09[anal add math cd]

1

Additional Mathematics SPM Chapter 9

© Penerbitan Pelangi Sdn. Bhd.

1. (a) limx → 1 (3x – 2) = 3(1) – 2

= 1

(b) limx → 4 1 5––––––

2x – 1 2 = 5–––––––2(4) – 1

= 5—7

(c) limx → 3 1 x2 – 9––––––

x – 3 2 = limx → 3 3 (x – 3)(x + 3)––––––––––––

(x – 3) 4 = limx → 3 (x + 3) = 3 + 3 = 6

(d) limx → 1 1 x2 – 3x + 2––––––––––

x – 1 2 = limx → 1 3 (x – 1)(x – 2)––––––––––––

(x – 1) 4 = limx → 1 (x – 2) = 1 – 2 = –1

(e) limx → ∞ 1 2x––––––

4x – 1 2 = limx → ∞ 1

2x–––x–––––––4x – 1––––––x

2 = lim

x → ∞ 1 2–––––––4 – 1—x

2 = lim

x → ∞ 1 2–––––4 – 0 2

= 2—4

= 1—2

(f) limx → 0 1x2 + 3x––––––

2x 2 = limx → 0 3 x(x + 3)–––––––

2x 4 = limx → 0 1 x + 3–––––

2 2 = 0 + 3–––––

2

= 3—2

2. Gradient of the chord AB = –4 – 1– 1—

4 2––––––––––

–2 – 1—2

= –4 + 1—

4––––––––2 1—

2

= 15–––4

× 2—5

= 3—2

3. Substitute x = 2, y = a into y = x2 + 1,a = 22 + 1 = 5

Gradient of the chord PQ = 5 – 2–––––––2 – (–1)

= 1

4. (a) y = 4x + 1 ............................ 1 y + dy = 4(x + dx) + 1 ................. 2

2 – 1, dy = (4x + 4dx + 1) – (4x + 1) = 4dx

dy

–––dx = 4

dy

–––dx = lim

dx → 0 1 dy–––dx 2

= limdx → 0 (4)

= 4

(b) y = x2 – 4x .............................................. 1 y + dy = (x + dx)2 – 4(x + dx) y + dy = x2 + 2xdx + (dx)2 – 4x – 4dx............ 2

2 – 1, dy = 2xdx + (dx)2 – 4dx

dy

–––dx = 2xdx–––––

dx + (dx)2

–––––dx

– 4dx––––dx

= 2x + dx – 4

dy

–––dx = lim

dx → 0 1 dy–––dx 2

= limdx → 0 (2x + dx – 4)

= 2x – 4

CHAPTER

9 Differentiation

2



5. (a) Let y = x3 + 2x ................................................. 1 y + dy = (x + dx)3 + 2(x + dx) = (x + dx)[x2 + 2xdx + (dx)2] + 2x + 2dx = x3 + 2x2dx + x(dx)2 + x2dx + 2x(dx)2

+ (dx)3 + 2x + 2dx y + dy = x3 + 3x2dx + 3x(dx)2 + (dx)3 + 2x

+ 2dx ................................................ 2

2 – 1, dy = 3x2dx + 3x(dx)2 + (dx)3 + 2dx

dy

–––dx = 3x2 + 3xdx + (dx)2 + 2

dy

–––dx = lim

dx → 0 1 dy–––dx 2

= 3x2 + 2

Therefore, f ′(x) = 3x2 + 2

(b) Let y = 1 – 2x + 3x2 .......................................1 y + dy = 1 – 2(x + dx) + 3(x + dx)2

= 1 – 2x – 2dx + 3[x2 + 2xdx + (dx)2] = 1 – 2x – 2dx + 3x2 + 6xdx + 3(dx)2

y + dy = 1 – 2x – 2dx + 3x2 + 6xdx + 3(dx)2 ...2

2 – 1, dy = –2dx + 6xdx + 3(dx)2

dy

–––dx = –2 + 6x + 3dx

dy

–––dx = lim

dx → 0 1 dy–––dx 2

= –2 + 6x

Therefore, f ′(x) = –2 + 6x

6. Gradient of the tangent at the point A = 2(1) = 2

7. (a) y = 2x3

dy

–––dx = 6x2

(b) y = 3–––x2

= 3x–2

dy

–––dx = (–2)(3)x–3

= – 6–––x3

(c) y = – x 4–––5

dy

–––dx = – 4x 3–––

5

(d) y = 1–––6x

= 1—6

x–1

dy

–––dx = (–1)1 1—

6 2x–2

= – 1–––6x2

8. (a) y = 8x2

dy

–––dx = 16x

When x = –1, dy

–––dx = 16(–1)

= –16(b) y = – 9–––

x3

= –9x–3

dy

–––dx = (–3)(–9)x–4

= 27–––x4

When x = 1, dy

–––dx = 27–––

14

= 27(c) y = – 1–––

2x = – 1—

2 x–1

dy

–––dx = (–1)1– 1—

2x–22

= 1–––2x2

When x = 2, dy

–––dx = 1–––––

2(2)2

= 1—8

9. (a) y = 4x2 – 3x + 5

dy

–––dx = 8x – 3

(b) y = 5x3 + 3—x – 4

= 5x3 + 3x–1 – 4

dy

–––dx = 15x2 – 3x–2

= 15x2 – 3–––x2

(c) y = x5 – 2–––3x2 + 1

= x5 – 2—3

x–2 + 1

dy

–––dx = 5x4 – (–2)1 2—

3 2x–3

= 5x4 + 4–––3x3

10. (a) f (x) = 4x2 + 5x f ′(x) = 8x + 5

(b) f (x) = 5x3 – 1 f ′(x) = 15x2

11. (a) d–––dx

(4x2 – 3x + 5)

= 8x – 3

3



(b) d–––dx 1 8—x – 4x + 32

= d–––dx

(8x –1 – 4x + 3)

= –8x –2 – 4

= – 8–––x2

– 4

12. (a) y = 8x2 – 1—4

x + 3

dy

–––dx = 16x – 1—

4

(b) y = x3 + 1–––2x2

= x3 + 1—2

x–2

dy

–––dx = 3x2 – x–3

= 3x2 – 1–––x3

(c) f (x) = 1—2

x2 – 8x + 1

f ′(x) = x – 8

13. (a) y = (4x – 1)(3x2)

dy

–––dx = (4x – 1)(6x) + (3x2)(4)

= 24x2 – 6x + 12x2

= 36x2 – 6x = 6x(6x – 1)

(b) y = (1 – 2x)(4x + 3)

dy

–––dx = (1 – 2x)(4) + (4x + 3)(–2)

= 4 – 8x – 8x – 6 = –2 – 16x = –2(1 + 8x)

(c) y = 2x1 1—x + 12(1 – 3x)

= (2 + 2x)(1 – 3x)

dy

–––dx = (2 + 2x)(–3) + (1 – 3x)(2)

= –6 – 6x + 2 – 6x = –4 – 12x = –4(1 + 3x)

(d) y = 4(x – 4)2

= 4(x2 – 8x + 16) = 4x2 – 32x + 64)

dy

–––dx = 8x – 32

= 8(x – 4)

14. (a) y = x––––––2x – 3

dy

–––dx = (2x – 3)(1) – x(2)–––––––––––––––

(2x – 3)2

= 2x – 3 – 2x––––––––––(2x – 3)2

= – 3––––––––(2x – 3)2

(b) y = x2 + 3––––––2x – 5

dy

–––dx =

(2x – 5)(2x) – (x2 + 3)(2)–––––––––––––––––––––

(2x – 5)2

= 4x2 – 10x – 2x2 – 6––––––––––––––––(2x – 5)2

= 2x2 – 10x – 6––––––––––––(2x – 5)2

= 2(x2 – 5x – 3)––––––––––––

(2x – 5)2

(c) y = 4x – 1––––––x2 + 1

dy

–––dx =

(x2 + 1)(4) – (4x – 1)(2x)–––––––––––––––––––––

(x2 + 1)2

= 4x2 + 4 – 8x2 + 2x–––––––––––––––– (x2 + 1)2

= –4x2 + 2x + 4––––––––––––(x2 + 1)2

= –2(2x2 – x – 2)–––––––––––––

(x2 + 1)2

15. (a) y = (2x – 1)10

dy

–––dx = 10(2x – 1)9 d–––

dx(2x – 1)

= 10(2x – 1)9(2) = 20(2x – 1)9

(b) y = (1 + 4x)7

dy

–––dx = 7(1 + 4x)6(4)

= 28(1 + 4x)6

(c) y = 2(x3 + 4)5

dy

–––dx = 5(2)(x3 + 4)4(3x2)

= 30x2(x3 + 4)4

(d) y = 3––––––––(2x + 1)4

= 3(2x + 1)–4

dy

–––dx = (–4)(3)(2x + 1)–5(2)

= – 24––––––––

(2x + 1)5

4



(e) y = 1––––––––4(x2 – 1)5

= 1—4

(x2 – 1)–5

dy

–––dx = (–5)1 1—

4 2(x2 – 1)–6(2x)

= – 5—2

x(x2 – 1)–6

= – 5x–––––––– 2(x2 – 1)6

16. (a) y = 3x(1 – 2x)5

dy

–––dx = 3x · 5(1 – 2x)4(–2) + (1 – 2x)5(3)

= –30x(1 – 2x)4 + 3(1 – 2x)5

= –3(1 – 2x)4[10x – (1 – 2x)] = –3(1 – 2x)4(12x – 1)

(b) y = x–––––––(x2 – 4)3

= x(x2 – 4)–3

dy

–––dx = x(–3)(x2 – 4)–4(2x) + (x2 – 4)–3(1)

= –6x2(x2 – 4)–4 + (x2 – 4)–3

= (x2 – 4)–4[–6x2 + (x2 – 4)] = (x2 – 4)–4(–5x2 – 4)

= –5x2 – 4––––––––(x2 – 4)4

(c) y = x2 + 1–––––– 4x

= x2–––4x

+ 1–––4x

= x—4

+ 1—4

x–1

dy

–––dx = 1—

4 – 1—

4x–2

= 1—4

– 1–––4x2

(d) y = (4x – 1)(x2 – 3)4

dy

–––dx = (4x – 1) · 4(x2 – 3)3(2x) + (x2 – 3)4(4)

= 8x(4x – 1)(x2 – 3)3 + 4(x2 – 3)4

= 4(x2 – 3)3[2x(4x – 1) + (x2 – 3)] = 4(x2 – 3)3(8x2 – 2x + x2 – 3) = 4(x2 – 3)3(9x2 – 2x – 3)

(e) y = x2 + 3x – 4–––––––––– x + 4

= (x + 4)(x – 1)–––––––––––– (x + 4)

= x – 1

dy

–––dx = 1

(f) y = x3(2x – 1)3

dy

–––dx = x3 · 3(2x – 1)2(2) + (2x – 1)3(3x2)

= 6x3(2x – 1)2 + (3x2)(2x – 1)3

= 3x2(2x – 1)2[2x + (2x – 1)] = 3x2(2x – 1)2(4x – 1)

17. (a) (i) y = 3x2 – 1

dy

–––dx = 6x

Gradient of tangent at (1, 2) = 6(1) = 6

(ii) Equation of tangent is y – 2 = 6(x – 1) y – 2 = 6x – 6 y = 6x – 4 (iii) Gradient of normal = – 1—

6 Equation of normal is

y – 2 = – 1—6

(x – 1)

= – 1—6

x + 1—6

y = – 1—6

x + 1—6

+ 2

y = – 1—6

x + 13–––6

(b) (i) y = 1–––x2 + 3

y = x–2 + 3

dy

–––dx = –2x–3

= – 2–––x3

Gradient of tangent at (–1, 4)

= – 2–––––(–1)3

= 2

(ii) Equation of tangent is y – 4 = 2(x + 1) y = 2x + 2 + 4 y = 2x + 6

(iii) Equation of normal is

y – 4 = – 1—2

(x + 1)

y = – 1—2

x – 1—2

+ 4

y = – 1—2

x + 7—2

5



(c) (i) y = 4(2x – 1)3

dy

–––dx = 12(2x – 1)2(2)

= 24(2x – 1)2

Gradient of tangent at (1, 4) = 24[2(1) – 1]2

= 24 (ii) Equation of tangent is y – 4 = 24(x – 1) y = 24x – 24 + 4 y = 24x – 20 (iii) Equation of normal is

y – 4 = – 1–––24

(x – 1)

y – 4 = – 1–––24

x + 1–––24

y = – 1–––24

x + 1–––24

+ 4

y = – 1–––24

x + 97–––24

(d) (i) y = x(2 – x)3

dy

–––dx = x(3)(2 – x)2(–1) + (2 – x)3(1)

= –3x(2 – x)2 + (2 – x)3

= (2 – x)2(–3x + 2 – x) = (2 – x)2(2 – 4x) = 2(2 – x)2(1 – 2x) Gradient of tangent at (2, 0) = 2(2 – 2)2[1 – 2(2)] = 0 (ii) Equation of tangent is y – 0 = 0(x – 2) y = 0 (iii) Equation of normal is x = 2

18. (a) y = 3x2 – 6x + 1 ...................................1

dy

–––dx = 6x – 6

For turning point, dy

–––dx = 0

6x – 6 = 0 x = 1

Substitute x = 1 into 1, y = 3 – 6 + 1 = –2 Hence, the turning point is (1, –2).

x 0 1 2

dy–––dx –6 0 6

Sketch of dy–––dx

(1, –2) is a minimum point.

(b) y = 1 – 3x – x2 .................. 1

dy

–––dx = –3 – 2x

dy

–––dx = 0,

–3 – 2x = 0 x = – 3—

2

Substitute x = – 3—2

into 1,

y = 1 – 31– 3—2 2 – 1– 3—

2 22

= 1 + 9—2

– 9—4

= 13–––4

Hence, the turning point is 1– 3—2

, 13–––4 2.

x –2 – 3—2 0

dy–––dx 1 0 –3


1– 3—2

, 13–––4 2 is a maximum point.

(c) y = x3 – 2x2 + 4 ....................... 1

dy

–––dx = 3x2 – 4x

When dy

–––dx = 0,

3x2 – 4x = 0 x(3x – 4) = 0 x = 0, 4—

3 Substitute x = 0 into 1, y = 4

Substitute x = 4—3

into 1,

y = 1 4—3 2

3 – 21 4—

3 22 + 4

= 64–––27

– 32–––9

+ 4

= 76–––27

The turning points are (0, 4) and 1 4—3

, 76–––27 2.

6



For (0, 4),

x –1 0 1

dy–––dx 7 0 –1


(0, 4) is a maximum point.

For 1 4—3

, 76–––27 2,

x 14—3 2

dy–––dx –1 0 4


1 4—3

, 76–––27 2 is a minimum point.

(d) y = –2x3 + 6x ....................1

dy

–––dx = –6x2 + 6

0 = –6x2 + 6 –6(x2 – 1) = 0 –6(x + 1)(x – 1) = 0 x = −1, 1

Substitute x = −1 into 1, y = −2(−1)3 + 6(−1) = 2 − 6 = −4

Substitute x = 1 into 1, y = −2(1)3 + 6(1) = −2 + 6 = 4

The turning points are (−1, −4) and (1, 4).

For (−1, −4),

x –2 –1 0

dy–––dx –18 0 6


(−1, −4) is a minimum point.

For (1, 4),

x 0 1 2

dy–––dx 6 0 –18


(1, 4) is a maximum point.

19. x + y = 50 y = 50 − x ................................1

Area, A = xy ..................................2

Substitute 1 into 2, A = x(50 − x) = 50x − x2

dA–––dx

= 50 − 2x

0 = 50 − 2x x = 25

y = 25d 2A––––dx2 = −2 , 0

Therefore, A is a maximum when x = 25 and y = 25.

Maximum area = xy = 25 × 25 = 625 unit2

20. Volume = 20 cm3

πr2h = 20

h = 20–––– πr2

Surface area, A = 2πr2 + 2πrh

= 2πr2 + 2πr1 20–––– πr2 2

= 2πr2 + 40––– r

dA–––dr

= 4πr – 40––– r2

0 = 4πr – 40––– r2

40––– r2 = 4πr

r3 = 40––– 4π

= 10––– π

= 10 × 7––– 22

= 70––– 22

r = 1.471

7



h = 20––– πr2

= 7 × 20–––––––––22(1.471)2

= 2.941

d 2A––––dr2 = 4π + 80–––

r3

= 4π + 80––––––1.4713 . 0

Hence, the area is a minimum when r = 1.471 and h = 2.941.

21. dr–––dt

= 0.1 cm s−1

A = πr 2

dA–––dr

= 2πr

dA–––dt

= dA–––dr

× dr–––dt

= 2πr × (0.1) = 2π(5) × (0.1) = π cm2 s−1

22. y = x2 − 2x

dy

–––dx = 2x – 2

dx–––dt

= 8 when x = 3,

dy–––dt

= dy

–––dx × dx–––

dt = (2x − 2)(8) = (2 × 3 − 2)(8) = 32 units s−1

23. dx–––dt

= 3.2 – 3.0––––––––2

= 0.1 cm s−1

y = 3x2 − 1

dy

–––dx = 6x

dy–––dt =

dy–––dx × dx–––

dt = (6x)(0.1) = (6 × 3.1)(0.1) = 1.86 cm s−1

24. dV–––dt

= −9 mm3 s−1

V = 4—3

πr3

dV–––dr

= 4—3

× 3πr2

= 4πr2

dr–––dt

= dr–––dV

× dV–––dt

= 1 1––––4πr2 2(−9)

= − 9––––4πr2

= − 9–––––––4π × 32

= – 1–––4π

cm s−1

25. y = 3x − 1 dy

–––dx = 3

dx = 2.01 – 2 = 0.01 unit

dy–––dx

≈ dy

–––dx

dy = dy

–––dx × dx

= 3(0.01) = 0.03 unit

26. y = x2 + 4dy

–––dx = 2x

dx = 1.9 – 2 = –0.1 unit

dy = dy

–––dx × dx

= 2x(−0.1) = 2(2)(−0.1) = –0.4 unit

27. y = 2—x

= 2x−1 dy

–––dx = − 2–––

x2

dy = 3.001 – 3 = 0.001 unit

When y = 3,

x = 2— y

= 2—3

8



dx

–––dy

≈ dx–––dy

dx = dx–––dy

× dy

= 1− x2––– 2 2(0.001)

= 3− 1 2—

3 22

–––––2 4(0.001)

= −0.00022 unit

28. y = x2

dy–––dx = 2x

When x = 3,

dy

–––dx = 2(3)

= 6

(a) 3.12 = 32 + dy

= 9 + dy

–––dx × dx dx = 3.1 − 3

= 0.1 = 9 + (6)(0.1) = 9.6

(b) 2.92 = 32 + dy

= 9 + dy

–––dx × dx dx = 2.9 − 3

= –0.1 = 9 + (6)(−0.1) = 9 − 0.6 = 8.4

29. (a) y = 1––– x2

y = x−2

dy

–––dx = − 2–––

x3

When x = 4,

dy

–––dx = − 2–––

43

= – 1––– 32

dx = 4.1 – 4 = 0.1

1–––– 4.12 = 1–––

42 + dy

= 1––– 16

+ dy

–––dx × dx

= 1––– 16

+ 1− 1––– 32 2(0.1)

= 1––– 16

− 0.1––– 32

= 0.05938

(b) y = 1––– x2

dy

–––dx = − 2–––

x3

When x = 4,

dy

–––dx = − 1–––

32 dx = 3.9 – 4.0 = −0.1

1–––– 3.92 = 1–––

42 + dy

= 1––– 16

+ dy

–––dx × dx

= 1––– 16

+ 1− 1––– 32 2(−0.1)

= 0.06563

30. (a) y = ABx

= x1—2

dy

–––dx = 1—

2x

− 1—2

= 1––––2ABx

When x = 4,

dy

–––dx = 1––––

2AB4

= 1— 4

dx = 4.1 – 4 = 0.1

ABB4.1 = AB4 + dy

–––dx × dx

= 2 + 1— 4

(0.1)

= 2.025

(b) ABB3.9 = AB4 + dy

–––dx × dx dx = 3.9 − 4

= –0.1

= AB4 + 1— 4

(– 0.1)

= 2 – 0.1––– 4

= 1.975

31. (a) y = 5x3 + 4x + 1

dy

–––dx = 15x2 + 4

d 2y

––––dx2 = 30x

9



(b) y = (4x2 − 1)5

dy

–––dx = 5(4x2 − 1)4(8x)

= 40x(4x2 − 1)4

d 2y

––––dx2 = 40x · 4(4x2 − 1)3(8x) + (4x2 − 1)4(40)

= 1280x2(4x2 − 1)3 + 40(4x2 − 1)4

(c) y = 2––– x3

= 2x−3

dy

–––dx = −3(2x−4)

= −6x−4

d 2y

––––dx2 = 24x−5

= 24––– x5

32. (a) f (x) = 4x3 − 1 f ′(x) = 12x2

f ′′(x) = 24x

(b) f ′(x) = 5x2 + 4x − 3 f ′′(x) = 10x + 4

(c) f (x) = 1––– 2x3 − 5

= x–3––– 2

− 5

f ′(x) = – 3— 2

x−4

f ′′(x) = 12–––2

x−5

= 6––– x5

33. (a) y = 4x2 − 4x + 1

dy

–––dx = 8x − 4

8x − 4 = 0

x = 4— 8

= 1— 2

y = 41 1— 2 2

2 − 41 1—

2 2 + 1

= 0

d 2y

––––dx2 = 8 . 0

The turning point 1 1— 2

, 02 is a minimum point.

(b) y = 5 − 2x2 + 4x

dy

–––dx = −4x + 4

−4x + 4 = 0 x = 1

y = 5 − 2(1)2 + 4(1) = 7

d 2y

––––dx2 = −4 , 0

The turning point (1, 7) is a maximum point.

(c) y = 1— 3

x3 − 2x2 + 50

dy

–––dx = x2 − 4x

x2 − 4x = 0 x(x − 4) = 0 x = 0, 4 When x = 0, y = 1—

3(0)3 − 2(0)2 + 50

= 50

When x = 4, y = 1— 3

(4)3 − 2(4)2 + 50

= 64–––3

− 32 + 50

= 39 1— 3

d 2y

––––dx2 = 2x − 4

For turning point (0, 50),

d 2y

––––dx2 = −4 , 0

Therefore, (0, 50) is a maximum point.

For turning point 14, 39 1— 3 2,

d 2y

––––dx2 = 2(4) − 4

= 4 . 0

Therefore, 14, 39 1— 3 2 is a minimum point.

(d) y = − 1— 3

x3 + x2 + 3x

dy

–––dx = −x2 + 2x + 3

−x2 + 2x + 3 = 0 x2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 x = 3, −1

When x = 3, y = − 1— 3

(3)3 + 32 + 3(3)

= −9 + 9 + 9 = 9

10



When x = −1, y = − 1— 3

(−1)3 + (−1)2 + 3(−1)

= 1— 3

+ 1 − 3

= −1 2— 3

= − 5— 3

d 2y

––––dx2 = −2x + 2

For turning point (3, 9),

d 2y

––––dx2 = −2(3) + 2

= −4 , 0 Therefore, (3, 9) is a maximum point.

For turning point 1−1, − 5— 3 2,

d 2y

––––dx2

= −2(−1) + 2 = 2 + 2 = 4 > 0

Therefore, 1−1, − 5— 3 2 is a minimum point.

34. y = x3 − 2x + 1

dy

–––dx = 3x2 − 2

d 2y

––––dx2 = 6x

d 2y

––––dx2 +

dy–––dx − y = −x3 − 8

6x + 3x2 − 2 − x3 + 2x − 1 = −x3 − 8 3x2 + 8x + 5 = 0 (3x + 5)(x + 1) = 0 x = − 5—

3, −1

1. ddx

1 x – 1x + 3 2 = f(x)

\ ∫0

1

f(x)2

dx = 12

∫0

1

f(x) dx

= 12

3 x – 1x + 3 4

0

1

= 12

1 1 – 11 + 3

– 0 – 10 + 3 2

= 12

10 + 13 2

= 16

2. Gradient function = px + k

dy

–––dx = px + k

Since ( 1— 2

, 0) is a turning point,

dy–––dx = 0 when x = 1—

2

0 = p1 1— 2 2 + k

p— 2

+ k = 0 .......................................... 1

Given gradient of normal at x = 2 is – 1––– 12

.

\ Gradient of tangent at x = 2 is 12.dy

–––dx = 12 when x = 2

12 = p(2) + k 2p + k = 12 ........................................ 2

2 – 1, 2p – p— 2

= 12

4p––– 2

– p— 2

= 12

3— 2

p = 12

p = 12 × 2— 3

= 8

Substitute p = 8 into 2, 2(8) + k = 12 k = 12 – 16 = – 4\ p = 8, k = –4.

3. f (x) = 1––––––––(2 − 3x)4

= (2 − 3x)−4

f ′(x) = −4(2 − 3x)−5(−3) = 12(2 − 3x)−5

f ′′(x) = −5(12)(2 − 3x)−6(−3) = 180(2 − 3x)−6

f ′′(1) = 180(−1)–6

= 180

4. dx = 1.9 – 2 = –0.1 y = 5x + x2

dy–––dx = 5 + 2x

dy = dy

–––dx dx

= (5 + 2x)dx = [5 + 2(2)](–0.1) = (9)(–0.1) = – 0.9

11



5. P = 100V

P = 100V–1

dPdV

= – 100V2

dP = (5 + P) – 5 = P

Using dPdV

= dPdV

dV = dPdPdV

= P

–100(5)2

= – P4

6. (a) y = –2x2 + 5x 0 = –2k2 + 5k k(–2k + 5) = 0 k = 0, k = 5—

2 Since k > 0, \k = 5—

2

(b) y = –2x2 + 5x

dy

–––dx = –4x + 5

= –41 5— 2 2 + 5

= –5

Gradient of normal at P 1 5— 2

, 02 is 1— 5

\ Equation of normal at point P

y – 0––––––x – 5—

2

= 1— 5

y = 1— 5 1x – 5—

2 2 y = 1—

5x – 1—

2

7. (a) y = 7x(x − 3) = 7x2 − 21x

dy

–––dx = 14x − 21

d 2y

––––dx2

= 14

For minimum y, dy

–––dx = 0

14x − 21 = 0 x = 21–––

14 = 3—

2

(b) Minimum y = 71 3— 2 21 3—

2 − 32

= 71 3— 2 21− 3—

2 2 = − 63–––

4

8. y = 3— x − 2x

y = 3x−1 − 2x

dy–––dx = −3x−2 − 2

= − 3––– x2 − 2

When x = 3, dy

–––dx = − 3—

9 − 2

= − 7— 3

dy–––dt = 5

dx–––dt

= dx–––dy

· dy

–––dt

= 1− 3— 7 2(5)

= − 15––– 7

units per second

9. y = 4kx2 + 6x

dy

–––dx = 8kx + 6

8k(3) + 6 = 10 24k = 4 k = 4–––

24

= 1— 6

10. V = 1— 3

h3 + 12h

dV–––dh

= h2 + 12

When h = 3, dV–––dh

= 9 + 12

= 21

Substitutex = 5

2

m1m2 = –1

12



Given dV–––dt

= 7 cm3 s−1

dh–––dt

= dh–––dV

× dV–––dt

= 1––– 21

× 7

= 1— 3

cm s−1

11.

h mr m

0.4 m

0.4 m

Let r be the radius, h be the height and V be the volume of oil.

Given dV–––dt

= p, dh–––dt

= 0.1

r— h

= 0.4––– 0.4

= 1Therefore, r = h

V = 1— 3

πr2h

V = 1— 3

πh3

dV–––dh

= πh2

dV–––dt

= dV–––dh

× dh–––dt

p = πh2 × 0.1

When h = 0.2,p = π(0.2)2(0.1) = 0.004π

1. limx → 0 1 x2 − 2x–––––––x 2 = limx → 0 (x − 2)

= −2

2. limx → ∞ 1 2x –––––1 + x 2 = lim

x → ∞ 12x–––x ––––––

1 + x–––––x2

= limx → ∞ 1 2––––––

1—x + 12 = 2–––––

0 + 1 = 2

3. limx → 2 1 4 − x2 ––––––

x − 2 2 = limx → 2 (2 − x)(2 + x)––––––––––––

(x − 2) = limx → 2 [−(2 + x)] = −4

4. dy = 2xdx + 4dx2

dy

–––dx

= 2x + 8dx

dy

–––dx = lim

dx → 0 (2x + 8dx)

= 2x

5. y = 4x(x2 − 1)5

dy–––dx = 4x d–––

dx(x2 − 1)5 + (x2 − 1)5 d–––

dx(4x)

= 4x · 5(x2 − 1)4(2x) + (x2 − 1)5(4) = 40x2(x2 − 1)4 + 4(x2 − 1)5

6. f (x) = 2––––––––(1 − 4x)3

= 2(1 − 4x)−3

f ′(x) = (−3)(2)(1 − 4x)−4(−4)

= 24––––––––(1 − 4x)4

\ f ′(0) = 24–––1

= 24

7. d–––dx 1 x − 1––––––

4 − x2 2

= (4 − x2) d–––

dx(x − 1) − (x − 1) d–––

dx(4 − x2)

––––––––––––––––––––––––––––––––––(4 − x2)2

= (4 − x2)(1) − (x − 1)(−2x)–––––––––––––––––––––– (4 − x2)2

= 4 − x2 + 2x2 − 2x––––––––––––––– (4 − x2)2

= x2 − 2x + 4––––––––––(4 − x2)2

8. d–––dx 1 x2 − 1––––––

x + 1 2

= d–––dx 3 (x + 1)(x − 1)––––––––––––

(x + 1) 4= d–––

dx(x − 1)

= 1

13



9. y = 5x3 − 1— x

+ 2

y = 5x3 − x−1 + 2

dy

–––dx = 15x2 + x−2

d 2y–––dx2 = 30x − 2x−3

= 30x − 2––– x3

10. f (x) = 2(3 − 4x)6

f ′(x) = 6⋅2(3 − 4x)5(−4) = (− 48)(3 − 4x)5

f ′′(x) = − 48 ⋅5(3 − 4x)4(−4) = 960(3 − 4x)4

f ′′(1) = 960(−1)4

= 960

11. y = 2x(x + 3) y = 2x2 + 6xdy

–––dx = 4x + 6

When x = 2,dy

–––dx = 4(2) + 6

= 14

12. y = 4x3 − 5x2 + 2x − 10dy

–––dx = 12x2 − 10x + 2

When dy

–––dx = 4,

12x2 − 10x + 2 = 412x2 − 10x − 2 = 0 6x2 − 5x − 1 = 0(6x + 1)(x − 1) = 0 x = − 1—

6, 1

13. y = (4 − 3x)5

dy–––dx = 5(4 − 3x)4(−3)

= −15(4 − 3x)4

The gradient function is −15(4 − 3x)4.

14. y = 1––––––––(1 + 2x)3

y = (1 + 2x)−3

dy–––dx = −3(1 + 2x)−4(2)

= −6(1 + 2x)−4

The gradient at the point (−1, −1) = − 6(1 – 2)−4

= –6––– 1

= −6

15. y = (x − 1)(x + 1) y = x2 − 1dy

–––dx = 2x

The gradient of the tangent at the point (1, 0)= 2(1)= 2

16. y = x2 + 4x dy

–––dx = 2x + 4

The gradient of the tangent at the point (1, 5) = 2(1) + 4= 6

Therefore, the gradient of the normal at the point

(1, 5) is – 1— 6

.

17. y = (2x + 5)2

dy–––dx = 2(2x + 5)(2)

= 4(2x + 5)

Given gradient of the tangent is −8.

dy

–––dx = −8

4(2x + 5) = −8 2x + 5 = –2 2x = −7 x = − 7—

2

Substitute x = − 7— 2

into y = (2x + 5)2,

y = 321− 7— 2 2 + 54

2

= 4

Therefore, the coordinates are 1− 7— 2

, 42.

18. y = 3(4x − 5)2 + 6 dy

–––dx = 6(4x − 5)(4)

= 24(4x – 5)

The gradient of the tangent for the point with gradient

of the normal 1— 2

is −2.

14



dy

–––dx = −2

24(4x – 5) = –2

4x – 5 = – 1––– 12

4x = 5 – 1––– 12

= 59––– 12

x = 59––– 48

Substitute x = 59––– 48

into y = 3(4x − 5)2 + 6,

y = 3341 59––– 48 2 − 54

2 + 6

= 31 59––– 12

− 522 + 6

= 31– 1––– 12 2

2 + 6

= 31 1–––– 144 2 + 6

= 1––– 48

+ 6

= 289–––– 48

Therefore, the coordinates are 1 59––– 48

, 289–––– 48 2.

19. y = p––– x2 + qx − 1 ......................1

y = px−2 + qx − 1dy

–––dx = −2px−3 + q

= − 2p––– x3 + q

Given the gradient of the tangent at the point (−1, −3) is 14.

− 2p–––––

(–1)3 + q = 14

2p + q = 14..........................2

Substitute x = −1, y = −3 into 1 since the point (−1, −3) lies on the curve 1,

−3 = p–––––

(–1)2 + q(−1) − 1

−3 = p − q − 1p − q = −2 .....................................3

2 + 3, 3p = 12 p = 4

Substitute p = 4 into 3, 4 − q = −2 q = 6

20. y = 2x2 + ax + b .......................1dy

–––dx = 4x + a

Given the gradient at the point (1, 5) is 8. 4(1) + a = 8 a = 4

Substitute x = 1, y = 5 and a = 4 into 1,5 = 2(1)2 + 4(1) + bb = −1

21. y = 2— x + 4x

= 2x−1 + 4xdy

–––dx = − 2–––

x2 + 4

The gradient of the tangent at (1, 6) = − 2––– 12 + 4

= 2

The equation of the tangent is y − 6 = 2(x − 1) y = 2x − 2 + 6 y = 2x + 4

22. y = 4–––––––– (3x − 1)2

= 4(3x − 1)−2

dy–––dx = −8(3x − 1)−3(3)

= –24–––––––– (3x – 1)3

At the point (0, 4), the gradient of the tangent

= –24––––– (–1)3

= 24

Therefore, the gradient of the normal is − 1––– 24

.

Hence, the equation of the normal is

y − 4 = − 1––– 24

(x − 0)

y = − 1––– 24

x + 4

23. y = x3 − 6x2 + 3dy

–––dx = 3x2 − 12x

For the tangent parallel to the x-axis, gradient = 0

Therefore, dy

–––dx = 0

3x2 − 12x = 0 3x(x − 4) = 0 x = 0, 4

15



When x = 0,y = 03 − 6(0)2 + 3 = 3

When x = 4,y = 43 − 6(4)2 + 3 = 64 – 96 + 3 = –29

Hence, the points are (0, 3) and (4, −29).

24. Substitute y = 8 into y = −x2 + 4x + 4, 8 = −x2 + 4x + 4x2 − 4x + 4 = 0 (x − 2)2 = 0 x = 2

Therefore, p = 2 and q = 8.

25. y = x2(x − 3) – 1 y = x3 − 3x2 − 1 ........................1dy

–––dx = 3x2 − 6x

For stationary point, dy

–––dx = 0

3x2 − 6x = 0 3x(x − 2) = 0 x = 0, 2

Substitute x = 0 and x = 2 into 1 respectively,When x = 0, y = −1When x = 2, y = 23 − 3(2)2 − 1 = 8 – 12 − 1 = –5

Therefore, the stationary points are (0, –1) and (2, –5).

26. y = px2 + qx + 4 .......................1dy

–––dx = 2px + q

dy–––dx = 0 at the point (−1, 5).

2p(−1) + q = 0 −2p + q = 0 q = 2p ............................2

Substitute x = −1, y = 5 into 1, 5 = p(−1)2 + q(−1) + 4p − q = 1 .......................................3

Substitute 2 into 3,p – 2p = 1 −p = 1 p = −1

Substitute p = −1 into 2,q = 2(−1) = −2

27. y = −x3 + 6x2 − 9x − 2 .........................1dy

–––dx = −3x2 + 12x − 9

For stationary points, dy

–––dx = 0

−3x2 + 12x − 9 = 0Divide by (−3), x2 − 4x + 3 = 0 (x − 1)(x − 3) = 0 x = 1, 3

Substitute x = 1 and x = 3 into 1 respectively,When x = 1, y = –1 + 6 – 9 – 2 = –6When x = 3, y = –(3)3 + 6(3)2 − 9(3) – 2 = –27 + 54 – 27 – 2 = –2

The stationary points are (1, –6) and (3, –2).d 2y

––––dx2 = –6x + 12

For point (1, –6), d 2y

––––dx2 = –6(1) + 12

= 6 . 0

For point (3, –2), d 2y

––––dx2 = –6(3) + 12

= –6 , 0

Therefore, the minimum point is (1, –6).

28. p = x2y and x + y = 10 y = 10 − x ........................ 1

Substitute 1 into p = x2y, p = x2(10 − x) = 10x2 − x3

dp–––dx = 20x − 3x2

When dp–––dx = 0,

20x − 3x2 = 0 x(20 − 3x) = 0 x = 0, x = 20–––

3d2p

––––dx2 = 20 − 6x

For x = 0, d2p–––dx2 = 20 − 6(0)

= 20 . 0

For x = 20––– 3

, d2p–––dx2 = 20 − 61 20–––

3 2 = 20 − 40 = −20 , 0

Therefore, for p to be maximum, x = 20––– 3

.

16



29. y = (2x − 1)3

dy–––dx = 3(2x − 1)2(2)

= 6(2x − 1)2

Given dx–––dt

= 2

dy–––dt =

dy–––dx ⋅ dx–––

dt = 6(2x − 1)2(2) = 12(2x − 1)2

When x = 1,dy

–––dt = 12(1)2

= 12 units per second

30. h = xy h = (1 − 2t)(1 + 3t)dh–––dt

= (1 − 2t)(3) + (1 + 3t)(−2)

= 3(1 − 2t) − 2(1 + 3t) = 3 − 6t − 2 − 6t = 1 − 12t

When t = 2,dh–––dt

= 1 − 12(2) = −23

31. 1 —v − 1 —u = 1— 7

u − v–––––vu = 1— 7

7u − 7v = uvuv + 7v = 7uv(u + 7) = 7u v = 7u ––––––

u + 7

dv–––du

= (u + 7)(7) − (7u)(1)–––––––––––––––––(u + 7)2

= 7u + 49 − 7u–––––––––––(u + 7)2

= 49–––––––(u + 7)2

Given du–––dt

= 12

dv–––dt

= dv–––du

⋅ du–––dt

= 49–––––––(u + 7)2

× 12

= 49––– 122 × 12

= 49––– 12

units s−1

32. pv = 20

p = 20––– v = 20v−1

dp–––dv = −20v−2

= − 20––– v2

When v = 2, dp–––dv = − 20–––

22 and dv = 2.01 − 2

= −5 = 0.01

dp = dp–––dv × dv

= (−5)(0.01) = −0.05

33. y = 2—x − 5

= 2x−1 − 5dy

–––dx = −2x−2

= − 2––x2

When x = 2, dy

–––dx = − 2–––

22 and dy = 1.9 − 2

= − 1—2

= – 0.1

dx–––dy ≈ dx–––

dy

dx = dx–––dy × dy

= (−2)(–0.1) = 0.2

34. y = 2x2 + 3dy

–––dx = 4x

When x = 4, dy

–––dx = 4(4) and dx = 4.1 − 4

= 16 = 0.1

dy = dy

–––dx × dx

= 16(0.1) = 1.6

dy–––y × 100 = 1.6––––––––

2(4)2 + 3 × 100

= 1.6–––35

× 100

= 4.57%

17



35. y = 27–––x3

= 27x−3

dy–––dx = −81x−4

= − 81–––x4

When x = 3, dy

–––dx = − 81–––

34 and dx = 3.01 − 3

= −1 = 0.01

27––––––(3.01)3

= 27–––33 + dy

= 1 + dy

–––dx × dx

= 1 + (−1)(0.01) = 1 − 0.01 = 0.99

36. y = 4–––x2

= 4x−2

dy–––dx = −8x−3

= − 8–––x3

When x = 2, dy

–––dx = − 8–––

23

= −1dx = 1.9 − 2 = – 0.1

4––––1.92 = 4–––

22 + dy

= 1 + dy

–––dx × dx

= 1 + (−1)(−0.1) = 1 + 0.1 = 1.1

37. y = 1–––x3

= x−3

dy–––dx = −3x−4

= − 3–––x4

When x = 1,dy

–––dx = −3

dx = 1.01 − 1 = 0.01

3––––––(1.01)3 = 33 1––––––

(1.01)3 4 = 31 1–––

13 + dy2 = 311 +

dy–––dx × dx2

= 3[1 + (−3)(0.01)] = 2.91

38. (a) y = 4t2 + t ...............................1

x = 1 − 2t

t = 1 − x ––––– 2 ...............................2

Substitute 2 into 1,

y = 43 1 − x ––––– 2 4

2 + 1 1 − x –––––

2 2 = 41 (1 − x)2

––––––– 4 2 + 1—

2 − 1—2 x

= (1 − x)2 + 1—2 − 1—

2 x

dy

–––dx = 2(1 − x)(−1) − 1—

2

= −2 + 2x − 1—2

= 2x − 5—2

(b) When t = 2, x = 1 − 2(2) = −3

When t = 2.01, x = 1 − 2(2.01) = −3.02 dx = –3 − (–3.02) = 0.02

When x = –3, dy

–––dx = 2(−3) − 5—

2 = −6 − 5—

2 = − 17–––

2

dy = dy

–––dx × dx

= 1− 17–––2 2(0.02)

= −0.17

39. (a) y = 2t2 + 1, x = 1 − 2t

dy

–––dt = 4t dx–––

dt = −2

dy

–––dx =

dy–––dt ⋅ dt–––

dx

= 4t1− 1—2 2

= −2t

18



(b) When x = 3, 3 = 1 − 2t

t = − 2—2

t = −1

dy

–––dx = −2(−1)

= 2

40. (a) y = x2 − 4x + 1

dy

–––dx = 2x − 4

d 2y

––––dx2

= 2

d 2y

––––dx2 + 1 dy

–––dx 2

2 − y = 2x + 1

2 + (2x − 4)2 − (x2 − 4x + 1) = 2x + 1 2 + 4x2 − 16x + 16 − x2 + 4x − 1 − 2x − 1 = 0 3x2 − 14x + 16 = 0 (x − 2)(3x − 8) = 0 x = 2, 8—

3

(b) (i) dy

–––dx = kx − 5

y + 7x − 5 = 0 y = −7x + 5 The gradient of the tangent at the point

(–1, 12) is −7.

Therefore, dy

–––dx = −7 when x = −1.

kx − 5 = −7 k(−1) − 5 = −7 −k = −2 k = 2

(ii) The gradient of the normal = 1—7

The equation of the normal is

y − 12 = 1—7 (x + 1)

y = 1—7 x + 1—

7 + 12

y = 1—7 x + 85–––

7

41. (a) y = 2x − 3 has a gradient of 2 at point P. y = x3 + 3x2 − 7x + 2

dy

–––dx = 3x2 + 6x − 7

dy

–––dx = 2

3x2 + 6x − 7 = 2

3x2 + 6x − 9 = 0 x2 + 2x – 3 = 0 (x − 1)(x + 3) = 0 x = 1, −3

When x = 1, y = x3 + 3x2 − 7x + 2 = 13 + 3(1)2 − 7(1) + 2 = 1 + 3 − 7 + 2 = −1

When x = −3, y = x3 + 3x2 − 7x + 2 = (−3)3 + 3(−3)2 − 7(−3) + 2 = −27 + 27 + 21 + 2 = 23

Therefore, the coordinates of P are (1, −1).

(b) Another point is (−3, 23).

42. (a) Area of the shaded region = Area of ∆OAB − Area of rectangle OPQR = 1—

2 × 5 × 10 − xy

= 25 − xy

Gradient of AB = Gradient of QA

10 − 0–––––– 0 − 5

= y − 0–––––x − 5

y = −2(x − 5) y = −2x + 10

Therefore, the area of the shaded region, A = 25 − x(−2x + 10) = 25 + 2x2 − 10x

(b) A = 2x2 −10x + 25

dA–––dx

= 4x − 10, d 2A––––dx2 = 4 . 0

4x − 10 = 0

x = 10–––4

= 5—2

Minimum area = 25 + 21 5—2 2

2 − 101 5—

2 2 = 25 + 25–––

2 − 50–––2

= 25 − 25–––2

= 25–––2

43. (a) Perimeter = 120 y + y + 8x + 2(60 − 9x) = 120 2y + 8x + 120 − 18x = 120 2y = 10x y = 5x

19



The area of the diagram = Area of rectangle ABCE + Area of ∆CDE = 8x(60 − 9x) + 1—

2 (8x)(ABBBBBBBy2 − 16x2 )

= 480x − 72x2 + 4xABBBBBBBB25x2 − 16x2

= 480x − 72x2 + 4x(3x) = 480x − 72x2 + 12x2

= 480x − 60x2

(b) A = 480x − 60x2

dA–––dx

= 480 − 120x

d 2A––––dx2 = −120 , 0

When dA–––dx

= 0,

480 − 120x = 0 x = 4

Therefore, x = 4 for the area to be maximum.

44. (a)

y cm

x cm

QMB CP

S R

A

16 cm

10 cm10 cm

In ∆ABM, AM 2 = AB2 − BM 2 = 102 − 82

= 36 cm AM = 6 cm Since ∆ABC and ∆ASR are similar, then

y

–––16 = 6 − x–––––

6

y = 16(6 − x)––––––––6

= 8(6 − x)––––––– 3

Area of rectangle PQRS, A = xy

= x3 8—3 (6 − x)4

= x116 − 8—3 x2

= 16x − 8—3 x2

(b) A = 16x − 8—3 x2

dA–––dx

= 16 − 16–––3 x

d 2A––––dx2 = − 16–––

3 , 0

When dA–––dx

= 0,

16 − 16–––3 x = 0

16–––3 x = 16

x = 3

Substitute x = 3 into y = 8—3 (6 − x),

y = 8—3 × 3

= 8 For the area of the rectangle to be largest, x = 3

and y = 8.

45. (a)

8x cm

10x cm

B

E

M F

A

(10 – 8x) cm

EM 2 = EF2 − MF2

= 100x2 − 64x2

= 36x2

EM = 6x cm EA = 6x + 10 − 8x = (10 − 2x) cm

Volume of the solid, V = Area of trapezium × BC

= 1—2 (8x)(10 − 8x + 10 − 2x) × 5

= 20x(20 − 10x) = 400x − 200x2

(b) (i) V = 400x − 200x2

dV–––dx

= 400 − 400x

dV–––dx

= 0

400 − 400x = 0 x = 1

d 2V––––dx2 = −400 , 0

Therefore, V is a maximum when x = 1.

(ii) Maximum value of V = 400(1) − 200(1)2

= 200

46. (a) Height of the box = 2 − x––––– 2

Volume of the box, V = (x)(x)1 2 − x––––– 2 2

= x211 − x—2 2

= x2 − 1—2 x3

20



(b) V = x2 − 1—2 x3

dV–––dx

= 2x − 3—2 x2

When dV–––dx

= 0,

2x − 3—2 x2 = 0

x12 − 3—2 x2 = 0

x = 0, 4—3

d 2V––––dx2 = 2 − 3x

When x = 0, d 2V––––dx2 = 2 . 0

When x = 4—3 , d 2V––––

dx2 = 2 − 31 4—3 2

= –2 , 0

Therefore, the volume is a maximum when x = 4—

3 .

47. (a) Volume of the cylinder = 81π cm3

πr2h = 81π

h = 81–––r2

Total surface area, A = πr2 + 2πrh + 14πr2––––

2 2 = πr2 + 2πrh + (2πr2) = 3πr2 + 2πrh

= 3πr2 + 2πr1 81–––r2 2

= 3πr2 + 162π––––r

(b) A = 3πr2 + 162πr –1

dA–––dr

= 6πr − 162πr –2

= 6πr − 162π––––r2

When dA–––dr

= 0,

6πr − 162π––––r2 = 0

6πr = 162π––––r2

r3 = 27 r = 3

d 2A––––dr2 = 6π + 324π–––––

r3

When r = 3,

d 2A––––dr2 = 18π . 0

Therefore, the total surface area is a minimum when r = 3.

48. (a) y = px2 − 4x + 1

dy

–––dx = 2px − 4

When x = 4, the gradient of the tangent is 0. Therefore, 0 = 2p(4) − 4 8p = 4 p = 1—

2

(b) When x = 4, y = 1—2 (4)2 − 4(4) + 1

= 8 – 16 + 1 = –7

Equation of the tangent is y = −7.

49.

A(4, –8)

P(a, b)P(a, b)

x0

y = x 2 – 8x + 12y

Let the point of contact between the tangent and the curve be P(a, b). y = x2 − 8x + 12dy

–––dx = 2x − 8

Gradient of the tangent at point P is 2a − 8.

Gradient of PA = b + 8––––– a − 4

b + 8––––– a − 4

= 2a − 8

b + 8 = (a − 4)(2a − 8) = 2a2 − 16a + 32 b = 2a2 − 16a + 24 ......................................1

Substitute x = a, y = b into y = x2 − 8x + 12,b = a2 − 8a + 12 ..................................................21 = 2, 2a2 – 16a + 24 = a2 – 8a + 12 a2 − 8a + 12 = 0 (a − 2)(a − 6) = 0 a = 2, 6

Substitute a = 2 into 2, b = 4 − 16 + 12 = 0

Substitute a = 6 into 2, b = 36 – 48 + 12 = 0

Gradient of the tangent PA = 2(2) − 8 = −4

\ Equation of the tangent is y − 0 = − 4(x − 2) y = − 4x + 8

21



Gradient of the another tangent = 2(6) − 8 = 4Equation of the second tangent is y − 0 = 4(x − 6) y = 4x − 24

50. y = x(x2 − 4) y = x3 − 4x

dy

–––dx = 3x2 − 4

d 2y

––––dx2 = 6x

xd 2y

––––dx2 +

dy–––dx = x(6x) + (3x2 − 4)

= 9x2 − 4

xd 2y

––––dx2 +

dy–––dx , 0

9x2 − 4 , 0(3x − 2)(3x + 2) , 0

x02– –32–3

y

Therefore, the range is – 2—3 , x , 2—

3 .

51. (a) dr–––dt

= – 0.3 cm s−1

Time taken = 6–––0.3

= 20 seconds

(b) A = πr2

dA–––dr

= 2πr

dA–––dt

= dA–––dr

× dr–––dt

= 2πr1 dr–––dt 2

= 2π(4)(–0.3) = −2.4π The area is decreasing at the rate of 2.4π cm2 s−1.

52. (a) x2y + 1 = 3y + x x2y – 3y = x − 1 y(x2 – 3) = x − 1 y = x − 1–––––

x2 − 3

dy

–––dx =

(x2 − 3)(1) − (x − 1)(2x)––––––––––––––––––––

(x2 − 3)2

= x2 − 3 − 2x2 + 2x––––––––––––––– (x2 − 3)2

= −x2 + 2x − 3–––––––––––(x2 − 3)2

(b) Given dp–––dt = 6 units s−1

p = 3x + 2

dp–––dx = 3

dx–––dt

= dx–––dp

⋅ dp–––dt

= 1—3 (6)

= 2 units s−1

53. dA–––dt

= 10π cm2 s−1

Area, A = πr2

dA–––dr

= 2πr

Let p be the perimeter, p = 2πrdp–––dr = 2π

dp–––dt =

dp–––dA ⋅ dA–––

dt

= dp–––dr × dr–––

dA × dA–––

dt

= (2π)1 1––––2πr 2(10π)

= 10π––––r

When r = 10, dp–––dt = 10π––––

10 = π cm s−1

54. Given dV–––dt

= 30π cm3 s−1

Volume of the sphere, V = 4—3 πr3

\ dV–––dr

= 4πr2

Surface area of the sphere, A = 4πr2

dA–––dr

= 8πr

dA–––dt

= dA–––dV

⋅ dV–––dt

= dA–––dr

⋅ dr–––dV

⋅ dV–––dt

= (8πr)1 1––––4πr2 2(30π)

= 60π––––r

When the volume = 36π,4—3 πr3 = 36π

22



r3 = 36π1 3–––4π 2

= 27 r = 3

The rate of change of the surface area

= 60π––––3

= 20π cm2 s−1

55. (a) 6 cm

r cm10 cm

h cm

Using properties of similar triangles,

r—6 = h–––

10 r = 6–––

10 h

r = 3—5 h

Volume of the unfilled space, V

= 1—3 π62(10) − 1—

3 πr2h

= 1—3 π(360) − 1—

3 π1 3—5 h2

2h

= 120π − 1—3 π1 9–––

25 2h3

= 120π − 3–––25 πh3

= 3–––25 π(1000 – h3)

(b) dV–––dt

= 10π cm3 s−1, dV–––dh

= − 9–––25 πh2

dh–––dt

= dh–––dV

⋅ dV–––dt

= − 25–––––9πh2 × (10π)

= − 250––––9h2

= − 250–––––––9 × (2)2

= − 250––––36

= – 125––––18

The height of water is decreasing at the rate of

125––––18

cm s−1.

56. (a)

5 cm

5 cm

5 cmr cm

A BO

h cm

r2 + h2 = 52

r2 = 52 − h2

The surface area of the water, A = πr2

= π(52 − h2) = π(25 − h2)

(b) A = 25π − πh2

dA–––dh

= −2πh

dh–––dt

= −0.1 cm s−1

dA–––dt

= dA–––dh

⋅ dh–––dt

= (–2πh)(–0.1) = 0.2πh

When h = 3, dA–––dt

= 0.6π cm2 s−1.

57. (a) (i) T = 20ABBBl–––10

T 2 = 4001 l–––10 2

l = T 2–––40

dl–––dT

= 2T–––40

= T–––20

(ii) dT = 2.2 − 2.0 = 0.2

When T = 2.0,

dl–––dT

= 2–––20

= 0.1

dl = dl–––dT

× dT

= (0.1) × (0.2) = 0.02

(b) dr–––dt

= −0.5

When r = 10,

Volume = 4—3 π(10)3

= 4000–––––3 π

23



Half of original volume = 2000–––––3 π

4—3 πr3 = 2000–––––

3 π

r3 = 500 r = 3ABBB500 r = 7.937

Time taken = 10 – 7.937–––––––––0.5

= 4.126 seconds

58. (a) x + x + y + y = 24 2x + 2y = 24 x + y = 12 y = 12 − x

Volume of 1 unit of cuboid = t × t × 1 = t2 cm3

Total surface area of the cuboid = 2t2 + 4(t × 1) = 2t2 + 4t

Number of cuboids = xy

–––––––2t2 + 4t = y

x = 2t2 + 4t

Volume of total cuboids, V = yt2

= (12 − x)t2

= (12 – 2t2 − 4t)t2

= t2(12 – 2t2 − 4t) = 2t2(6 – t2 − 2t)

(b) V = 2t2(6 – t2 − 2t) = 12t2 – 2t4 − 4t3

dV–––dt

= 24t – 8t3 − 12t2

When dV–––

dt = 0,

24t – 8t3 − 12t2 = 0 8t3 + 12t2 − 24t = 0 4t(2t2 + 3t − 6) = 0

t = –3 ± ABBBBBBBBB32 – 4(2)(–6)–––––––––––––––––2(2)

= –3 ± ABB57–––––––––4

= –2.637, 1.137

Since t . 0 for the length, then t = 1.137

d 2V––––dt2 = 24 − 24t2 − 24t

When t = 1.137,

d 2V––––dt2 = 24 − 24(1.137)2 − 24(1.137)

= 24(1 – 1.1372 − 1.137) = −34.31 , 0

Hence, the volume is a maximum when t = 1.137.

1. 1—u + 1—v = 1–––12

v + u–––––uv = 1–––12

12(v + u) = uv 12v + 12u = uv uv – 12v = 12u v(u – 12) = 12u

v = 12u––––––u – 12

dv

–––du =

(u – 12) d–––du

(12u) – 12u d–––du

(u – 12)––––––––––––––––––––––––––––––

(u – 12)2

= (u – 12)(12) – 12u(1)–––––––––––––––––––

(u – 12)2

= 12u – 144 – 12u––––––––––––––(u – 12)2

= – 144––––––––(u – 12)2

2. 4xy = y + x 4xy – y = x y(4x – 1) = x y = x––––––

4x – 1

dy

–––dx =

(4x – 1) d–––dx

(x) – x d–––dx

(4x – 1)–––––––––––––––––––––––––

(4x – 1)2

= (4x – 1) – x(4)–––––––––––––(4x – 1)2

= 4x – 1 – 4x––––––––––(4x – 1)2

= – 1––––––––(4x – 1)2

3. y

–––x2 = (5 – 2x)4

y = x2(5 – 2x)4

dy

–––dx = x2 d–––

dx(5 – 2x)4 + (5 – 2x)4 d–––

dx(x2)

= x2 · 4(5 – 2x)3(–2) + (5 – 2x)4(2x) = –8x2(5 – 2x)3 + 2x(5 – 2x)4

= 2x(5 – 2x)3[–4x + (5 – 2x)] = 2x(5 – 2x)3(5 – 6x)

When dy

–––dx = 0,

2x(5 – 2x)3(5 – 6x) = 0x = 0, 5 – 2x = 0, 5 – 6x = 0 x = 5—

2 , x = 5—6

24



4. Given the gradient of the normal is –1,therefore, the gradient of the tangent is 1.

y = a—x + bx

dy

–––dx = – a–––

x2 + b .....................1

Substitute dy

–––dx = 1 for x = 1 into 1,

1 = – a + b – a + b = 1 ..............................2

Substitute x = 1, y = 7 into y = a—x + bx, 7 = a + b a + b = 7 ................................3

2 + 3, 2b = 8 b = 4

Substitute b = 4 into 3, a + 4 = 7 a = 3

5. y = x(x – 5)2

dy–––dx = x d–––

dx(x – 5)2 + (x – 5)2 d–––

dx(x)

= x · 2(x – 5)(1) + (x – 5)2(1) = (x – 5)[2x + (x – 5)] = (x – 5)(3x – 5)


–––dx = 0

(x – 5)(3x – 5) = 0 x = 5, 5—

3When x = 5,y = 5(5 – 5)2

= 0

When x = 5—3 ,

y = 5—3 1 5—

3 – 522

= 5—3 1– 10–––

3 22

= 500–––27

The stationary points are (5, 0) and ( 5—3 , 500––––

27).

dy

–––dx = (x – 5)(3x – 5)

d 2y––––dx2 = (x – 5)(3) + (3x – 5)(1)

= 3x – 15 + 3x – 5 = 6x – 20

For (5, 0),d 2y

––––dx2 = 6(5) – 20

= 10 . 0Therefore, (5, 0) is a minimum point.

For 1 5—3 , 500––––

27 2,d 2y

––––dx2 = 61 5—

3 2 – 20

= 10 – 20 = –10 , 0

Therefore, 1 5—3 , 500––––

27 2 is a maximum point.

When x = 0, y = 0(0 – 5)2

= 0When y = 0, x = 0, 5

0x

y

5

�–, –�

5–3

500–27

53

50027

6. (a) Area of rectangle ABCD = (2k)(k) = (2k2) cm2

Area of ∆ABE = 1—2 (2k)(x)

= (kx) cm2

Area of ∆FEC = 1—2 (2x)(k – x)

= x(k – x) = (kx – x2) cm2

Area of ∆ADF = 1—2 k(2k – 2x)

= (k2 – kx) cm2

Area of ∆AEF = 2k2 – [kx + (kx – x2) + (k2 – kx)] = 2k2 – kx – kx + x2 – k2 + kx = (k2 – kx + x2) cm2

(b) Let the area of ∆AEF be y y = k2 – kx + x2

dy

–––dx = –k + 2x

When dy

–––dx = 0,

–k + 2x = 0 x = 1—

2 k

d 2y

––––dx2 = 2 . 0

Therefore, y is a minimum when x = 1—2 k.

25



(c) Minimum area of ∆AEF

= k2 – k1 1—2 k2 + 1 1—

2 k22

= k2 – 1—2 k2 + 1—

4 k2

= 3—4 k2 cm2

7.

y cm

x cm

Perimeter of the circle = Perimeter of the rectangle 2p(7) = 2(x + y) x + y = 7p y = 7p – x ......................1

Area of the rectangle, A = xy = x(7p – x) = 7px – x2

dA–––dx

= 7p – 2x

= 71 22–––7 2 – 2x

= 22 – 2x

When dA–––dx

= 0,

22 – 2x = 0 x = 11

y = 7p – 11

= 71 22–––7 2 – 11

= 11

d 2A––––dx2 = –2 , 0

Therefore, A is a maximum when x = 11 and y = 11.

Maximum area of the rectangle = 11 × 11 = 121 cm2

8. Let V be the volume of the sphere, r be the radius and A be the surface area. t is the time in second.

dV–––dt

= 30p cm3 s–1

dA–––dt

= dA–––dV

· dV–––dt

= 1 dA–––dr

× dr–––dV 2 dV–––

dt

A = 4pr2 and V = 4—3 pr3

dA–––dr

= 8pr dV–––dr

= 4pr2

Therefore, dA–––dt

= 8pr × 1 1––––4πr2 2 × 30p

= 60p–––r

When r = 4,dA–––dt

= 60p––––4

= 15p cm2 s–1

9. (a) Given dV–––dt

= 8 cm3 s–1

Let the length of the side be x. V = x3

dV–––dx

= 3x2

dx–––dV

= 1––––3x2

dx–––dt

= dx–––dV

· dV–––dt

= 1 1––––3x2 2(8)

= 8––––3x2

When x = 2,

dx–––dt

= 8–––––3(2)2

= 2—3 cm s–1

(b) Total surface area, A = 6x2

dA–––dx

= 12x

dA–––dt

= dA–––dx

· dx–––dt

= 12x1 2—3 2

= 12(2)1 2—3 2

= 16 cm2 s–1

10. Given dV–––dt

= k cm3 s–1

Let h be the height of the water level.

Given dh–––dt

= 2 cm s–1

(a) V = p(5)2h = 25ph dV–––

dh = 25p

dV–––dt

= dV–––dh

· dh–––dt

k = (25p)(2) = 50p

26



(b) Area of the curved surface, A = 2p(5)h A = 10ph

dA–––dh

= 10p

dA–––dt

= dA–––dh

· dh–––dt

= (10p)(2) = 20p cm2 s–1

(c) Time taken = 1000p––––––50p

= 20 seconds

11. y = 1–––x3

dy

–––dx = – 3–––

x4

When x = 1,dy

–––dx = – 3–––

14

= –32–––––

0.993 = 23 1–––––0.993 4

= 23 1–––13 +

dy–––dx · dx4 Where dx = 0.99 – 1

= –0.01 = 2[1 + (–3)(–0.01)] = 2.06

12. (a) dy

–––dx = 3x2 – kx

Given equation of the normal is x – 7y + 13 = 0 7y = x + 13

y = 1—7 x + 13–––

7 Gradient of the tangent at (1, 2) is –7.

Therefore, dy

–––dx = –7 when x = 1

–7 = 3(1)2 – k(1) –7 = 3 – k k = 10

(b) Equation of the tangent at (1, 2) is y – 2 = –7(x – 1) y = –7x + 7 + 2 y = –7x + 9

13. y = x2 + nx + 2

dy

–––dx = 2x + n

(3, –7) is a minimum point,

therefore, dy

–––dx = 0 when x = 3

0 = 2(3) + nn = – 6

14. (a) y = ax3 + bx2 + 9x

dy

–––dx = 3ax2 + 2bx + 9


–––dx = 0 when x = 1 and

x = 3

When x = 1, 3a + 2b + 9 = 0 ...................1

When x = 3, 27a + 6b + 9 = 0 9a + 2b + 3 = 0 ...................2 2 – 1, 6a = 6 a = 1

Substitute a = 1 into 2, 9(1) + 2b + 3 = 0 2b = –12 b = –6

(b) y = x3 – 6x2 + 9x When x = 1, y = 1 – 6 + 9 = 4 \ (1, 4)

When x = 3, y = 33 – 6(3)2 + 9(3) = 0 \ (3, 0)

The distance between the two stationary points

= ABBBBBBBBBBBB(3 – 1)2 + (0 – 4)2

= ABBBBB4 + 16

= 4.472 units

15. (a) BD2 = x2 + y2

BD = ABBBBBx2 + y2

Radius of the circle is 1—2

ABBBBBx2 + y2

Area of the shaded region, A = Area of the circle – Area of the rectangle

= p1 1—2

ABBBBBx2 + y2 22 – xy

= p1 1—4 2(x2 + y2) – xy

= 1—4 p(x2 + y2) – xy

(b) A = 1—4 px2 + 1—

4 py2 – xy

= 1—4 px2 + 1—

4 p(10)2 – x(10)

= 1—4 px2 + 25p – 10x

dA–––dx

= 1—2 px – 10

27



When dA–––dx

= 0,

1—2 px – 10 = 0

x = 20–––p

d 2A––––dx2 = 1—

2 p . 0

Therefore, A is a minimum when x = 20–––p .

16. PQ = 2x – x2

Let PQ = s\ s = 2x – x2

ds–––dx

= 2 – 2x

When ds–––dx

= 0,

2 – 2x = 0 x = 1d 2s–––dx2 = –2 , 0

\ s is a maximum when x = 1.

Maximum distance of PQ = 2(1) – 12

= 1 unit

17. y = 2x + 1––––––x – 4

dy

–––dx =

(x – 4)(2) – (2x + 1)(1)––––––––––––––––––––

(x – 4)2

= 2x – 8 – 2x – 1–––––––––––––(x – 4)2

= –9–––––––(x – 4)2

dy

–––dx = –9(x – 4)–2

d 2y

––––dx2 = 18(x – 4)–3

= 18–––––––(x – 4)3

\ d 2y

––––dx2 +

dy–––dx = 18–––––––

(x – 4)3 + –9–––––––(x – 4)2

= 18 – 9(x – 4)–––––––––––(x – 4)3

= 18 – 9x + 36–––––––––––(x – 4)3

= 54 – 9x–––––––(x – 4)3

d 2y––––dx2 +

dy–––dx = 0

54 – 9x–––––––(x – 4)3 = 0

54 – 9x = 0 9x = 54 x = 6

18. (a) f (x) = 5––––––1 – 4x

= 5(1 – 4x)–1

f ′(x) = –5(1 – 4x)–2(–4) = 20(1 – 4x)–2

f ′′(x) = – 40(1 – 4x)–3(–4) = 160(1 – 4x)–3

f ′′(0) = 160(1 – 0)–3

= 160

(b) (i) y = x(x2 – 12) = x3 – 12x

dy

–––dx = 3x2 – 12

dx–––dy

= 1––––dy

–––dx

= 1––––––––3x2 – 12

(ii) dy

–––dx = 0

3x2 – 12 = 0 3x2 = 12 x2 = 4 x = ±2

When x = 2, y = 2(22 – 12) = –16

When x = –2, y = –2[(–2)2 – 12] = 16 \y = ±16

19. (a) y = 4x2 – 8x + 1

dy

–––dx = 8x – 8

d 2y

––––dx2 = 8

When d 2y

––––dx2 =

dy–––dx

8 = 8x – 8 8x = 16 x = 2 y = 4(2)2 – 8(2) + 1 = 1

28



(b) dy

–––dx = 0

8x – 8 = 0 x = 1

y = 4(1)2 – 8(1) + 1 = –3 Therefore, the stationary point is (1, –3).

d 2y

––––dx2 = 8 . 0

\ (1, –3) is a minimum point.

When x = 0, y = 1

When y = 0, 4x2 – 8x + 1 = 0

x = 8 ± 64 – 4(4)(1)ABBBBBBBBB––––––––––––––––

2(4)

= 8 ± ABB48––––––––8

= 8 ± 4AB3–––––––8

= 1 ± AB3–––2

1 – –x

y

023

1

(1, –3)

��1 + –2

3��

20. Given dx–––dt

= –0.01 cm s–1

x cm2x cm

16 cm

Total surface area,A = 2(2x2) + 2(16x) + 2(2x)(16) = 4x2 + 32x + 64x = 4x2 + 96x

dA–––dx

= 8x + 96

dA–––dt

= dA–––dx

· dx–––dt

= (8x + 96)(–0.01)

When volume = 162, (2x)(x)16 = 162

x2 = 162––––32

= 81–––16

x = 9—4 x . 0

\ dA–––dt

= 18 × 9—4 + 962(–0.01)

= –1.14 cm2 s–1

21.

10 cm

r cm

10 cm

12 cml cm

h cm

Let h be the height of water level in the cylinder and l be the height of water level in the cone.

l–––12 = r–––

10

r = 10–––12 l

= 5—6 l

V = 1—3 pr2l

= 1—3 p1 5—

6 l22l

= 25––––108 pl3

dV–––dl

= 25––––108 × 3pl2

= 25–––36 pl2

V = p102h = 100phdV–––dh

= 100p

dh–––dt

= 0.2 cm s–1

29



dl–––dt

= dl–––dh

· dh–––dt

= 1 dl–––dV

· dV–––dh 2(0.2)

= 1 36–––––25pl2 2(100p)(0.2)

= 36–––––––25p(6)2 (100p)(0.2)

= 0.8 cm s–1

Therefore, the water level in the cone is decreasing at a constant rate of 0.8 cm s–1.

22. (a) Given dh–––dt

= 10–––t m s–1

dV–––dt

= – 20–––t p cm3 s–1

= – 20–––t p × 10–6 m3 s–1

dV–––dh

= dV–––dt

× dt–––dh

= 1– 20–––t p2(10–6)1 t–––10 2

= –2 × 10–6 p

(b) dV–––dt

= dV–––dh

× dh–––dt

= (–2 × 10–6 p)(2) = – 4 × 10–6 p m3 s–1

= – 4 × 10–6 p × 106 cm3 s–1

= – 4p cm3 s–1

23. (a) p = 3x + 2, dp–––dt

= 6 units per second

dp–––dx

= 3

dx–––dt

= dx–––dp

× dp–––dt

= 1 1—3 2(6)

= 2 units s–1

(b) y = – 5–––p2

= – 5––––––––(3x + 2)2

= –5(3x + 2)–2

dy

–––dx

= 10(3x + 2)–3(3)

= 30––––––––(3x + 2)3

(c) dx = 1.01 – 1 = 0.01

dy = dy

–––dx

· dx

= 3 30––––––––(3x + 2)3 4 dx

= 30–––––––––

[3(1) + 2]3 × (0.01)

= 30–––53 × 0.01

= 0.0024

24. y = 3x2 – 4x + 3

dy

–––dx

= 6x – 4

At point (1, 2),dy

–––dx

= 6(1) – 4 = 2

(a) dy = 2.01 – 2 = 0.01

dx = dx–––dy

· dy

= 1 1—2 2(0.01)

= 0.005

(b) dx–––dt

= dx–––dy

· dy

–––dt

= 1 1—2 2(0.4)

= 0.2 unit s–1

25. (a) Given dx–––dt

= 0.1 cm s–1

Radius, r = 1—2 x

dr–––dx

= 1—2

dr–––dt

= dr–––dx

· dx–––dt

= 1 1—2 2(0.1)

= 0.05 cm s–1

(b) Area of the metal, A = x2 – pr2

= x2 – p1 1—2 x2

2

= x2 – 1—4 px2

30



dA–––dx

= 2x – 1—2 px

dA–––dt

= 1 dA–––dx 21 dx–––

dt 2 = 12x – 1—

2 px2(0.1)

= 32(2) – 1—2 p(2)4(0.1)

= (4 – p)(0.1) = 0.08584 cm2 s–1

(c) dx = 2.1 – 2 = 0.1

dA = dA–––dx

· dx

= 12x – 1—2 px2dx

= 32(2) – 1—2 p(2)4(0.1)

= (4 – p)(0.1) = 0.08584 cm2

09[anal add math cd]

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x dx3

x 12x2

x dxx2

substitute x

22x2 x 2x2

3x2 1x3c f x

dx dydx

dx y dy