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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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
(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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
(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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
(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
Sketch of dy–––dx
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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
For (0, 4),
x –1 0 1
dy–––dx 7 0 –1
Sketch of dy–––dx
(0, 4) is a maximum point.
For 1 4—3
, 76–––27 2,
x 14—3 2
dy–––dx –1 0 4
Sketch of dy–––dx
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
Sketch of dy–––dx
(−1, −4) is a minimum point.
For (1, 4),
x 0 1 2
dy–––dx 6 0 –18
Sketch of dy–––dx
(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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
(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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
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
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Additional Mathematics SPM Chapter 9
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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
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Additional Mathematics SPM Chapter 9
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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
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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.
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Additional Mathematics SPM Chapter 9
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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
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Additional Mathematics SPM Chapter 9
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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
.
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Additional Mathematics SPM Chapter 9
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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%
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Additional Mathematics SPM Chapter 9
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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
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(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
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Additional Mathematics SPM Chapter 9
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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
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(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
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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
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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
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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
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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)
For stationary points, dy
–––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.
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(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
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(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
For stationary points, dy
–––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
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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
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(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
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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
Additional Mathematics SPM Chapter 9
© Penerbitan Pelangi Sdn. Bhd.
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