Mathematics Form3

Fast Track to Success: Success Maths PMR (Form 3) Paper 2 Questions (Fully-Worked (Fully-W orked Solutions) 5 (a) 3p – 18 pq = 3 p(1 – 6q) (b) 3e2 – 108 = 3( e2 – 36) = 3( e2 – 6 2) 3(ee – 6)(e 6)(e + 6) = 3(

Chapter 5: Indices 1 42 x – 1 = (2 x)(23) 22(2 x – 1) = 2 x + 3 2(2 x–1) =  x + 3 4 x – 2 =  x + 3 4 x –  x = 3 + 2 3 x = 5 5  x = — 3 1 — 2

2 2 × 12 × 3 3

5 — 2

1 — 2

= 2 × (2 × 3) × 3 3

6 20 – 5m2 = 5(4 – m2) = 5(22 – m2) = 5(2 + m)(2 – m)

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1 —

7 (a) 3 x + 9 = 3( x + 3) (b) 32 – 2y 2 = 2(16 – y 2) = 2(42 – y 2) )(4 –  y ) = 2(4 +  y )(4

5 — 2

5 —

= 23 × 2 × 3 2 × 3 2 1 + 5 — —

= 23 + 1 × 3 2 2 = 24 × 33 = 16 × 27 = 432 m5 × m2 m5 + 2 3 ———— = ——— –3 m m–3 = m7 – (–3) = m10 2 —

5 —

4 (a) 6 3 ÷ 6 3 2 5 —–— 3

= 63 = 6–1 1 =— 6

1 1 – —p  5 3 8 —– – ————— 3m

mp



(b) (pq5)2 × p3 = p2q10 × p3 = p2+3q10 =  p5q10

Chapter 6: Algebraic Expressions (III) 1 (2 x – 3) 2 – (5 x + 4) = 4 x2 – 12 x + 9 – 5 x – 4 17 x   x  + 5 = 4 x 2 – 17 2 (a – 2 b)2 + b(2a – 3 b) = (a2 – 4 ab + 4 b2) + (2ab – 3b2) = a2 – 2ab + b2 = (a – b)(a – b) = (a – b)2 3 (a) p(7 + q) = 7 p + pq (b) (5m – n)2 = (5 m – n)(5m – n) = 25 m2 – 10mn + n2 4 (a) 10ab – 5 a2 (2b = 5a(2 b – a) 2 (b)  x – 5( x + 2) – (5 – 7 x) =  x2 – 5 x – 10 – 5 + 7 x =  x2 + 2 x – 15 3)( x  + 5) = ( x  – 3)( x 



1 5p – 3 1 – —p 3 = ———————— 3mp 5p – 3 + p = ————– 3mp 6p – 3 = ———– 3mp 1 3(2p – 1) = ————– 1 3mp 2 p – 1 = ———– mp m+5 2 6m – m – 5 9 –— – –——– = –——–——– 5m 15m2 15m2 5m – 5 = –——–– 15m2 1 5(m – 1) = –——–— 2 3 15m m–1 = –——– 3m2

1 4–v 3v – 4 + v 10 –— – –——– = –——–—— 4m 12mv 12mv 4v – 4 = –——–– 12mv 1 4(v – 1) = –——–— 3 12mv v  – 1 = –——– 3mv 

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Chapter 7: Algebraic Formulae 3(p – 1) 1 –——–— = 7 k 3(p – 1) = 7k 7k p – 1 = –— 3 7k p = –— + 1 3 7k + 3 p = –——–– 3 6y  – 1 2 –——–– = 4 y  + 2 p 6y  – 1 = 4y  + 8 p 2y = 8 p + 1 8p + 1 y = –——–– 2 1 y = —(8 p + 1) 2 8n – nk 3 F  = –——–– 3 3F  = n(8 – k) 3F  n = –——– 8– k 2 4 p +3 = q p2 = q – 3

Chapter 10: Transformation (II) 2 40°

85°

55°

85°

40°

55°

(a) The corresponding  angle to ∠HFG is RPQ.. RPQ

(b) The corresponding  side to FG is  PQ  PQ..

Chapter 12: Linear Inequalities 4 5 – 3 x  4 –  x –3 x +  x  4 – 5 –2 x  –1 1  x  — 2 5 (a) m – 5  2 m 2+ 5 7 m (b) 7 + 4n  10 – 2n 4n + 2 n  10 – 7 6n  3 1 — n 2

p = √q – 3

6 (a) 2 +  x  7  x  5  x (b) — – 4  1 3  x —5 3  x  15

Chapter 9: Scale Drawings 1

9

10

9 –  x ∴  x

11

12

13

 

0 9

14

15

1 cm

∴ 9   x  15 ⇒  x  = 10, 11, 12, 13, 14, 15

1 cm

2

 x 7 — 3  x



1



3

1 – 2 x –2 x  x

–3

–2

–1

0

∴ –3   x  3 ⇒  x  = –2, –1, 0, 1, 2, 3

2

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2

  

7 6 –3

3

Chapter 13: Graphs of Functions 7  y 

 AB = 8 cm  AC = √82 + 62 = 10 cm 8 DE sin y ° = –— = –— 17  AE

12

 AD = √172 – 8 2 = 15 cm ∴ CD =  AD –  AC = 15 – 10 = 5 cm

10

8

6

4

2

–2

–1

0

1

2

–2 –4 –6

–8

Chapter 15: Trigonometry   AB 1 tan  x° = –— BM BC = √172 – 82 = 15 cm 15 BM = –— 2 16 8 tan  x° = ——– = —– 15 15 —– 2 5 2 cos PRQ = — 7 5 ∴ cos  x° = – — 7 3 3 tan  x° = — 4 BC = –—  AB 6 = –—  AB 3 6 ⇒ — = –— 4  AB

 

3

4

 x

3  AD 4 (a) tan  x° = — = –— 4 BD 3 9 ⇒ — = –— 4 BD BD = 12 cm BD cos  x° = –—  AB  AB = √122 + 92 = 15 cm 4 12 ⇒ cos  x° = –— = — 5 15 4 BD (b) sin y ° = — = –— 7 CD 4 12 ⇒ — = –— 7 CD CD = 21 cm LM 5 5 (a) cos  x° = –— = –— 13 26 5 LM ⇒ –— = –— 13 26 LM = 10 cm KM = √262 – 102 = 24 cm 12 KM 24 ∴ tan  x° = –— = –— = –— 5 LM 10 8 16 (b) sin y ° = –— = –— 17 PN  ⇒ PN  = 34 cm PM = √342 – 16 2 = 30 cm ⇒ KMP  = KM + PM = 24 + 30 = 54 cm

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