – 5x + c (b) 1 3 e3x + 1 4 e4x + c (c) – 1 2 e–2x – 4e–x + 4x + c (d) 1 2 e2x – 1 2 e–2x + 2x + c (e) ex – 1 4 e–4x + c 3. (a) 1 2 lnx + c (b) 1 3 ln|3x + 2| + c

Mathematics Semester 2 STPM Answers 203 (c) – 1 2 ln|3 – 2x| + c (d) 1 2 ln|x 2 + 2x + 5| + c (e) 1 3 ln|x 3 + 1| + c (f) –ln |2 – sin x| + c (g) 1 3 ln |e3x + 1| + c (h) ln |ln x| + c (i) ln |1 + tan x| + c (j) –ln |sin x + cos x| + c 4. (a) etan x + c (b) –e 1 —x + c (c) 2ex + c (d) 1 2 ex2 + c 5. ln|1 + ex | + c, x – ln|1 + ex | + c Exercise 3.6 1. (a) u = x, dv dx = ex x ex – ex + c (b) u = x, dv dx = 3x 1 3 x sin 3x + 1 9 cos 3x + c (c) u = x, dv dx = e–2x – 1 2 x e–2x – 1 4 e–2x + c (d) u = 2x + 1, dv dx = cos x (2x + 1) sin x + 2 cos x + c (e) u = lnx, dv dx = 1 x ln x – x + c (f) u = ln x, dv dx = x – 1 4 x 2 + 1 2 x 2 lnx + c (g) u = x, dv dx = sin x sin x – x cos x + c (h) u = x, dv dx = sec2 x x tan x + ln |cos x| + c (i) u = ln3x, dv dx = 1 x ln3x – x + c 2. (a) ex (x 2 – 2x + 2) + c (b) – 1 4 e–2x (2x 2 + 2x + 1) (c) (2 – x2 ) cos x + 2x sin x + c (d) 1 2 (–ex cos x + ex sin x) + c (e) x2 2 (ln x) 2 – x2 2 ln x + x2 4 + c 3. 2 15 (1 + x) 3 —2 (3x – 2) 4. 1 15 (x – 2)5 (5x + 2) Exercise 3.7 1. (a) 114 (b) – 4 3 (c) 1 2 ln 3 (d) 5 2 –2 ln4 (e) 0.95 (f) ln 32 27 (g) 3 1 2 (h) 5 32 e4 – 1 32 (i) π2 – 4 (j) e2 (k) ln 2 2. (a) 35 72 (b) 4(7 – 3 ) (c) 2 (d) – 2 3 (e) 1 3 (f) π 48 3. (b) – 1 34 4. (b) 1 4 (5e4 – 1) 6. x e2x ; e2 + 1 4 Exercise 3.8 1. (a) 14 1 4 units2 (b) 500 3 units2 (c) 123 units2 2. (a) y x –1 0 area = 2 1 6 units2 (b) y x 0 area = 4 15 units2 (c) y x –3 0 2 area = 11 1 12 units2

204 Mathematics Semester 2 STPM Answers (d) y x –2 0 3 area = 8 1 6 units2 3. (a) 21 1 3 units2 (b) 791 units2 (c) 45 1 3 units2 4. 1 2 units2 5. y x 0 y = x 2 y = 4x – x 2 (0, 0), (2, 4); 2 2 3 units2 6. (a) y y = 2x – 3 y = x (4 –x) x 0 (–1, –5), (3, 3), 10 3 2 units2 7. y y = 9 – 3x x 0 y = 6 – x 8. (ln 3, 3), 2 ln 3 units2 Exercise 3.9 1. (a) 104 3 π units3 (b) 56 3 π units3 (c) 8π units3 2. (a) 7π units3 (b) 234π units3 (c) 18π units3 3. (a) 243 5 π units3 (b) 2 3 π units3 (c) 96 5 π units3 4. 188 15 π units3 5. (a) 2 000 3 π units3 (b) 1 408 3 π units3 6. y x 0 16 2 4 337 1 15 π units3 7. 2.15 units3 8. π(6 – 4e–1 – 2e–2) units3 9. 40 cm3 10. π1 9 4 – 2 ln 22 units3 11. 2π units3 STPM Practice 3 1. (a) 1 2 ln|2x – a| + c (b) 1 3 ln|3x 2 + 9x – 1| + c (c) ex – 2e–2x + c (d) (2x – 1)2 (x + 1) 3 + c (e) –4 – x2 + c (f) – 1 3 e–x3 + c (g) ln x 2x + 1 + c (h) x 2 – x + ln|x + 1| + c 2. (a) 1 2 ln 32 27 (b) 4 15 (c) 1 9 (1 – 4e–3) (d) 4 15 (1 + 2 ) (e) 1 2 (e2 + 3) (f) 1 2 ln 10 (g) 2 1 4 (h) ln 2 3 3. 18 ln(3) – 8 ln(2) – 5 4 4. –√25 – x2 25x + c 6. –π, π 8. π2 18 9. (a) 2 ln 7 3 (b) 4(7 – 3 ) (c) 2 – 3 2 ln 7 3

Mathematics Semester 2 STPM Answers 205 10. (b) π 6 12. 0.0587 13. 6.8 units2 15. 16 3 units2 , 8π units3 16. (a) V = π∫ 0 12 x 2 dy (b) 3 litres (c) ∫ 0 10 π(y + 4) dy = 90π = 3 4 (120π) 17. (a) y x 0 10 25 62.5 y = 10 (b) 45 – 9t 18. A(–3, 4), B(3, 4); 36π units3 19. ln 2 units2 , 1 2 π units3 20. (a) 0.5 unit2 (b) 1 3 π2 unit3 21. π 6 (10 – 3π) units3 22. 12 – 2 e 2 units2 ; 2π11 – 1 e2 2 units3 23. sin–1 1 1 2 2 24. 4 3 units2 , 16 15 π units3 26. (b) A(1, 4), B(2, 7) (c) 1 6 units2 (d) 9 5 π units3 27. 0.1811 28. √3 9 tan–1 1 3x + 2 √3 2 + c 30. (a) 6 2 4 O y x y = x2 + 2 y = 6 – x2 (b) (– 2 , 4), ( 2 , 4) (c) 4 units3 4 Differential Equations Exercise 4.1 1. (a) order: 2, degree: 1 (b) order: 2, degree: 1 (c) order: 1, degree: 2 (d) order: 2, degree: 1 (e) order: 3, degree: 2 2. (a) 10 (b) 2 (c) 2 3. y = 3x – 2 x 4. y + 2 = 2x + x 2 Exercise 4.2 1. y = x3 3 – 1 x + C 2. r = 3 ln  t + t 2 2 + C 3. y = 1 2 ln  x 2 – 1 + C 4. x = ±t 2 + 4t + C 5. y = 2(1 + Ae 4x ) 1 – Ae 4x 6. y = 1 ln x + C 7. y = ±Ae2x – 2 8. y = A(x – 2) 9. y = xAex 10. y = ±–2e–x + C 11. y = 1 3 2 x 2 ln  x – 3 4 x2 + C2 1 —3 12. y = 1 – 1 ln |x| + C 13. ln  1 + y 1 – y  – y = x 2 + C 14. y = ±Ae x2 —2 – 1 15. y = A x 16. y = ln |x – 1| + C 17. y = Ae x – 2 18. y = A(x 2 – 2) 19. y = A(x – 1) 20. y = Ae x2 + x 21. y = ±x2 + 4x – 1 22. y = ln |x| + x2 2 + 1 2 23. y = 1 2 (e 2x + 3) 24. y = x + 2 ln |x – 1| – 1 25. y = ln 1 + x 2 2  26. y = ±1 3 (2ex + 1 + 1) 27. y = ex – 2 28. y = ± ln |x|

206 Mathematics Semester 2 STPM Answers 29. y = 2(e 4x – 2) e 4x + 2 30. y = ln |–xe –x – e–x + e + 1| 31. y = ±2x (3 – 2x) 32. y = 1 3 (x 2 + 9) 3 —2 33. y = ln 2 2 – x  34. y3 3 – y2 2 = x2 2 – ln |x| – 1 2 35. y = 2(x 2 – 1) x 2 + 1 Exercise 4.3 1. (a) y = 1 3 (1 + x2 ) + A(1 + x2 ) – 1 —2 (b) y = 1 2 x e–x – 1 4 e–x + Ae–3x (c) y = x + Ax–1 (d) y = 2x2 – 4x + 4 + Ae–x (e) y = e3x + Ae–5x 2. (a) y = – 1 2 x + 1 2 x3 + x4 (b) y = 1 2 3(1 + x2 ) + (1 + x2 ) – 1 —2 4 (c) y = e5x (d) y = – 1 2 + 3 2 ex2 – 1 Exercise 4.4 1. (a) y = x ln |x | + Cx (b) y = 1 3 x + A x2 (c) y = ±xAx2 – 1 (d) y = ±x2 ln|x| + C (e) y = ±x2 2 – A x2 (f) ln |xy | + y 2 2x 2 + C = 0 2. ln  y 2  = x2 2y2 – 1 8 3. (y – x) 2 2 + 5y – x + C = 0 4. (x + y) 2 2 + y + C = 0 5. x – y = ln |2x – y + 7| + C 6. y 2 = (A – 4x)x 7. v3 = ln ex3 8. v = A(x – 1)3 9. dz dx – 2z = –2x 10. y2 = 4x(a – x) 11. y2 = 1 x(–ln x + 1) Exercise 4.5 1. dh dt = k(H – h) , h = H11 – 9 10e t —T In 8 —9 2 , t = 56 days 2. dn dt = kn , n = 1 4 1 N T t + 2N2 2 , 2T 3. dx dt = –k(x – 30), 52.7 minutes, 64.8°C 4. dn dt = λn, λ = ln2 10 (a) 4 N (b) 15.8 5. dm dt = k(M – m), 78.4 g 6. dx dt = αx – β (a) 0.805 (b) 680 7. dv dt = –(kv + c), v = 1 k {(kV + c)e –kt – c}, t = 1 k ln 11 + k c V 2 8. 171 days, 61.47% 9. 4.74 minutes 10. dx dt = –k(a – x)(b – x) (i) x = a2 kt 1 – akt (ii) x = ab(e –(a – b)kt – 1) ae –(a – b)kt – b 11. dx dt = 1 A (W – kx), x = W k (1 – e– k —A t ) 12. (a) dv dt = – k v2 (b) v = u2 – 2ct (c) v = 3 u3 – 3cx 13. (b) 1.25 (c) No birth of turtle (d) x = 100 1 + 3e–0.05t x O t 100 25 (e) (i) 60 turtles (ii) 22 years 14. (a) dx dt = –kx (b) x = Ae–kt (c) x = 1 2 3 2 t Q (d) x = 2 11 27 Q STPM Practice 4 1. y 2 = x2 ln | x | 2. y = ln x 4 – x  3. y = 2x 2 – x2

Mathematics Semester 2 STPM Answers 207 4. y = ±e x 2 – e 2 2ex 5. y = sin–1 1 ln x + c x 2 6. (2x – y) 2 (x + y) = 16 7. y = x cos–1 (x–2) 2 8. (a) y = 40 (b) t = 53 667 hours 9. dx dt = kx (a) x = 90 017 (b) x = 1 808 042 1.07 hours 10. dy dx = y x , y = 2x, dx dt = –2x 3 , x = 1 1 + 4t 11. v = 100e –kt, d = 131.0 m 12. dx dt = kx, 6 510 thousands 13. 266 seconds 14. 73% 15. 3 hours 48 minutes 16. y = 1 4 (x 4 —3 – 1) 17. (c) 5 days 18. y = sin–1 (A ln x) 19. y = [x4 (4 ln x + 1)] 1 —4 20. ln y x 2 = 1 2 1 x2 y2 – 12 21. y = x sin–1 x 22. y = x – 3 x 23. y = x – 2 ln|2x + y + 3| + 2 ln 6 24. y = c – cos 2x cos x 25. y = Ax(y + 3) – 3 26. y = x2 ln x – x2 + 4x 5 Maclaurin Series Exercise 5.1 1. (a) 1 – 1 2! x2 + 1 4! x4 – 1 6! x6 + ··· (b) 1 + x + x2 + x3 + ··· (c) 1 + nx + n(n – 1) 2! x2 + n(n – 1)(n – 2) 3! x3 + ··· 2. (a) 1 – 5x + 25 2 x2 – 125 6 x3 + ···, (–∞, ∞) (b) –2x + 1 3 x3 – 1 60 x5 + 1 2 520 x7 + ···, (–∞, ∞) (c) 1 – 1 2 x + 1 24 x2 – 1 720 x3 + ···, (–∞, ∞) (d) 2 3 + 2 9 x + 2 27 x2 + 2 81 x3 + ···, (–3, 3) (e) x2 – x3 + x4 – x5 + ···, (–1, 1) 3. sin–1 x = x + 1 6 x3 + 3 40 x5 + 5 112 x7 + ··· = ∞ ∑ n=0 (2n)! x2n + 1 (2n n!)(2n + 1) , (–1, 1) 4. 1 + x ln a + (x ln a) 2 2! + (x ln a) 3 3! + ··· Exercise 5.2 1. (a) 1 4 – 1 16 x2 + 1 64 x4 – 1 256x8 + ···, (–2, 2) (b) 1 + x3 + x6 + x9 + x12 + ···, (–1, 1) (c) 1 – 2x2 + 2x4 – 4 3 x6 + ···, (–∞, ∞) (d) 2x – 4 3 x3 + 4 15 x5 – 8 315x7 + ···, (–∞, ∞) (e) x2 – 1 3 x6 + 1 5 x10 – 1 7 x14 + ···, [–1, 1] (f) 2x4 – 2x8 + 8 3 x12 – 4x16 + ···, 1– 1 42 , 1 42 2 (g) x4 – 1 6 x12 + 1 120x20 – 1 5 040 x28 + ···, (–∞, ∞) 2. 1 – 3x + 6x2 – 10x3 + ···, (–1, 1) Exercise 5.3 1. (a) –x2 – 1 2 x4 – 1 3 x6 – ··· (b) 2(x + 1 3 x3 + 1 5 x5 + ··· (c) 7x + 1 2 x2 + 43 3 x3 + ··· (d) x + 1 6 x3 + 1 120x5 + 1 5 040 x7 + ··· (e) 1 + 1 2 x2 + 1 24 x4 + 1 720x6 + ··· 2. (a) 1 – 3 2 x2 + 25 24 x4 – 331 720 x6 + ··· (b) x3 – x4 + 1 2 x5 – 1 6 x6 + ··· (c) x + 1 2 x2 – 1 6 x3 – 1 6 x4 + ··· (d) 1 2 x3 – 1 4 x4 + 1 8 x5 – 1 16 x6 + ··· (e) x – x2 + 1 3 x3 – 1 30 x5 + ··· 3. x2 – 1 3 x4 + 2 45 x6 – 1 315x8 4. 1 – 1 4 x2 – 1 96 x4 + ··· 5. 1 + ax + 1 2 (a2 – b2 )x2 + 1 6 a(a2 – 3b2 )x3 + ··· Exercise 5.4 1. (a) 1 (b) –1 (c) 1 6 (d) 0 (e) 3 2 (f) – 1 4 (g) 1 3 (h) – 1 6 (i) 5

208 Mathematics Semester 2 STPM Answers STPM Practice 5 1. (a) 2 + 1 4 x2 – 19 192x4 + ··· (b) 3 + 4 3 x + 10 27 x2 + ··· (c) x2 + 2 3 x4 + 17 45 x6 + ··· (d) x2 + 1 3 x4 + 8 45 x6 + ··· 2. x – 1 2 x2 + 2 3 x3 + …; 1 4 3. 1 + x + 1 2 x2 + ··· 4. (a) x + 1 2 x2 – 1 6 x3 + … (b) x + 1 2 x2 + 1 3 x3 + … 5. 1 – 3x + 9x2 2 – 27 6 x3 + …; 1 + 2x + 2x2 + 8 3 x3 + …; 1 – x + 1 2 x2 + 7 6 x3 + … 6. x – x3 3 + x5 5 + …; x + 2 3 x3 + 11 30 x5 + … 7. –2, 48; 1 – 2x + 2x2 – 2x4 + ··· 8. x2 + x4 3 + ··· 9. 1 – 1 4 x2 – 1 96 x4 + … 10. (b) 3x – 6x2 + 3 2 x3 + … (c) (i) 3x – 9x2 + 21 2 x3 + … (ii) 3 11. 1 – 2x + 3x2 – 4x3 + ···; (a) –x + 7 2 x2 – 31 3 x3 + ···; – 1 2 , x , 1 2 (b) –0.0148 12. n = 9 4 13. 1 – 3x + 9 2 x2 – 27 6x3 + …; x + 1 2 x3 + …; x – 3x2 + 29 6 x3 + … 14. 1 – 8x2 + 32 3 x4 + 4 096 720 x6 + …; 1 3 15. 2x + 2x3 3 + …; –1 , x , 1 (a) 1 + 2x + 2x2 + 2x3 (b) 0.412 16. 1.0164 17. (b) 1 2 x2 + 1 12 x4 (c) 0.011 18. (a) 1 9 (b) 3 19. (a) 1 6 (b) – 1 2 20. – 1 6 21. 1 9 22. e(1 + x + x2 2 + …) (a) 1 3 e (b) 0.150e 23. – 1 2 24. (b) x2 + 1 3 x4 + …; –1 , x , 1 (c) 3.122 6 Numerical Methods Exercise 6.1 1. (c) (i) 1.435 (ii) 1.841 (iii) diverges 2. (a) x3 – 2x2 + 1 = 0 (b) 2.160, 2.214, 2.204, 2.206 3. (a) x3 + 2x2 – 5x – 1 = 0 (b) 0.185 4. (a) 0.236 (b) x1 is not defined (c) –4.236 5. –1.67 Exercise 6.2 1. (a) 0.20 (b) 0.22 (c) 0.64 2. 2.51 3. 1.035 4. n = 2, n + 1 = 3, 2.20, 2.19 5. 2.15 y x y = 1 x _ y = sin πx 1 2 (2.13, 0.46) (2.89, 0.35) y = 2, 1, 2 3 , 1 2 , 2 5 y = 1, 0, –1, 0, 1 6. –1.21 7. 2.999 8. 3.104 9. 2, 1.90 10. 1 , a , 2, 4 , b , 5, 1.450 y x (1, 0)(2, 0)

Mathematics Semester 2 STPM Answers 209 11. 3 y x (0, 0) (–2.2, 0) 12. x = 2, y = –2, 2 real roots, 1.67 13. n = 1, 1.31 Exercise 6.3 1. (a) 2.5643, over (b) 5.1667, over (c) 0.9436, over 2. (a) 1.5 (b) 0.92 (c) 10.58 (d) 1.03 (e) 8.827 3. 3.1312 4. (b) 12.6598, too small 5. 0.879 STPM Practice 6 1. (a) 0.21311 (b) A = 2; 1.1231 (c) Because x – 1 —2 e–2x is not define at x = 0. 2. 2.208 3. 0 1 3 –3 y y = e y = 3 sin 2x x – 4 π – 2 1 – x 2 π 3 – 4 π π 1.237 4. 2.145 5. 2.926 6. x2 = x1 – (x1 + 1)5 – (x1 + 2)3 – 4 5(x1 + 1)4 – 3(x1 + 2)2 , 0.981 7. (a) 2x ln 2 (b) –2.82, 2.45 (c) 2.445 (e) 2.03 8. xn + 1 = 1 8 (3 – xn 3 ); 0.369 9. y y = 4 y = 2ex x 0 2 1 2 2 3 4 5 6 46 –– – – x2 0.614 10. 1.9469 11. 0.945 13. k = 10; f(x) = (x – 2)(2x – 5)(x + 1); 18 14. (a) 0.70 (b) 17 15. 2.419 16. 1.910 17. 0 1 y y = e–x y = 3x – 4 x –4 4 – 3 y = e–x and y = 3x – 4 intersect at only one point. 1.414 18. 4.552 STPM Model Paper (954/2) 1. f(x) = 1 – x 3 + x is defined for 1 – x 3 + x > 0. (a) –3 1 (1 – x) + + – (3 + x) – + + 1 – x 3 + x – + – For 1 – x 3 + x > 0, –3 , x < 1 → (–3, 1]

210 Mathematics Semester 2 STPM Answers (b) Let c [ (–3, 1) lim x→c f(x) = lim x→c 1 – x 3 + x = 1 – c 3 + c = f(c) \ f is continuous on the interval (–3, 1). At x = 1; lim x→1– f(x) = lim x→1– 1 – x 3 + x = 0 = f(1) \ f(x) is continuous from the left at x = 1. Hence, f(x) is continuous on the interval (–3, 1]. 2. (a) Let u = ex when x = 0, u = e0 = 1 du dx = ex when x = 1, u = e1 = e \ du = ex dx ∫0 1 2ex ex + 2e–x dx = ∫1 e 2 u + 2 u du = ∫1 e 2u u2 + 2 du = [ln|u2 + 2|] e 1 = ln|2 + e2 | – ln 3 ∫0 1 2ex ex + 2e–x dx = ln u 2 + e2 3 u (b) 1 0 x5 ex3 + 1 dx Let u = x3 and dV dx dx = x2 ex3+1 dx du dx = 3x2 \ V = 1 3 ex3 + 1 du = 3x2 dx 1 0 x5 ex3 + 1 dx = 3x3 1 1 3 ex3 + 124 1 0 – 1 0 1 3 ex3 + 1(3x2 )dx = 1 1 3 e2 2 – 1 3 3ex3 + 14 1 0 = 1 3 e2 – 1 3 3ex3 + 14 1 0 = 1 3 e2 – 1 3 [e2 – e1 ] \ 1 0 x5 ex3 + 1 = 1 3 e 3. (a) y 2 = x – 2 → x = y 2 + 2 Volume generated = π 3 0 x2 dy = π 3 0 (y2 + 2)2 dy = π 3 0 (y4 + 4y2 + 4) dy = π3 1 5 y5 + 4 3 y3 + 4y4 3 0 = π1 243 5 + 36 + 122 Volume generared = 483 5 π unit3 (b) 0 3 y y2 = x – 2 x 2 11 Volume generated = π(3)2 (11) – π 11 2 (x – 2) dx = 99π – π3 x2 2 – 2x4 11 2 = 99π – π31 121 2 – 222 – (2 – 4)4 = 99π – π31 121 2 – 202 Volume generated = 59.5π unit3 4. Given x2 dy dx + xy – y2 = 0 … (1) u = xy Differentiate with respect to x, du dx = x dy dx + y x dy dx = du dx – y … (2) Subtitute (2) into (1), x( du dx – y) + xy – y2 = 0 x du dx – y2 = 0 x3 du dx – x2 y2 = 0 \ x3 du dx = u2 du u2 = dx x3 → – 1 u = – 1 2x2 + c When y = x = 1, u = xy = 1 – 1 1 = – 1 2 + c c = – 1 2 \ 1 u = 1 2x2 + 1 2 → 1 xy = 1 2x2 + 1 2 1 y = 1 2 1x + 1 x 2 y = 2x 1 + x2 5. f(x) = e√(1 + x) f '(x) = e√(1 + x) · 1 2 (1 + x) – 1 —2 2√(1 + x) · f '(x) = e√(1 + x) Differentiate both sides with respect to x 2√(1 + x) · f "(x) + 2f '(x)3 1 2 (1 + x) – 1 —2 4 = 1 2 (1 + x) – 1 —2 e√(1 + x) 4(1 + x)f "(x) + 2f '(x) = e√(1 + x) → 4(1 + x)f "(x) + 2f '(x) = f(x) (shown)

Mathematics Semester 2 STPM Answers 211 Differentiate both sides again with respect to x: 4(1 + x)f "'(x) + 4f "(x) + 2f "(x) = f '(x) When x = 0; f(0) = e f'(0) = 1 2 e f"(0) = 0 f"'(0) = 1 8 e By Maclaurin theorem, f(x) = f(0) + f '(0)x + f "(0) x2 2! + f "'(0) · x3 3! + … = e + 1 1 2 e2x + 0 + 1 1 6 2x3 1 1 8 e2 + … \ f(x) = e11 + 1 2 x + 1 48x3 + …2 6. 0 –1 y g(x) = –(x + 1) f(x) = In (x – 1) x –1 2 Let f(x) = ln(x – 1) + x + 1 f(1.1) = ln(1.1 – 1) + 1.1 + 1 = –0.2026 , 0 f(2) = ln(2 – 1) + 2 + 1= 3 . 0 Since f(1.1) and f(2) have opposite signs, f(x) has one real root in the interval [1.1, 2]. f(x) = ln(x – 1) + x + 1 f '(x) = 1 x – 1 + 1 x1 = x0 – f(x0 ) f '(x0 ) x1 = 1.1 – f(1.1) f '(1.1) = 1.1 – ln(1.1 – 1) + 1.1 + 1 1 1.1 – 1 + 1 x1 = 1.118 x2 = 1.118 – ln(1.118 – 1) + 1.118 + 1 1 1.118 – 1 + 1 x2 = 1.120 x3 = 1.120 The root is 1.120, correct to three decimal places. 7. dy dx + 2xy = 2x(1 + x2 ) Integrating factor = e∫ 2x dx = ex2 Multiply both sides by ex2 . ex2 dy dx + 2xy · ex2 = 2x · ex2 (1 + x2 ) d(yex2 ) dx = 2xex2 (1 + x2 ) Integrating both sides with respect to x: yex2 = (2xex2 + 2x3 ex2 ) dx yex2 = 2xex2 dx + 2x3 ex2 dx yex2 = ex2 + (x2 ex2 – 2xex2 dx) yex2 = ex2 + x2 ex2 – ex2 + c where c is the arbitrary constant. \ yex2 = x2 ex2 + c \ y = x2 + ce–x2 (a) At stationary point, dy dx = 2x – 2xce–x2 dy dx = 2x(1 – ce–x2 ) = 0 ce–x2 = 1 e–x2 = 1 c ex2 = c x2 = ln c x = ±√ln c When c < 1, ln c < 0 → dy dx = 0 for 1 value of x, \ x = 0 Hence, the curve has only 1 stationary point when c < 1. When c . 1, ln c . 0 → dy dx = 0 for 3 values of x, \ x = 0 and x = ±√ln c Hence, the curve has three stationary points when c . 1. (b) y = x + ce–x2 If c . 1, x → ±∞, y → (x2 ) + If c , 0, x → ±∞, y → (x2 ) – When c = 2; y = x2 + 2e–x2 (c . 1) When c = –1; y = x2 – e–x2 (c , 1) 0 –1 y f(x) = x2 + 2e–x2 (c > 1) (c < 1) g(x) = x2 – e–x 2 x 2 –0.75 0.75

212 Mathematics Semester 2 STPM Answers 8. (a) dv dt = 3(5 – v) (i) dv 5 – v = 3 dt –ln(5 – v) = 3t + c When t = 0, v = 0; \ –ln 5 = c –ln(5 – v) = 3t – ln 5 ln(5 – v) – ln 5 = –3t ln1 5 – v 5 2 = –3t 5 – v 5 = e–3t 5 – v = 5e–3t \ v = 5(1 – e–3t ) v = ds dt = 5(1 – e–3t ) ds = 5 (1 – e–3t ) dt s = 51t + e–3t 3 2 + c When t = 0, s = 0; 0 = 510 + 1 3 2 + c \ c = – 5 3 \ s = 51t + e–3t 3 2 – 5 3 (ii) When t = 4 s; s = 514 + e–12 3 2 – 5 3 s = 20 – 5 3 + 5 3e12 s = 18 1 3 m, since 5 3e12 ≈ 0 (b) Volume of a right circular cylinder, πr 2 h = 432π \ h = 432 r 2 (i) Total surface area of the cylinder s = 2πr 2 + 2πrh s = 2πr 2 + 2πr1 432 r 2 2 \ s = 2πr 2 + 864 r π (ii) ds dr = 4πr – 864 r 2 π For stationary value, ds dr = 0 4πr – 864 r 2 π = 0 4π r 2 – (r 3 – 216) = 0 r 3 = 216 r = 6 cm d2 s dr 2 = 4π + 1 728 r 3 π When r = 6, d2 s dr 2 . 0 \ r = 6 cm makes the area a minimum.

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