mathematics t coursework stpm sem 2

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

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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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STPM Mathematics (T) Semester 2

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Stpm math t coursework 2021-2022 (semester 1, 2 & 3), stpm math t coursework 2022 (semester 3).

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What will you get? 1) Introduction: 5 pages 2) Methodology: 2 pages 3) Results: 17 page 4) Conclusion: 2 pages Total: 26 pages

To get the full solutions, you can buy it from us . just follow the following steps: 1) bank in rm50 to cimb bank account: 8600470918, account name: pusat tuisyen teman pintar. 2) sms/whatapps your payment receipt/transaction slip to 0163551988. tell us you need semester 2 math t full solutions. let us know your email address as well. 3) we will email the full solution to you within 24 hours and text you once the email is sent., 4)  special offer :  bank in rm40 to cimb bank account: 8600470918, account name: pusat tuisyen teman pintar if you can press"subscribe" on our youtube channel (diagram 1 below). steps: open youtube and search lee chee kin eric and then click subscribe. snap a photo and send to 0163551988 through whatapps. then  sms/whatapps your payment receipt/transaction slip to 0163551988. tell us you need semester 2 math t full solutions. let us know your email address as well. we will email the full solution to you within 24 hours and text you once the email is sent.  alternative: click the link below to find the site in youtube that you can click subscibe. youtube link to click subscribe click subsribe as below:, 5) bundle deal:  purchase assignment solution together with online video lessons for semester 2 math t (all chapters) by eric lee at promotional price of rm80 (assignment +online video lessons) if you can press like and subscribe to the sample video (click the sample video below and press like and subscribe, screenshot it and send to 0163551988 through whatapps) (without like and subscription the bundle deal is sold at rm90). you will get a list of links through email. the links will channel you to youtube for the online lessons.  if you wanted to buy the online video lessons without the assignment, the price is rm70. sample video:  7. limits and continuity 6) special offer 2: share and save: if you have assignment solution for other subjects (example: chemistry, physics etc, from mpm only).  in term 2 2022 to share. you may send a message to 0163551988 to find out whether the solution rebate is still available (we are looking for 1 solution for each subject). if it is available, you may send it through whatapps to 0163551988. checking will be done in 24 hours. rm40 rebate will be given so that math t assignment can be bought at no cost. this special offer comes at first come first serve basis.  why buy our math t assignment as your reference  save your time on this assignment. quickly settle it and focus on your exam. please do not copy 100% of the sample answer. the sample answer serves as reference for students., stpm math t coursework 2021-2022 (semester 1).

What will you get? 1) Introduction: 3 pages 2) Methodology: 4 pages 3) Results: 10 page 4) Conclusion: 2 pages Total: 19 pages

To get the full solutions, you can buy it from us . just follow the following steps: 1) bank in rm50 to cimb bank account: 8600470918, account name: pusat tuisyen teman pintar. 2) sms/whatapps your payment receipt/transaction slip to 0163551988. tell us you need semester 1 math t full solutions. let us know your email address as well. 3) we will email the full solution to you within 24 hours and text you once the email is sent., 4)  special offer :  bank in rm40 to cimb bank account: 8600470918, account name: pusat tuisyen teman pintar if you can "like/follow" on facebook (diagram 1 below). steps: login to facebook and search pro a tuition centre kajang. click "like/follow" and snap a photo and send to 0163551988 through whatapps. then  sms/whatapps your payment receipt/transaction slip to 0163551988. tell us you need semester 1 math t full solutions. let us know your email address as well. we will email the full solution to you within 24 hours and text you once the email is sent.  facebook link: https://www.facebook.com/proatuition like this (diagram 1) or follow us 5) bundle deal: purchase assignment solution together with online video lessons for semester 1 math t (all chapters) by eric lee at promotional price of rm80 (assignment +online video lessons) if you can press like and subscribe to the sample video (click the sample video below and press like and subscribe, screenshot it and send to 0163551988 through whatapps) (without like and subscription the bundle deal is sold at rm90). you will get a list of links through email. the links will channel you to youtube for the online lessons.  if you wanted to buy the online video lessons without the assignment, the price is rm70. details on the total duration of the videos in each chapters  1) functions: 7 hours 41 minutes 34 seconds 2) sequences and series: 3 hours 16 minutes 13 seconds 3) matrices: 2 hours 31 minutes 7 seconds  4) complex numbers: 2 hours 3 minutes 30 seconds 5) analytical geometry: 2 hours 50 minutes 9 seconds 6) vectors: 3 hours 43 minutes 10 seconds total duration: 19 hours 15 minutes 34 seconds sample video:  1.1 functions click “like” and “subscribe”: * the online video lessons exclude asking question or asking solution for any particular question., why buy our math t assignment as your reference  save your time on this assignment. quickly settle it and focus on your exam. please do not copy 100% of the sample answer. the sample answer serves as reference for students., no comments:, post a comment.

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STPM MATHEMATICS T

(NEW SYLLABUS - STARTING 2012)

MATHEMATICS T

Sequences and series.

"Right clip" then "Open link in new tab": ENABLE YOU TO GET ENLARGED IMAGE

To print out: SAVE IMAGE TO PAINT AND INVERT THE COLOR

Solutions of FOCUS ON EXAM is available at the website of the Publisher: cw.oxfordfajar.com.my

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CHU MATHS T OFFICE

Sunday 9 april 2023, stpm past year papers term 3.

 STPM PAST YEAR PAPERS TERM 3

2022-3-ACTUAL

2022-3-ACTUAL-SUGGESTED ANSWERS

2021-3-ACTUAL

2021-3-ACTUAL-SUGGESTED ANSWERS

2020-3-ACTUAL

2020-3-ACTUAL-SUGGESTED ANSWERS

2019-3-ACTUAL

2019-3-ACTUAL-SUGGESTED ANSWERS

Thursday 1 September 2022

Stpm past year papers term 2.

 STPM PAST YEAR PAPERS TERM 2

2023-2-ULANGAN

2023-2-ULANGAN-SUGGESTED ANSWERS

2023-2-ACTUAL

2023-2-ACTUAL-SUGGESTED ANSWERS

2022-2-ULANGAN

2022-2-ULANGAN-SUGGESTED ANSWERS

2022-2-ACTUAL

2022-2-ACTUAL-SUGGESTED ANSWERS

2021-2-ACTUAL

2021-2-ACTUAL-SUGGESTED ANSWERS

2021-2-ULANGAN

2021-2-ULANGAN-SUGGESTED ANSWERS

2020-2-ACTUAL

2020-2-ACTUAL-SUGGESTED ANSWERS

2020-2-ULANGAN

2020-2-ULANGAN-SUGGESTED ANSWERS

Wednesday 25 May 2022

Term 2 past year questions (chapter wise).

Updated until STPM 2023 REPEAT 2

  PAST YEAR QUESTIONS-CHAPTER 7-QUESTIONS  

Friday 22 April 2022

Stpm maths t paper 3 trial papers collecction 1.

 2021-3-954 PAPERS COLLECTION 1   

Saturday 9 April 2022

Term 3 past year questions (chapter wise).

  PAST YEAR QUESTIONS-CHAPTER 13-QUESTIONS  

Tuesday 25 January 2022

Term 1 maths t exam papers with solutions -ox (8 sets).

 TERM 1 EXAM PAPERS 954-1 OX (8 SETS) 

Saturday 18 December 2021

Kalender persekolahan tahun 2022/2023.

 STPM PAST YEAR PAPERS TERM 3 2022-3-ACTUAL 2022-3-ACTUAL-SUGGESTED ANSWERS 2021-3-ACTUAL 2021-3-ACTUAL-SUGGESTED ANSWERS 2020-3-ACTUAL 2020...

• TERM 2 PAST YEAR QUESTIONS (CHAPTER WISE) Updated until STPM 2023 REPEAT 2   PAST YEAR QUESTIONS-CHAPTER 7-QUESTIONS   PAST YEAR QUESTIONS-CHAPTER 7-SOLUTIONS   PAST YEAR QUESTIONS-C...
• TERM 1 PAST YEAR QUESTIONS (CHAPTER WISE) PAST YEAR QUESTIONS-CHAPTER 1-QUESTIONS   PAST YEAR QUESTIONS-CHAPTER 1-SOLUTIONS   PAST YEAR QUESTIONS-CHAPTER 2-QUESTIONS  PAST YEAR QUEST...
• STPM PAST YEAR PAPERS TERM 2  STPM PAST YEAR PAPERS TERM 2 2023-2-ULANGAN 2023-2-ULANGAN-SUGGESTED ANSWERS 2023-2-ACTUAL 2023-2-ACTUAL-SUGGESTED ANSWERS 2022-2-ULANGAN 2...

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Question: MATHEMATICS (T) COURSEWORK STPM 2021 2.4 Assignments Assignment A: Mathematical Investigation The general equation for any conic section is Ar +By+y+De+ Ey+F=0 where 4, B, C, D, E and F are constants. MATHEMATICS (T) COURSEWORK STPM 2021 1 3. For the three cases below, determine the angle, through which the conic with each equation should be rotated and

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To start transforming the general conic equation into its standard form, group the and terms and then apply the method of completing the square for both variables.

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