stpm assignment mathematics t 2022

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Wednesday 25 may 2022, term 2 past year questions (chapter wise).

Updated until STPM 2023 REPEAT 2

  PAST YEAR QUESTIONS-CHAPTER 7-QUESTIONS  

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 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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Tuesday 16 August 2022

Stpm math t coursework 2022-2023 (semester 1, 2 and 3)), stpm math t coursework 2022-2023 (semester 3).

1) Introduction: 3 pages 2) Methodology: 3 pages 3) Results: 8 pages (+original Excel /Apple Number file for editing purpose) 4) Conclusion: 2 pages Total: 16 pages (What if your raw data is different from mine? Don’t worry. You can use the Excel/Number file given to change the data and everything will be calculated automatically)

To get the full solutions, you can buy it from us . here are the 4 options: a) assignment full solution + assignment result explanation video + online video lessons for semester 3 math t (all chapters): rm80 b)  assignment full solution + assignment result explanation video: rm60 c) assignment full solution +  online video lessons for semester 3 math t (all chapters): rm70 d) assignment full solution only: rm50, ** the assignment full solution and the online video lessons excludes explanation (or asking question) for any part.  click this for the payment details:  payment link (optional)   special offer : if you can press “follow” on our blog (www.proatuition.blogspot.com, follow button is on the right. if you can’t see it, scroll down the page and click web version and you should be able to find the “follow” button), then  screenshot it and send to  0163551988   through whatapps), then you can get rm10 discount on the price above. click the “follow” button as below: steps after you made payment:  1) send us message using whatsapps ( 0163551988 ) and tell us which options  you are buying (for example: math t assignment full solution,  semester 3 , option a) 2) send us your email address 3) send us payment receipt/screenshot. 4) send us the screenshot of “follow” to enjoy rm10 discount (optional). we shall send you what you purchase within 12 hours. details about assignment result explanation video: you will get a link. the link will channel you to the video through youtube. the duration of the video is about 45 minutes to 120 minutes.  details about  online video lessons for semester 3 math t (all chapters): you will get a list of links through email. the links will channel you to youtube for the online lessons.  sample video:   data description 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., we need your support to continue our service. please support us..

Hints: 

What will you get 1) introduction: 7 pages 2) methodology: 3 pages 3) results: 49 page 4) conclusion: 2 pages total: 61 pages, 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 assignment result explanation video  and online video lessons for semester 2 math t (all chapters) by eric lee. price: a) assignment solution + assignment result explanation video (57minutes 12 seconds) + online video lessons for semester 2 math t (all chapters): rm90 b)  assignment solution + assignment result explanation video (57 minutes 12 sec): rm60 c) assignment solution +  online video lessons for semester 2 math t (all chapters): rm80 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), then you can get rm10 discount on the price listed in bundle deal above (for a: rm80. for b: rm50. for c: rm70) 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) limits and continuity: 57 minutes 37 seconds 2) differentiation: 4 hours 55 minutes 8 seconds 3) integration: 3 hours 13 minutes 49 seconds  4) differential equations : 1 hours 12 minutes 16 seconds 5) maclaurin series: 1hour 44 minutes 17 seconds 6) numerical methods: 1hour 10 minutes 44 seconds total duration: 13 hours 13 minutes 51 seconds sample video:  limits and continuity 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., stpm math t coursework 2022-2023 (semester 1).

Task 1: a) k=1, p=1, q=4

b) -(x-1)^2+4

d) (-1,0), (3,0)

e) (1,-2), (1,2)

i) a) k=2, p=-1, q=3

b) -2(x+1)^2+3

c) -2.2247, 0.2247

d) (-2.2247,0), (0.2247,0)

e) (-1, -1.2247), (-1, 1.2247)

ii) a) k=1, p=2, q=3

b) -(x-2)^2+3

c) 0.2679, 3.7321

d) (0.2679,0), (3.7321, 0)

e) (2, -1.732), (2, 1.732)

iii) a) k=-1, p=2, q=-3

b) -(x-2)^2-3

c) 0..2679, 3,7321

d) (0.2679,0), (3.7321,0)

iv) a) k=3, p=1/3,q=2/3

b) -3(x-1/3)^2+2/3

c) -0.1381, 0.8047

d) (-0.1381,0), (0.8047, 0)

e) (1/3, -0.4714), (1/3, 0.4714)

v) a) k=-1, p=2, q=-5

b) (x-2)^2-5

c) -0.2361, 4.2361

d) (-0.2361,0), (4.2361,0)

e) (2, -2.2361), (2, 2.2361)

(Table): no hint

Conclusion: ..."same"...

Task 2: 

a) 4+i, 4-i

b) beta=25, 4+i, 4-i.

c) graph, use Microsoft Excel and ..."same"...

Repeat for other functions: no hint

What will you get? 1) Introduction: 2 pages 2) Methodology: 2 pages 3) Results: 23 pages+ 8 excel files 4) Conclusion: 2 pages Total: 29 pages + 8 excel files

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 assignment result explanation video  and online video lessons for semester 1 math t (all chapters) by eric lee. price: a) assignment solution + assignment result explanation video + online video lessons for semester 1 math t (all chapters): rm90 b)  assignment solution + assignment result explanation video: rm60 c) assignment solution +  online video lessons for semester 1 math t (all chapters): rm80 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), then you can get rm10 discount on the price listed in bundle deal above (for a: rm80. for b: rm50. for c: rm70) 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., no comments:, post a comment.

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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.

FEATURES W.M: RM31.95 / E.M: RM32.95 CC039332b ISBN: 978-967-2779-96-4 SEMESTER 2 SPRE-U Mathematics (T) PRE-U STPM Text Mathematics (T) Semester 2 specially designed for students who are sitting for the STPM examination. The comprehensive notes and practices are based on the latest syllabus and exam format set by Majlis Peperiksaan Malaysia. This book will provide you with the necessary skills and strategies to excel in the subject. Our Pre-U & STPM Titles: › Success with MUET › MUET My Way › Pengajian Am Semester 1, 2, & 3 › Bahasa Melayu Semester 1, 2, & 3 › Biology Semester 1, 2, & 3 › Physics Semester 1, 2, & 3 › Chemistry Semester 1, 2, & 3 › Mathematics (T) Semester 1, 2, & 3 › Sejarah Semester 1, 2, & 3 › Geografi Semester 1, 2, & 3 › Ekonomi Semester 1, 2, & 3 › Pengajian Perniagaan Semester 1, 2, & 3 Purchase eBook here! ■ Comprehensive Notes and Practices ■ Useful Features like Bilingual Keywords, Learning Outcomes and Worked Examples ■ Summary ■ STPM Practices ■ STPM Model Paper Semester 2 ■ Complete Answers PELANGI STPM Text

Understanding the Latest STPM Mathematics syllabus

1 June 2023

6 minutes to read

• 01. What is the Difference Between STPM Mathematics (T) & STPM Mathematics (M)?
• 02. STPM Maths (T) Syllabus
• 03. STPM Maths (M) Syllabus
• 04. Benefits of Choosing STPM

The Sijil Tinngi Persekolahan Malaysia (STPM), or Malaysian Higher School Certificate, is a pre-university program available to students in Malaysia who have completed SPM. It is one of several options for students to consider before pursuing a university degree. The Malaysian Examination Council administers the program, which implemented a new modular format in 2012. As part of this format, a UK's Cambridge Assessment representative will oversee and endorse the results to maintain the program's high standards and quality.

Although many students may find mathematics dull, excessively challenging, and irrelevant to their everyday lives, it is important to recognise that mathematics is actually crucial in everyone's life. It enhances an individual's intellectual and personal growth and fosters rational and systematic reasoning.

If you are a student preparing for the STPM examinations this year and want to learn more about mathematics, you have come to the right place. This article provides an overview of the mathematics syllabus for STPM, helps you identify the difference between the two mathematics papers and discuss the benefits of pursuing STPM.

What is the Difference Between STPM Mathematics (T) & STPM Mathematics (M)?

The STPM program offers two different mathematics subjects, STPM Maths (T) and STPM Maths (M). Although both subjects are mathematics-based, they have some notable differences.

STPM Maths (T) is a theoretical mathematics subject that focuses on the abstract concepts and principles of mathematics. It covers topics such as calculus, algebra, geometry, and probability theory. It is a compulsory subject for science stream students. The subject requires students to have a strong foundation in mathematics as it involves complex problem-solving and mathematical reasoning. Students who excel in this subject often have a natural aptitude for mathematics and enjoy the challenge of solving complex problems.

On the other hand, STPM Maths (M) is a more practical-oriented mathematics subject that focuses on the application of mathematical concepts and principles in real-world situations. It is an optional subject taken by Art stream students. It covers topics such as statistics, mechanics, and decision mathematics. This subject is designed for students who want to pursue careers in fields such as finance, and science, where the ability to apply mathematical concepts to practical problems is essential.

While both subjects are challenging, STPM Maths (T) is considered to be more difficult than STPM Maths (M) due to its theoretical nature and the level of mathematical abstraction involved. However, STPM Maths (M) is still challenging, requiring strong mathematical skills and a good understanding of mathematical concepts. There are plenty of tips and resources to help you master them all!

In terms of university admission requirements, some universities may require students to take specific mathematics subjects, depending on the course they are applying for.

STPM Maths (T) Syllabus

The Mathematics (T) syllabus has been created to establish a structure for a pre-university program that allows students to enhance their comprehension of mathematical ideas and reasoning, as well as acquire proficiency in tackling problems and applying mathematical principles to areas related to science and technology.

If you have opted for the Science stream in your Form 6 studies, then it is expected that you will include STPM Maths (T) in your course selection. The syllabus is divided into two semesters, with each semester comprising six chapters that centre around a distinct themes.

Please find the topics covered below:

First Term: Algebra and Geometry

• Chapter 1 Functions
• Chapter 2 Sequences and Series
• Chapter 3 Matrices
• Chapter 4 Complex Numbers
• Chapter 5 Analytic Geometry
• Chapter 6 Vectors

Second Term: Calculus

• Chapter 7 Limits and Continuity
• Chapter 8 Differentiation
• Chapter 9 Integration
• Chapter 10 Differential Equations
• Chapter 11 Maclaurin Series
• Chapter 12 Numerical Methods

Third Term: Statistics

• Chapter 13 Data Description
• Chapter 14 Probability
• Chapter 15 Probability Distributions
• Chapter 16 Sampling and Estimation
• Chapter 17 Hypothesis Testing
• Chapter 18 Chi-squared Tests

STPM Maths (M) Syllabus

The purpose of the Mathematics (M) syllabus is to establish a structure for a pre-university program that helps students enhance their comprehension of mathematical principles and reasoning, and gain proficiency in solving problems and applying mathematics in fields such as social sciences and management.

First Term: Algebra and Calculus

• Chapter 4 Differentiation
• Chapter 5 Integration
• Chapter 6 Differential Equations

Second Term: Statistics

• Chapter 7 Data Description
• Chapter 8 Probability
• Chapter 9 Probability Distributions
• Chapter 10 Correlation and Regression
• Chapter 11 Index Numbers
• Chapter 12 Time Series

Third Term: Financial and Decision Mathematics

• Chapter 13 Interest, Annuity and Depreciation
• Chapter 14 Cost, Revenue and Profit
• Chapter 15 Linear Programming
• Chapter 16 Critical Path Analysis
• Chapter 17 Inventory Models
• Chapter 18 Game Theory

Now that you are familiar with the syllabus, find out about the format and marking scheme as well.

Benefits of Choosing STPM

Are you unsure if you want pursue STPM? If you're considering your pre-university options, here are some reasons why you should consider STPM.

Firstly, STPM is an extremely affordable pre-U option. The cost of tuition and exam fees is significantly lower compared to other pre-U programmes, making it accessible to many students who may not have the financial means to pursue other options.

Furthermore, you don't have to wear the school uniform unless you want to. This may seem like a small thing, but it can make a big difference in how comfortable and confident you feel in your learning environment.

STPM is recognised both locally and internationally as a pre-university qualification. This means that it can open doors to a wide range of universities and courses, both in Malaysia and abroad.

There is a huge number of subject choices in STPM, allowing you to choose subjects that align with your interests and career aspirations. This can help you stay motivated and engaged throughout your studies.

Moreover, T he new STPM modular format is more appealing than the old format. Instead of one big exam at the end of 18 months, you will be assessed at the end of each of the three terms. This can help reduce the stress and pressure of exam season and allow you to focus on smaller, more manageable chunks of learning.

Most subjects have a School-Based Assessment (SBA) component. This means that your project work, field study, and practical work will help contribute towards your final grade. This can help you develop important skills such as research, critical thinking, and communication.

Another positive point is that you will progress to the next term regardless of your academic performance. This means that you won't get held back if you fail a paper, giving you the opportunity to learn from your mistakes and improve in future terms.

Finally, you can improve your subject grade point by re-sitting for a paper. This can give you a second chance to improve your understanding and performance in a particular subject, which can be especially helpful if you plan to pursue a course that requires a high grade point average.

STPM offers many benefits that make it a compelling pre-U option for Malaysian students. From affordability and flexibility to a wide range of subject choices and internationally recognised qualifications, STPM is definitely worth considering.

Click here to check out Revision Techniques to Score in STPM Mathematics

Do you need help comprehending the concepts of mathematics? Superprof can assist you in finding a STPM mathematics tutor for this purpose. Superprof is a platform that connects students and tutors across Malaysia to achieve their academic and professional goals.

Private tutors for maths is available on this platform, but it is essential to have clear objectives beforehand. Communicating your goals to your tutor can save you time and money. Determine whether you require a tutor for a few weeks at the beginning of each semester to get started, only for exam preparation, or for an entire academic year or semester.

It is important to have a clear understanding of your learning objectives as it will help you determine the type of maths tutor that is suitable for you. You may require a tutor with an academic approach or someone who adopts a more practical method. This decision is solely yours. In addition to this, it is advisable to review the tutor's experience, professional background, education, and feedback from previous students who have worked with them. This will provide you with first-hand insight into what to expect from their lessons. Other factors to consider include their teaching techniques, learning materials, availability for remote or online classes, and their knowledge of a specific branch of mathematics that you may be interested in learning more about.

It is vital to keep cost in mind when choosing a tutor since their fees are often based on their level of experience. Once you have found a suitable tutor, you can request a free first lesson to evaluate whether you and the tutor are compatible and to determine what to expect from future lessons. From there, it is up to you to decide. Prepare to learn maths with a Superprof tutor!

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Sutha Ramasamy

As a communications graduate, I have always had a passion for writing. I love to read and strongly believe that one can never stop learning.

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