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Additional Mathematics SPM Chapter 9
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1. (a) limx β 1 (3x β 2) = 3(1) β 2
= 1
(b) limx β 4 1 5ββββββ
2x β 1 2 = 5βββββββ2(4) β 1
= 5β7
(c) limx β 3 1 x2 β 9ββββββ
x β 3 2 = limx β 3 3 (x β 3)(x + 3)ββββββββββββ
(x β 3) 4 = limx β 3 (x + 3) = 3 + 3 = 6
(d) limx β 1 1 x2 β 3x + 2ββββββββββ
x β 1 2 = limx β 1 3 (x β 1)(x β 2)ββββββββββββ
(x β 1) 4 = limx β 1 (x β 2) = 1 β 2 = β1
(e) limx β β 1 2xββββββ
4x β 1 2 = limx β β 1
2xβββxβββββββ4x β 1ββββββx
2 = lim
x β β 1 2βββββββ4 β 1βx
2 = lim
x β β 1 2βββββ4 β 0 2
= 2β4
= 1β2
(f) limx β 0 1x2 + 3xββββββ
2x 2 = limx β 0 3 x(x + 3)βββββββ
2x 4 = limx β 0 1 x + 3βββββ
2 2 = 0 + 3βββββ
2
= 3β2
2. Gradient of the chord AB = β4 β 1β 1β
4 2ββββββββββ
β2 β 1β2
= β4 + 1β
4ββββββββ2 1β
2
= 15βββ4
Γ 2β5
= 3β2
3. Substitute x = 2, y = a into y = x2 + 1,a = 22 + 1 = 5
Gradient of the chord PQ = 5 β 2βββββββ2 β (β1)
= 1
4. (a) y = 4x + 1 ............................ 1 y + dy = 4(x + dx) + 1 ................. 2
2 β 1, dy = (4x + 4dx + 1) β (4x + 1) = 4dx
dy
βββdx = 4
dy
βββdx = lim
dx β 0 1 dyβββdx 2
= limdx β 0 (4)
= 4
(b) y = x2 β 4x .............................................. 1 y + dy = (x + dx)2 β 4(x + dx) y + dy = x2 + 2xdx + (dx)2 β 4x β 4dx............ 2
2 β 1, dy = 2xdx + (dx)2 β 4dx
dy
βββdx = 2xdxβββββ
dx + (dx)2
βββββdx
β 4dxββββdx
= 2x + dx β 4
dy
βββdx = lim
dx β 0 1 dyβββdx 2
= limdx β 0 (2x + dx β 4)
= 2x β 4
CHAPTER
9 Differentiation
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Additional Mathematics SPM Chapter 9
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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 β 1, dy = 3x2dx + 3x(dx)2 + (dx)3 + 2dx
dy
βββdx = 3x2 + 3xdx + (dx)2 + 2
dy
βββdx = lim
dx β 0 1 dyβββdx 2
= 3x2 + 2
Therefore, f β²(x) = 3x2 + 2
(b) Let y = 1 β 2x + 3x2 .......................................1 y + dy = 1 β 2(x + dx) + 3(x + dx)2
= 1 β 2x β 2dx + 3[x2 + 2xdx + (dx)2] = 1 β 2x β 2dx + 3x2 + 6xdx + 3(dx)2
y + dy = 1 β 2x β 2dx + 3x2 + 6xdx + 3(dx)2 ...2
2 β 1, dy = β2dx + 6xdx + 3(dx)2
dy
βββdx = β2 + 6x + 3dx
dy
βββdx = lim
dx β 0 1 dyβββdx 2
= β2 + 6x
Therefore, f β²(x) = β2 + 6x
6. Gradient of the tangent at the point A = 2(1) = 2
7. (a) y = 2x3
dy
βββdx = 6x2
(b) y = 3βββx2
= 3xβ2
dy
βββdx = (β2)(3)xβ3
= β 6βββx3
(c) y = β x 4βββ5
dy
βββdx = β 4x 3βββ
5
(d) y = 1βββ6x
= 1β6
xβ1
dy
βββdx = (β1)1 1β
6 2xβ2
= β 1βββ6x2
8. (a) y = 8x2
dy
βββdx = 16x
When x = β1, dy
βββdx = 16(β1)
= β16(b) y = β 9βββ
x3
= β9xβ3
dy
βββdx = (β3)(β9)xβ4
= 27βββx4
When x = 1, dy
βββdx = 27βββ
14
= 27(c) y = β 1βββ
2x = β 1β
2 xβ1
dy
βββdx = (β1)1β 1β
2xβ22
= 1βββ2x2
When x = 2, dy
βββdx = 1βββββ
2(2)2
= 1β8
9. (a) y = 4x2 β 3x + 5
dy
βββdx = 8x β 3
(b) y = 5x3 + 3βx β 4
= 5x3 + 3xβ1 β 4
dy
βββdx = 15x2 β 3xβ2
= 15x2 β 3βββx2
(c) y = x5 β 2βββ3x2 + 1
= x5 β 2β3
xβ2 + 1
dy
βββdx = 5x4 β (β2)1 2β
3 2xβ3
= 5x4 + 4βββ3x3
10. (a) f (x) = 4x2 + 5x f β²(x) = 8x + 5
(b) f (x) = 5x3 β 1 f β²(x) = 15x2
11. (a) dβββdx
(4x2 β 3x + 5)
= 8x β 3
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(b) dβββdx 1 8βx β 4x + 32
= dβββdx
(8x β1 β 4x + 3)
= β8x β2 β 4
= β 8βββx2
β 4
12. (a) y = 8x2 β 1β4
x + 3
dy
βββdx = 16x β 1β
4
(b) y = x3 + 1βββ2x2
= x3 + 1β2
xβ2
dy
βββdx = 3x2 β xβ3
= 3x2 β 1βββx3
(c) f (x) = 1β2
x2 β 8x + 1
f β²(x) = x β 8
13. (a) y = (4x β 1)(3x2)
dy
βββdx = (4x β 1)(6x) + (3x2)(4)
= 24x2 β 6x + 12x2
= 36x2 β 6x = 6x(6x β 1)
(b) y = (1 β 2x)(4x + 3)
dy
βββdx = (1 β 2x)(4) + (4x + 3)(β2)
= 4 β 8x β 8x β 6 = β2 β 16x = β2(1 + 8x)
(c) y = 2x1 1βx + 12(1 β 3x)
= (2 + 2x)(1 β 3x)
dy
βββdx = (2 + 2x)(β3) + (1 β 3x)(2)
= β6 β 6x + 2 β 6x = β4 β 12x = β4(1 + 3x)
(d) y = 4(x β 4)2
= 4(x2 β 8x + 16) = 4x2 β 32x + 64)
dy
βββdx = 8x β 32
= 8(x β 4)
14. (a) y = xββββββ2x β 3
dy
βββdx = (2x β 3)(1) β x(2)βββββββββββββββ
(2x β 3)2
= 2x β 3 β 2xββββββββββ(2x β 3)2
= β 3ββββββββ(2x β 3)2
(b) y = x2 + 3ββββββ2x β 5
dy
βββdx =
(2x β 5)(2x) β (x2 + 3)(2)βββββββββββββββββββββ
(2x β 5)2
= 4x2 β 10x β 2x2 β 6ββββββββββββββββ(2x β 5)2
= 2x2 β 10x β 6ββββββββββββ(2x β 5)2
= 2(x2 β 5x β 3)ββββββββββββ
(2x β 5)2
(c) y = 4x β 1ββββββx2 + 1
dy
βββdx =
(x2 + 1)(4) β (4x β 1)(2x)βββββββββββββββββββββ
(x2 + 1)2
= 4x2 + 4 β 8x2 + 2xββββββββββββββββ (x2 + 1)2
= β4x2 + 2x + 4ββββββββββββ(x2 + 1)2
= β2(2x2 β x β 2)βββββββββββββ
(x2 + 1)2
15. (a) y = (2x β 1)10
dy
βββdx = 10(2x β 1)9 dβββ
dx(2x β 1)
= 10(2x β 1)9(2) = 20(2x β 1)9
(b) y = (1 + 4x)7
dy
βββdx = 7(1 + 4x)6(4)
= 28(1 + 4x)6
(c) y = 2(x3 + 4)5
dy
βββdx = 5(2)(x3 + 4)4(3x2)
= 30x2(x3 + 4)4
(d) y = 3ββββββββ(2x + 1)4
= 3(2x + 1)β4
dy
βββdx = (β4)(3)(2x + 1)β5(2)
= β 24ββββββββ
(2x + 1)5
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(e) y = 1ββββββββ4(x2 β 1)5
= 1β4
(x2 β 1)β5
dy
βββdx = (β5)1 1β
4 2(x2 β 1)β6(2x)
= β 5β2
x(x2 β 1)β6
= β 5xββββββββ 2(x2 β 1)6
16. (a) y = 3x(1 β 2x)5
dy
βββdx = 3x Β· 5(1 β 2x)4(β2) + (1 β 2x)5(3)
= β30x(1 β 2x)4 + 3(1 β 2x)5
= β3(1 β 2x)4[10x β (1 β 2x)] = β3(1 β 2x)4(12x β 1)
(b) y = xβββββββ(x2 β 4)3
= x(x2 β 4)β3
dy
βββdx = x(β3)(x2 β 4)β4(2x) + (x2 β 4)β3(1)
= β6x2(x2 β 4)β4 + (x2 β 4)β3
= (x2 β 4)β4[β6x2 + (x2 β 4)] = (x2 β 4)β4(β5x2 β 4)
= β5x2 β 4ββββββββ(x2 β 4)4
(c) y = x2 + 1ββββββ 4x
= x2βββ4x
+ 1βββ4x
= xβ4
+ 1β4
xβ1
dy
βββdx = 1β
4 β 1β
4xβ2
= 1β4
β 1βββ4x2
(d) y = (4x β 1)(x2 β 3)4
dy
βββdx = (4x β 1) Β· 4(x2 β 3)3(2x) + (x2 β 3)4(4)
= 8x(4x β 1)(x2 β 3)3 + 4(x2 β 3)4
= 4(x2 β 3)3[2x(4x β 1) + (x2 β 3)] = 4(x2 β 3)3(8x2 β 2x + x2 β 3) = 4(x2 β 3)3(9x2 β 2x β 3)
(e) y = x2 + 3x β 4ββββββββββ x + 4
= (x + 4)(x β 1)ββββββββββββ (x + 4)
= x β 1
dy
βββdx = 1
(f) y = x3(2x β 1)3
dy
βββdx = x3 Β· 3(2x β 1)2(2) + (2x β 1)3(3x2)
= 6x3(2x β 1)2 + (3x2)(2x β 1)3
= 3x2(2x β 1)2[2x + (2x β 1)] = 3x2(2x β 1)2(4x β 1)
17. (a) (i) y = 3x2 β 1
dy
βββdx = 6x
Gradient of tangent at (1, 2) = 6(1) = 6
(ii) Equation of tangent is y β 2 = 6(x β 1) y β 2 = 6x β 6 y = 6x β 4 (iii) Gradient of normal = β 1β
6 Equation of normal is
y β 2 = β 1β6
(x β 1)
= β 1β6
x + 1β6
y = β 1β6
x + 1β6
+ 2
y = β 1β6
x + 13βββ6
(b) (i) y = 1βββx2 + 3
y = xβ2 + 3
dy
βββdx = β2xβ3
= β 2βββx3
Gradient of tangent at (β1, 4)
= β 2βββββ(β1)3
= 2
(ii) Equation of tangent is y β 4 = 2(x + 1) y = 2x + 2 + 4 y = 2x + 6
(iii) Equation of normal is
y β 4 = β 1β2
(x + 1)
y = β 1β2
x β 1β2
+ 4
y = β 1β2
x + 7β2
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Additional Mathematics SPM Chapter 9
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(c) (i) y = 4(2x β 1)3
dy
βββdx = 12(2x β 1)2(2)
= 24(2x β 1)2
Gradient of tangent at (1, 4) = 24[2(1) β 1]2
= 24 (ii) Equation of tangent is y β 4 = 24(x β 1) y = 24x β 24 + 4 y = 24x β 20 (iii) Equation of normal is
y β 4 = β 1βββ24
(x β 1)
y β 4 = β 1βββ24
x + 1βββ24
y = β 1βββ24
x + 1βββ24
+ 4
y = β 1βββ24
x + 97βββ24
(d) (i) y = x(2 β x)3
dy
βββdx = x(3)(2 β x)2(β1) + (2 β x)3(1)
= β3x(2 β x)2 + (2 β x)3
= (2 β x)2(β3x + 2 β x) = (2 β x)2(2 β 4x) = 2(2 β x)2(1 β 2x) Gradient of tangent at (2, 0) = 2(2 β 2)2[1 β 2(2)] = 0 (ii) Equation of tangent is y β 0 = 0(x β 2) y = 0 (iii) Equation of normal is x = 2
18. (a) y = 3x2 β 6x + 1 ...................................1
dy
βββdx = 6x β 6
For turning point, dy
βββdx = 0
6x β 6 = 0 x = 1
Substitute x = 1 into 1, y = 3 β 6 + 1 = β2 Hence, the turning point is (1, β2).
x 0 1 2
dyβββdx β6 0 6
Sketch of dyβββdx
(1, β2) is a minimum point.
(b) y = 1 β 3x β x2 .................. 1
dy
βββdx = β3 β 2x
dy
βββdx = 0,
β3 β 2x = 0 x = β 3β
2
Substitute x = β 3β2
into 1,
y = 1 β 31β 3β2 2 β 1β 3β
2 22
= 1 + 9β2
β 9β4
= 13βββ4
Hence, the turning point is 1β 3β2
, 13βββ4 2.
x β2 β 3β2 0
dyβββdx 1 0 β3
Sketch of dyβββdx
1β 3β2
, 13βββ4 2 is a maximum point.
(c) y = x3 β 2x2 + 4 ....................... 1
dy
βββdx = 3x2 β 4x
When dy
βββdx = 0,
3x2 β 4x = 0 x(3x β 4) = 0 x = 0, 4β
3 Substitute x = 0 into 1, y = 4
Substitute x = 4β3
into 1,
y = 1 4β3 2
3 β 21 4β
3 22 + 4
= 64βββ27
β 32βββ9
+ 4
= 76βββ27
The turning points are (0, 4) and 1 4β3
, 76βββ27 2.
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Additional Mathematics SPM Chapter 9
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For (0, 4),
x β1 0 1
dyβββdx 7 0 β1
Sketch of dyβββdx
(0, 4) is a maximum point.
For 1 4β3
, 76βββ27 2,
x 14β3 2
dyβββdx β1 0 4
Sketch of dyβββdx
1 4β3
, 76βββ27 2 is a minimum point.
(d) y = β2x3 + 6x ....................1
dy
βββdx = β6x2 + 6
0 = β6x2 + 6 β6(x2 β 1) = 0 β6(x + 1)(x β 1) = 0 x = β1, 1
Substitute x = β1 into 1, y = β2(β1)3 + 6(β1) = 2 β 6 = β4
Substitute x = 1 into 1, y = β2(1)3 + 6(1) = β2 + 6 = 4
The turning points are (β1, β4) and (1, 4).
For (β1, β4),
x β2 β1 0
dyβββdx β18 0 6
Sketch of dyβββdx
(β1, β4) is a minimum point.
For (1, 4),
x 0 1 2
dyβββdx 6 0 β18
Sketch of dyβββdx
(1, 4) is a maximum point.
19. x + y = 50 y = 50 β x ................................1
Area, A = xy ..................................2
Substitute 1 into 2, A = x(50 β x) = 50x β x2
dAβββdx
= 50 β 2x
0 = 50 β 2x x = 25
y = 25d 2Aββββdx2 = β2 , 0
Therefore, A is a maximum when x = 25 and y = 25.
Maximum area = xy = 25 Γ 25 = 625 unit2
20. Volume = 20 cm3
Οr2h = 20
h = 20ββββ Οr2
Surface area, A = 2Οr2 + 2Οrh
= 2Οr2 + 2Οr1 20ββββ Οr2 2
= 2Οr2 + 40βββ r
dAβββdr
= 4Οr β 40βββ r2
0 = 4Οr β 40βββ r2
40βββ r2 = 4Οr
r3 = 40βββ 4Ο
= 10βββ Ο
= 10 Γ 7βββ 22
= 70βββ 22
r = 1.471
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h = 20βββ Οr2
= 7 Γ 20βββββββββ22(1.471)2
= 2.941
d 2Aββββdr2 = 4Ο + 80βββ
r3
= 4Ο + 80ββββββ1.4713 . 0
Hence, the area is a minimum when r = 1.471 and h = 2.941.
21. drβββdt
= 0.1 cm sβ1
A = Οr 2
dAβββdr
= 2Οr
dAβββdt
= dAβββdr
Γ drβββdt
= 2Οr Γ (0.1) = 2Ο(5) Γ (0.1) = Ο cm2 sβ1
22. y = x2 β 2x
dy
βββdx = 2x β 2
dxβββdt
= 8 when x = 3,
dyβββdt
= dy
βββdx Γ dxβββ
dt = (2x β 2)(8) = (2 Γ 3 β 2)(8) = 32 units sβ1
23. dxβββdt
= 3.2 β 3.0ββββββββ2
= 0.1 cm sβ1
y = 3x2 β 1
dy
βββdx = 6x
dyβββdt =
dyβββdx Γ dxβββ
dt = (6x)(0.1) = (6 Γ 3.1)(0.1) = 1.86 cm sβ1
24. dVβββdt
= β9 mm3 sβ1
V = 4β3
Οr3
dVβββdr
= 4β3
Γ 3Οr2
= 4Οr2
drβββdt
= drβββdV
Γ dVβββdt
= 1 1ββββ4Οr2 2(β9)
= β 9ββββ4Οr2
= β 9βββββββ4Ο Γ 32
= β 1βββ4Ο
cm sβ1
25. y = 3x β 1 dy
βββdx = 3
dx = 2.01 β 2 = 0.01 unit
dyβββdx
β dy
βββdx
dy = dy
βββdx Γ dx
= 3(0.01) = 0.03 unit
26. y = x2 + 4dy
βββdx = 2x
dx = 1.9 β 2 = β0.1 unit
dy = dy
βββdx Γ dx
= 2x(β0.1) = 2(2)(β0.1) = β0.4 unit
27. y = 2βx
= 2xβ1 dy
βββdx = β 2βββ
x2
dy = 3.001 β 3 = 0.001 unit
When y = 3,
x = 2β y
= 2β3
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dx
βββdy
β dxβββdy
dx = dxβββdy
Γ dy
= 1β x2βββ 2 2(0.001)
= 3β 1 2β
3 22
βββββ2 4(0.001)
= β0.00022 unit
28. y = x2
dyβββdx = 2x
When x = 3,
dy
βββdx = 2(3)
= 6
(a) 3.12 = 32 + dy
= 9 + dy
βββdx Γ dx dx = 3.1 β 3
= 0.1 = 9 + (6)(0.1) = 9.6
(b) 2.92 = 32 + dy
= 9 + dy
βββdx Γ dx dx = 2.9 β 3
= β0.1 = 9 + (6)(β0.1) = 9 β 0.6 = 8.4
29. (a) y = 1βββ x2
y = xβ2
dy
βββdx = β 2βββ
x3
When x = 4,
dy
βββdx = β 2βββ
43
= β 1βββ 32
dx = 4.1 β 4 = 0.1
1ββββ 4.12 = 1βββ
42 + dy
= 1βββ 16
+ dy
βββdx Γ dx
= 1βββ 16
+ 1β 1βββ 32 2(0.1)
= 1βββ 16
β 0.1βββ 32
= 0.05938
(b) y = 1βββ x2
dy
βββdx = β 2βββ
x3
When x = 4,
dy
βββdx = β 1βββ
32 dx = 3.9 β 4.0 = β0.1
1ββββ 3.92 = 1βββ
42 + dy
= 1βββ 16
+ dy
βββdx Γ dx
= 1βββ 16
+ 1β 1βββ 32 2(β0.1)
= 0.06563
30. (a) y = ABx
= x1β2
dy
βββdx = 1β
2x
β 1β2
= 1ββββ2ABx
When x = 4,
dy
βββdx = 1ββββ
2AB4
= 1β 4
dx = 4.1 β 4 = 0.1
ABB4.1 = AB4 + dy
βββdx Γ dx
= 2 + 1β 4
(0.1)
= 2.025
(b) ABB3.9 = AB4 + dy
βββdx Γ dx dx = 3.9 β 4
= β0.1
= AB4 + 1β 4
(β 0.1)
= 2 β 0.1βββ 4
= 1.975
31. (a) y = 5x3 + 4x + 1
dy
βββdx = 15x2 + 4
d 2y
ββββdx2 = 30x
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(b) y = (4x2 β 1)5
dy
βββdx = 5(4x2 β 1)4(8x)
= 40x(4x2 β 1)4
d 2y
ββββdx2 = 40x Β· 4(4x2 β 1)3(8x) + (4x2 β 1)4(40)
= 1280x2(4x2 β 1)3 + 40(4x2 β 1)4
(c) y = 2βββ x3
= 2xβ3
dy
βββdx = β3(2xβ4)
= β6xβ4
d 2y
ββββdx2 = 24xβ5
= 24βββ x5
32. (a) f (x) = 4x3 β 1 f β²(x) = 12x2
f β²β²(x) = 24x
(b) f β²(x) = 5x2 + 4x β 3 f β²β²(x) = 10x + 4
(c) f (x) = 1βββ 2x3 β 5
= xβ3βββ 2
β 5
f β²(x) = β 3β 2
xβ4
f β²β²(x) = 12βββ2
xβ5
= 6βββ x5
33. (a) y = 4x2 β 4x + 1
dy
βββdx = 8x β 4
8x β 4 = 0
x = 4β 8
= 1β 2
y = 41 1β 2 2
2 β 41 1β
2 2 + 1
= 0
d 2y
ββββdx2 = 8 . 0
The turning point 1 1β 2
, 02 is a minimum point.
(b) y = 5 β 2x2 + 4x
dy
βββdx = β4x + 4
β4x + 4 = 0 x = 1
y = 5 β 2(1)2 + 4(1) = 7
d 2y
ββββdx2 = β4 , 0
The turning point (1, 7) is a maximum point.
(c) y = 1β 3
x3 β 2x2 + 50
dy
βββdx = x2 β 4x
x2 β 4x = 0 x(x β 4) = 0 x = 0, 4 When x = 0, y = 1β
3(0)3 β 2(0)2 + 50
= 50
When x = 4, y = 1β 3
(4)3 β 2(4)2 + 50
= 64βββ3
β 32 + 50
= 39 1β 3
d 2y
ββββdx2 = 2x β 4
For turning point (0, 50),
d 2y
ββββdx2 = β4 , 0
Therefore, (0, 50) is a maximum point.
For turning point 14, 39 1β 3 2,
d 2y
ββββdx2 = 2(4) β 4
= 4 . 0
Therefore, 14, 39 1β 3 2 is a minimum point.
(d) y = β 1β 3
x3 + x2 + 3x
dy
βββdx = βx2 + 2x + 3
βx2 + 2x + 3 = 0 x2 β 2x β 3 = 0 (x β 3)(x + 1) = 0 x = 3, β1
When x = 3, y = β 1β 3
(3)3 + 32 + 3(3)
= β9 + 9 + 9 = 9
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Additional Mathematics SPM Chapter 9
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When x = β1, y = β 1β 3
(β1)3 + (β1)2 + 3(β1)
= 1β 3
+ 1 β 3
= β1 2β 3
= β 5β 3
d 2y
ββββdx2 = β2x + 2
For turning point (3, 9),
d 2y
ββββdx2 = β2(3) + 2
= β4 , 0 Therefore, (3, 9) is a maximum point.
For turning point 1β1, β 5β 3 2,
d 2y
ββββdx2
= β2(β1) + 2 = 2 + 2 = 4 > 0
Therefore, 1β1, β 5β 3 2 is a minimum point.
34. y = x3 β 2x + 1
dy
βββdx = 3x2 β 2
d 2y
ββββdx2 = 6x
d 2y
ββββdx2 +
dyβββdx β y = βx3 β 8
6x + 3x2 β 2 β x3 + 2x β 1 = βx3 β 8 3x2 + 8x + 5 = 0 (3x + 5)(x + 1) = 0 x = β 5β
3, β1
1. ddx
1 x β 1x + 3 2 = f(x)
\ β«0
1
f(x)2
dx = 12
β«0
1
f(x) dx
= 12
3 x β 1x + 3 4
0
1
= 12
1 1 β 11 + 3
β 0 β 10 + 3 2
= 12
10 + 13 2
= 16
2. Gradient function = px + k
dy
βββdx = px + k
Since ( 1β 2
, 0) is a turning point,
dyβββdx = 0 when x = 1β
2
0 = p1 1β 2 2 + k
pβ 2
+ k = 0 .......................................... 1
Given gradient of normal at x = 2 is β 1βββ 12
.
\ Gradient of tangent at x = 2 is 12.dy
βββdx = 12 when x = 2
12 = p(2) + k 2p + k = 12 ........................................ 2
2 β 1, 2p β pβ 2
= 12
4pβββ 2
β pβ 2
= 12
3β 2
p = 12
p = 12 Γ 2β 3
= 8
Substitute p = 8 into 2, 2(8) + k = 12 k = 12 β 16 = β 4\ p = 8, k = β4.
3. f (x) = 1ββββββββ(2 β 3x)4
= (2 β 3x)β4
f β²(x) = β4(2 β 3x)β5(β3) = 12(2 β 3x)β5
f β²β²(x) = β5(12)(2 β 3x)β6(β3) = 180(2 β 3x)β6
f β²β²(1) = 180(β1)β6
= 180
4. dx = 1.9 β 2 = β0.1 y = 5x + x2
dyβββdx = 5 + 2x
dy = dy
βββdx dx
= (5 + 2x)dx = [5 + 2(2)](β0.1) = (9)(β0.1) = β 0.9
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Additional Mathematics SPM Chapter 9
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5. P = 100V
P = 100Vβ1
dPdV
= β 100V2
dP = (5 + P) β 5 = P
Using dPdV
= dPdV
dV = dPdPdV
= P
β100(5)2
= β P4
6. (a) y = β2x2 + 5x 0 = β2k2 + 5k k(β2k + 5) = 0 k = 0, k = 5β
2 Since k > 0, \k = 5β
2
(b) y = β2x2 + 5x
dy
βββdx = β4x + 5
= β41 5β 2 2 + 5
= β5
Gradient of normal at P 1 5β 2
, 02 is 1β 5
\ Equation of normal at point P
y β 0ββββββx β 5β
2
= 1β 5
y = 1β 5 1x β 5β
2 2 y = 1β
5x β 1β
2
7. (a) y = 7x(x β 3) = 7x2 β 21x
dy
βββdx = 14x β 21
d 2y
ββββdx2
= 14
For minimum y, dy
βββdx = 0
14x β 21 = 0 x = 21βββ
14 = 3β
2
(b) Minimum y = 71 3β 2 21 3β
2 β 32
= 71 3β 2 21β 3β
2 2 = β 63βββ
4
8. y = 3β x β 2x
y = 3xβ1 β 2x
dyβββdx = β3xβ2 β 2
= β 3βββ x2 β 2
When x = 3, dy
βββdx = β 3β
9 β 2
= β 7β 3
dyβββdt = 5
dxβββdt
= dxβββdy
Β· dy
βββdt
= 1β 3β 7 2(5)
= β 15βββ 7
units per second
9. y = 4kx2 + 6x
dy
βββdx = 8kx + 6
8k(3) + 6 = 10 24k = 4 k = 4βββ
24
= 1β 6
10. V = 1β 3
h3 + 12h
dVβββdh
= h2 + 12
When h = 3, dVβββdh
= 9 + 12
= 21
Substitutex = 5
2
m1m2 = β1
12
Additional Mathematics SPM Chapter 9
Β© Penerbitan Pelangi Sdn. Bhd.
Given dVβββdt
= 7 cm3 sβ1
dhβββdt
= dhβββdV
Γ dVβββdt
= 1βββ 21
Γ 7
= 1β 3
cm sβ1
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 dVβββdt
= p, dhβββdt
= 0.1
rβ h
= 0.4βββ 0.4
= 1Therefore, r = h
V = 1β 3
Οr2h
V = 1β 3
Οh3
dVβββdh
= Οh2
dVβββdt
= dVβββdh
Γ dhβββdt
p = Οh2 Γ 0.1
When h = 0.2,p = Ο(0.2)2(0.1) = 0.004Ο
1. limx β 0 1 x2 β 2xβββββββx 2 = limx β 0 (x β 2)
= β2
2. limx β β 1 2x βββββ1 + x 2 = lim
x β β 12xβββx ββββββ
1 + xβββββx2
= limx β β 1 2ββββββ
1βx + 12 = 2βββββ
0 + 1 = 2
3. limx β 2 1 4 β x2 ββββββ
x β 2 2 = limx β 2 (2 β x)(2 + x)ββββββββββββ
(x β 2) = limx β 2 [β(2 + x)] = β4
4. dy = 2xdx + 4dx2
dy
βββdx
= 2x + 8dx
dy
βββdx = lim
dx β 0 (2x + 8dx)
= 2x
5. y = 4x(x2 β 1)5
dyβββdx = 4x dβββ
dx(x2 β 1)5 + (x2 β 1)5 dβββ
dx(4x)
= 4x Β· 5(x2 β 1)4(2x) + (x2 β 1)5(4) = 40x2(x2 β 1)4 + 4(x2 β 1)5
6. f (x) = 2ββββββββ(1 β 4x)3
= 2(1 β 4x)β3
f β²(x) = (β3)(2)(1 β 4x)β4(β4)
= 24ββββββββ(1 β 4x)4
\ f β²(0) = 24βββ1
= 24
7. dβββdx 1 x β 1ββββββ
4 β x2 2
= (4 β x2) dβββ
dx(x β 1) β (x β 1) dβββ
dx(4 β x2)
ββββββββββββββββββββββββββββββββββ(4 β x2)2
= (4 β x2)(1) β (x β 1)(β2x)ββββββββββββββββββββββ (4 β x2)2
= 4 β x2 + 2x2 β 2xβββββββββββββββ (4 β x2)2
= x2 β 2x + 4ββββββββββ(4 β x2)2
8. dβββdx 1 x2 β 1ββββββ
x + 1 2
= dβββdx 3 (x + 1)(x β 1)ββββββββββββ
(x + 1) 4= dβββ
dx(x β 1)
= 1
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Additional Mathematics SPM Chapter 9
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9. y = 5x3 β 1β x
+ 2
y = 5x3 β xβ1 + 2
dy
βββdx = 15x2 + xβ2
d 2yβββdx2 = 30x β 2xβ3
= 30x β 2βββ x3
10. f (x) = 2(3 β 4x)6
f β²(x) = 6β 2(3 β 4x)5(β4) = (β 48)(3 β 4x)5
f β²β²(x) = β 48 β 5(3 β 4x)4(β4) = 960(3 β 4x)4
f β²β²(1) = 960(β1)4
= 960
11. y = 2x(x + 3) y = 2x2 + 6xdy
βββdx = 4x + 6
When x = 2,dy
βββdx = 4(2) + 6
= 14
12. y = 4x3 β 5x2 + 2x β 10dy
βββdx = 12x2 β 10x + 2
When dy
βββdx = 4,
12x2 β 10x + 2 = 412x2 β 10x β 2 = 0 6x2 β 5x β 1 = 0(6x + 1)(x β 1) = 0 x = β 1β
6, 1
13. y = (4 β 3x)5
dyβββdx = 5(4 β 3x)4(β3)
= β15(4 β 3x)4
The gradient function is β15(4 β 3x)4.
14. y = 1ββββββββ(1 + 2x)3
y = (1 + 2x)β3
dyβββdx = β3(1 + 2x)β4(2)
= β6(1 + 2x)β4
The gradient at the point (β1, β1) = β 6(1 β 2)β4
= β6βββ 1
= β6
15. y = (x β 1)(x + 1) y = x2 β 1dy
βββdx = 2x
The gradient of the tangent at the point (1, 0)= 2(1)= 2
16. y = x2 + 4x dy
βββdx = 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 β 1β 6
.
17. y = (2x + 5)2
dyβββdx = 2(2x + 5)(2)
= 4(2x + 5)
Given gradient of the tangent is β8.
dy
βββdx = β8
4(2x + 5) = β8 2x + 5 = β2 2x = β7 x = β 7β
2
Substitute x = β 7β 2
into y = (2x + 5)2,
y = 321β 7β 2 2 + 54
2
= 4
Therefore, the coordinates are 1β 7β 2
, 42.
18. y = 3(4x β 5)2 + 6 dy
βββdx = 6(4x β 5)(4)
= 24(4x β 5)
The gradient of the tangent for the point with gradient
of the normal 1β 2
is β2.
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Additional Mathematics SPM Chapter 9
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dy
βββdx = β2
24(4x β 5) = β2
4x β 5 = β 1βββ 12
4x = 5 β 1βββ 12
= 59βββ 12
x = 59βββ 48
Substitute x = 59βββ 48
into y = 3(4x β 5)2 + 6,
y = 3341 59βββ 48 2 β 54
2 + 6
= 31 59βββ 12
β 522 + 6
= 31β 1βββ 12 2
2 + 6
= 31 1ββββ 144 2 + 6
= 1βββ 48
+ 6
= 289ββββ 48
Therefore, the coordinates are 1 59βββ 48
, 289ββββ 48 2.
19. y = pβββ x2 + qx β 1 ......................1
y = pxβ2 + qx β 1dy
βββdx = β2pxβ3 + q
= β 2pβββ x3 + q
Given the gradient of the tangent at the point (β1, β3) is 14.
β 2pβββββ
(β1)3 + q = 14
2p + q = 14..........................2
Substitute x = β1, y = β3 into 1 since the point (β1, β3) lies on the curve 1,
β3 = pβββββ
(β1)2 + q(β1) β 1
β3 = p β q β 1p β q = β2 .....................................3
2 + 3, 3p = 12 p = 4
Substitute p = 4 into 3, 4 β q = β2 q = 6
20. y = 2x2 + ax + b .......................1dy
βββdx = 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 = β1
21. y = 2β x + 4x
= 2xβ1 + 4xdy
βββdx = β 2βββ
x2 + 4
The gradient of the tangent at (1, 6) = β 2βββ 12 + 4
= 2
The equation of the tangent is y β 6 = 2(x β 1) y = 2x β 2 + 6 y = 2x + 4
22. y = 4ββββββββ (3x β 1)2
= 4(3x β 1)β2
dyβββdx = β8(3x β 1)β3(3)
= β24ββββββββ (3x β 1)3
At the point (0, 4), the gradient of the tangent
= β24βββββ (β1)3
= 24
Therefore, the gradient of the normal is β 1βββ 24
.
Hence, the equation of the normal is
y β 4 = β 1βββ 24
(x β 0)
y = β 1βββ 24
x + 4
23. y = x3 β 6x2 + 3dy
βββdx = 3x2 β 12x
For the tangent parallel to the x-axis, gradient = 0
Therefore, dy
βββdx = 0
3x2 β 12x = 0 3x(x β 4) = 0 x = 0, 4
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Additional Mathematics SPM Chapter 9
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When x = 0,y = 03 β 6(0)2 + 3 = 3
When x = 4,y = 43 β 6(4)2 + 3 = 64 β 96 + 3 = β29
Hence, the points are (0, 3) and (4, β29).
24. Substitute y = 8 into y = βx2 + 4x + 4, 8 = βx2 + 4x + 4x2 β 4x + 4 = 0 (x β 2)2 = 0 x = 2
Therefore, p = 2 and q = 8.
25. y = x2(x β 3) β 1 y = x3 β 3x2 β 1 ........................1dy
βββdx = 3x2 β 6x
For stationary point, dy
βββdx = 0
3x2 β 6x = 0 3x(x β 2) = 0 x = 0, 2
Substitute x = 0 and x = 2 into 1 respectively,When x = 0, y = β1When x = 2, y = 23 β 3(2)2 β 1 = 8 β 12 β 1 = β5
Therefore, the stationary points are (0, β1) and (2, β5).
26. y = px2 + qx + 4 .......................1dy
βββdx = 2px + q
dyβββdx = 0 at the point (β1, 5).
2p(β1) + q = 0 β2p + q = 0 q = 2p ............................2
Substitute x = β1, y = 5 into 1, 5 = p(β1)2 + q(β1) + 4p β q = 1 .......................................3
Substitute 2 into 3,p β 2p = 1 βp = 1 p = β1
Substitute p = β1 into 2,q = 2(β1) = β2
27. y = βx3 + 6x2 β 9x β 2 .........................1dy
βββdx = β3x2 + 12x β 9
For stationary points, dy
βββdx = 0
β3x2 + 12x β 9 = 0Divide by (β3), x2 β 4x + 3 = 0 (x β 1)(x β 3) = 0 x = 1, 3
Substitute x = 1 and x = 3 into 1 respectively,When x = 1, y = β1 + 6 β 9 β 2 = β6When x = 3, y = β(3)3 + 6(3)2 β 9(3) β 2 = β27 + 54 β 27 β 2 = β2
The stationary points are (1, β6) and (3, β2).d 2y
ββββdx2 = β6x + 12
For point (1, β6), d 2y
ββββdx2 = β6(1) + 12
= 6 . 0
For point (3, β2), d 2y
ββββdx2 = β6(3) + 12
= β6 , 0
Therefore, the minimum point is (1, β6).
28. p = x2y and x + y = 10 y = 10 β x ........................ 1
Substitute 1 into p = x2y, p = x2(10 β x) = 10x2 β x3
dpβββdx = 20x β 3x2
When dpβββdx = 0,
20x β 3x2 = 0 x(20 β 3x) = 0 x = 0, x = 20βββ
3d2p
ββββdx2 = 20 β 6x
For x = 0, d2pβββdx2 = 20 β 6(0)
= 20 . 0
For x = 20βββ 3
, d2pβββdx2 = 20 β 61 20βββ
3 2 = 20 β 40 = β20 , 0
Therefore, for p to be maximum, x = 20βββ 3
.
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Additional Mathematics SPM Chapter 9
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29. y = (2x β 1)3
dyβββdx = 3(2x β 1)2(2)
= 6(2x β 1)2
Given dxβββdt
= 2
dyβββdt =
dyβββdx β dxβββ
dt = 6(2x β 1)2(2) = 12(2x β 1)2
When x = 1,dy
βββdt = 12(1)2
= 12 units per second
30. h = xy h = (1 β 2t)(1 + 3t)dhβββdt
= (1 β 2t)(3) + (1 + 3t)(β2)
= 3(1 β 2t) β 2(1 + 3t) = 3 β 6t β 2 β 6t = 1 β 12t
When t = 2,dhβββdt
= 1 β 12(2) = β23
31. 1 βv β 1 βu = 1β 7
u β vβββββvu = 1β 7
7u β 7v = uvuv + 7v = 7uv(u + 7) = 7u v = 7u ββββββ
u + 7
dvβββdu
= (u + 7)(7) β (7u)(1)βββββββββββββββββ(u + 7)2
= 7u + 49 β 7uβββββββββββ(u + 7)2
= 49βββββββ(u + 7)2
Given duβββdt
= 12
dvβββdt
= dvβββdu
β duβββdt
= 49βββββββ(u + 7)2
Γ 12
= 49βββ 122 Γ 12
= 49βββ 12
units sβ1
32. pv = 20
p = 20βββ v = 20vβ1
dpβββdv = β20vβ2
= β 20βββ v2
When v = 2, dpβββdv = β 20βββ
22 and dv = 2.01 β 2
= β5 = 0.01
dp = dpβββdv Γ dv
= (β5)(0.01) = β0.05
33. y = 2βx β 5
= 2xβ1 β 5dy
βββdx = β2xβ2
= β 2ββx2
When x = 2, dy
βββdx = β 2βββ
22 and dy = 1.9 β 2
= β 1β2
= β 0.1
dxβββdy β dxβββ
dy
dx = dxβββdy Γ dy
= (β2)(β0.1) = 0.2
34. y = 2x2 + 3dy
βββdx = 4x
When x = 4, dy
βββdx = 4(4) and dx = 4.1 β 4
= 16 = 0.1
dy = dy
βββdx Γ dx
= 16(0.1) = 1.6
dyβββy Γ 100 = 1.6ββββββββ
2(4)2 + 3 Γ 100
= 1.6βββ35
Γ 100
= 4.57%
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Additional Mathematics SPM Chapter 9
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35. y = 27βββx3
= 27xβ3
dyβββdx = β81xβ4
= β 81βββx4
When x = 3, dy
βββdx = β 81βββ
34 and dx = 3.01 β 3
= β1 = 0.01
27ββββββ(3.01)3
= 27βββ33 + dy
= 1 + dy
βββdx Γ dx
= 1 + (β1)(0.01) = 1 β 0.01 = 0.99
36. y = 4βββx2
= 4xβ2
dyβββdx = β8xβ3
= β 8βββx3
When x = 2, dy
βββdx = β 8βββ
23
= β1dx = 1.9 β 2 = β 0.1
4ββββ1.92 = 4βββ
22 + dy
= 1 + dy
βββdx Γ dx
= 1 + (β1)(β0.1) = 1 + 0.1 = 1.1
37. y = 1βββx3
= xβ3
dyβββdx = β3xβ4
= β 3βββx4
When x = 1,dy
βββdx = β3
dx = 1.01 β 1 = 0.01
3ββββββ(1.01)3 = 33 1ββββββ
(1.01)3 4 = 31 1βββ
13 + dy2 = 311 +
dyβββdx Γ dx2
= 3[1 + (β3)(0.01)] = 2.91
38. (a) y = 4t2 + t ...............................1
x = 1 β 2t
t = 1 β x βββββ 2 ...............................2
Substitute 2 into 1,
y = 43 1 β x βββββ 2 4
2 + 1 1 β x βββββ
2 2 = 41 (1 β x)2
βββββββ 4 2 + 1β
2 β 1β2 x
= (1 β x)2 + 1β2 β 1β
2 x
dy
βββdx = 2(1 β x)(β1) β 1β
2
= β2 + 2x β 1β2
= 2x β 5β2
(b) When t = 2, x = 1 β 2(2) = β3
When t = 2.01, x = 1 β 2(2.01) = β3.02 dx = β3 β (β3.02) = 0.02
When x = β3, dy
βββdx = 2(β3) β 5β
2 = β6 β 5β
2 = β 17βββ
2
dy = dy
βββdx Γ dx
= 1β 17βββ2 2(0.02)
= β0.17
39. (a) y = 2t2 + 1, x = 1 β 2t
dy
βββdt = 4t dxβββ
dt = β2
dy
βββdx =
dyβββdt β dtβββ
dx
= 4t1β 1β2 2
= β2t
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Additional Mathematics SPM Chapter 9
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(b) When x = 3, 3 = 1 β 2t
t = β 2β2
t = β1
dy
βββdx = β2(β1)
= 2
40. (a) y = x2 β 4x + 1
dy
βββdx = 2x β 4
d 2y
ββββdx2
= 2
d 2y
ββββdx2 + 1 dy
βββdx 2
2 β y = 2x + 1
2 + (2x β 4)2 β (x2 β 4x + 1) = 2x + 1 2 + 4x2 β 16x + 16 β x2 + 4x β 1 β 2x β 1 = 0 3x2 β 14x + 16 = 0 (x β 2)(3x β 8) = 0 x = 2, 8β
3
(b) (i) dy
βββdx = kx β 5
y + 7x β 5 = 0 y = β7x + 5 The gradient of the tangent at the point
(β1, 12) is β7.
Therefore, dy
βββdx = β7 when x = β1.
kx β 5 = β7 k(β1) β 5 = β7 βk = β2 k = 2
(ii) The gradient of the normal = 1β7
The equation of the normal is
y β 12 = 1β7 (x + 1)
y = 1β7 x + 1β
7 + 12
y = 1β7 x + 85βββ
7
41. (a) y = 2x β 3 has a gradient of 2 at point P. y = x3 + 3x2 β 7x + 2
dy
βββdx = 3x2 + 6x β 7
dy
βββdx = 2
3x2 + 6x β 7 = 2
3x2 + 6x β 9 = 0 x2 + 2x β 3 = 0 (x β 1)(x + 3) = 0 x = 1, β3
When x = 1, y = x3 + 3x2 β 7x + 2 = 13 + 3(1)2 β 7(1) + 2 = 1 + 3 β 7 + 2 = β1
When x = β3, y = x3 + 3x2 β 7x + 2 = (β3)3 + 3(β3)2 β 7(β3) + 2 = β27 + 27 + 21 + 2 = 23
Therefore, the coordinates of P are (1, β1).
(b) Another point is (β3, 23).
42. (a) Area of the shaded region = Area of βOAB β Area of rectangle OPQR = 1β
2 Γ 5 Γ 10 β xy
= 25 β xy
Gradient of AB = Gradient of QA
10 β 0ββββββ 0 β 5
= y β 0βββββx β 5
y = β2(x β 5) y = β2x + 10
Therefore, the area of the shaded region, A = 25 β x(β2x + 10) = 25 + 2x2 β 10x
(b) A = 2x2 β10x + 25
dAβββdx
= 4x β 10, d 2Aββββdx2 = 4 . 0
4x β 10 = 0
x = 10βββ4
= 5β2
Minimum area = 25 + 21 5β2 2
2 β 101 5β
2 2 = 25 + 25βββ
2 β 50βββ2
= 25 β 25βββ2
= 25βββ2
43. (a) Perimeter = 120 y + y + 8x + 2(60 β 9x) = 120 2y + 8x + 120 β 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 βCDE = 8x(60 β 9x) + 1β
2 (8x)(ABBBBBBBy2 β 16x2 )
= 480x β 72x2 + 4xABBBBBBBB25x2 β 16x2
= 480x β 72x2 + 4x(3x) = 480x β 72x2 + 12x2
= 480x β 60x2
(b) A = 480x β 60x2
dAβββdx
= 480 β 120x
d 2Aββββdx2 = β120 , 0
When dAβββdx
= 0,
480 β 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 βABM, AM 2 = AB2 β BM 2 = 102 β 82
= 36 cm AM = 6 cm Since βABC and βASR are similar, then
y
βββ16 = 6 β xβββββ
6
y = 16(6 β x)ββββββββ6
= 8(6 β x)βββββββ 3
Area of rectangle PQRS, A = xy
= x3 8β3 (6 β x)4
= x116 β 8β3 x2
= 16x β 8β3 x2
(b) A = 16x β 8β3 x2
dAβββdx
= 16 β 16βββ3 x
d 2Aββββdx2 = β 16βββ
3 , 0
When dAβββdx
= 0,
16 β 16βββ3 x = 0
16βββ3 x = 16
x = 3
Substitute x = 3 into y = 8β3 (6 β x),
y = 8β3 Γ 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 β 8x) cm
EM 2 = EF2 β MF2
= 100x2 β 64x2
= 36x2
EM = 6x cm EA = 6x + 10 β 8x = (10 β 2x) cm
Volume of the solid, V = Area of trapezium Γ BC
= 1β2 (8x)(10 β 8x + 10 β 2x) Γ 5
= 20x(20 β 10x) = 400x β 200x2
(b) (i) V = 400x β 200x2
dVβββdx
= 400 β 400x
dVβββdx
= 0
400 β 400x = 0 x = 1
d 2Vββββdx2 = β400 , 0
Therefore, V is a maximum when x = 1.
(ii) Maximum value of V = 400(1) β 200(1)2
= 200
46. (a) Height of the box = 2 β xβββββ 2
Volume of the box, V = (x)(x)1 2 β xβββββ 2 2
= x211 β xβ2 2
= x2 β 1β2 x3
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(b) V = x2 β 1β2 x3
dVβββdx
= 2x β 3β2 x2
When dVβββdx
= 0,
2x β 3β2 x2 = 0
x12 β 3β2 x2 = 0
x = 0, 4β3
d 2Vββββdx2 = 2 β 3x
When x = 0, d 2Vββββdx2 = 2 . 0
When x = 4β3 , d 2Vββββ
dx2 = 2 β 31 4β3 2
= β2 , 0
Therefore, the volume is a maximum when x = 4β
3 .
47. (a) Volume of the cylinder = 81Ο cm3
Οr2h = 81Ο
h = 81βββr2
Total surface area, A = Οr2 + 2Οrh + 14Οr2ββββ
2 2 = Οr2 + 2Οrh + (2Οr2) = 3Οr2 + 2Οrh
= 3Οr2 + 2Οr1 81βββr2 2
= 3Οr2 + 162Οββββr
(b) A = 3Οr2 + 162Οr β1
dAβββdr
= 6Οr β 162Οr β2
= 6Οr β 162Οββββr2
When dAβββdr
= 0,
6Οr β 162Οββββr2 = 0
6Οr = 162Οββββr2
r3 = 27 r = 3
d 2Aββββdr2 = 6Ο + 324Οβββββ
r3
When r = 3,
d 2Aββββdr2 = 18Ο . 0
Therefore, the total surface area is a minimum when r = 3.
48. (a) y = px2 β 4x + 1
dy
βββdx = 2px β 4
When x = 4, the gradient of the tangent is 0. Therefore, 0 = 2p(4) β 4 8p = 4 p = 1β
2
(b) When x = 4, y = 1β2 (4)2 β 4(4) + 1
= 8 β 16 + 1 = β7
Equation of the tangent is y = β7.
49.
A(4, β8)
P(a, b)P(a, b)
x0
y = x 2 β 8x + 12y
Let the point of contact between the tangent and the curve be P(a, b). y = x2 β 8x + 12dy
βββdx = 2x β 8
Gradient of the tangent at point P is 2a β 8.
Gradient of PA = b + 8βββββ a β 4
b + 8βββββ a β 4
= 2a β 8
b + 8 = (a β 4)(2a β 8) = 2a2 β 16a + 32 b = 2a2 β 16a + 24 ......................................1
Substitute x = a, y = b into y = x2 β 8x + 12,b = a2 β 8a + 12 ..................................................21 = 2, 2a2 β 16a + 24 = a2 β 8a + 12 a2 β 8a + 12 = 0 (a β 2)(a β 6) = 0 a = 2, 6
Substitute a = 2 into 2, b = 4 β 16 + 12 = 0
Substitute a = 6 into 2, b = 36 β 48 + 12 = 0
Gradient of the tangent PA = 2(2) β 8 = β4
\ Equation of the tangent is y β 0 = β 4(x β 2) y = β 4x + 8
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Additional Mathematics SPM Chapter 9
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Gradient of the another tangent = 2(6) β 8 = 4Equation of the second tangent is y β 0 = 4(x β 6) y = 4x β 24
50. y = x(x2 β 4) y = x3 β 4x
dy
βββdx = 3x2 β 4
d 2y
ββββdx2 = 6x
xd 2y
ββββdx2 +
dyβββdx = x(6x) + (3x2 β 4)
= 9x2 β 4
xd 2y
ββββdx2 +
dyβββdx , 0
9x2 β 4 , 0(3x β 2)(3x + 2) , 0
x02β β32β3
y
Therefore, the range is β 2β3 , x , 2β
3 .
51. (a) drβββdt
= β 0.3 cm sβ1
Time taken = 6βββ0.3
= 20 seconds
(b) A = Οr2
dAβββdr
= 2Οr
dAβββdt
= dAβββdr
Γ drβββdt
= 2Οr1 drβββdt 2
= 2Ο(4)(β0.3) = β2.4Ο The area is decreasing at the rate of 2.4Ο cm2 sβ1.
52. (a) x2y + 1 = 3y + x x2y β 3y = x β 1 y(x2 β 3) = x β 1 y = x β 1βββββ
x2 β 3
dy
βββdx =
(x2 β 3)(1) β (x β 1)(2x)ββββββββββββββββββββ
(x2 β 3)2
= x2 β 3 β 2x2 + 2xβββββββββββββββ (x2 β 3)2
= βx2 + 2x β 3βββββββββββ(x2 β 3)2
(b) Given dpβββdt = 6 units sβ1
p = 3x + 2
dpβββdx = 3
dxβββdt
= dxβββdp
β dpβββdt
= 1β3 (6)
= 2 units sβ1
53. dAβββdt
= 10Ο cm2 sβ1
Area, A = Οr2
dAβββdr
= 2Οr
Let p be the perimeter, p = 2Οrdpβββdr = 2Ο
dpβββdt =
dpβββdA β dAβββ
dt
= dpβββdr Γ drβββ
dA Γ dAβββ
dt
= (2Ο)1 1ββββ2Οr 2(10Ο)
= 10Οββββr
When r = 10, dpβββdt = 10Οββββ
10 = Ο cm sβ1
54. Given dVβββdt
= 30Ο cm3 sβ1
Volume of the sphere, V = 4β3 Οr3
\ dVβββdr
= 4Οr2
Surface area of the sphere, A = 4Οr2
dAβββdr
= 8Οr
dAβββdt
= dAβββdV
β dVβββdt
= dAβββdr
β drβββdV
β dVβββdt
= (8Οr)1 1ββββ4Οr2 2(30Ο)
= 60Οββββr
When the volume = 36Ο,4β3 Οr3 = 36Ο
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Additional Mathematics SPM Chapter 9
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r3 = 36Ο1 3βββ4Ο 2
= 27 r = 3
The rate of change of the surface area
= 60Οββββ3
= 20Ο cm2 sβ1
55. (a) 6 cm
r cm10 cm
h cm
Using properties of similar triangles,
rβ6 = hβββ
10 r = 6βββ
10 h
r = 3β5 h
Volume of the unfilled space, V
= 1β3 Ο62(10) β 1β
3 Οr2h
= 1β3 Ο(360) β 1β
3 Ο1 3β5 h2
2h
= 120Ο β 1β3 Ο1 9βββ
25 2h3
= 120Ο β 3βββ25 Οh3
= 3βββ25 Ο(1000 β h3)
(b) dVβββdt
= 10Ο cm3 sβ1, dVβββdh
= β 9βββ25 Οh2
dhβββdt
= dhβββdV
β dVβββdt
= β 25βββββ9Οh2 Γ (10Ο)
= β 250ββββ9h2
= β 250βββββββ9 Γ (2)2
= β 250ββββ36
= β 125ββββ18
The height of water is decreasing at the rate of
125ββββ18
cm sβ1.
56. (a)
5 cm
5 cm
5 cmr cm
A BO
h cm
r2 + h2 = 52
r2 = 52 β h2
The surface area of the water, A = Οr2
= Ο(52 β h2) = Ο(25 β h2)
(b) A = 25Ο β Οh2
dAβββdh
= β2Οh
dhβββdt
= β0.1 cm sβ1
dAβββdt
= dAβββdh
β dhβββdt
= (β2Οh)(β0.1) = 0.2Οh
When h = 3, dAβββdt
= 0.6Ο cm2 sβ1.
57. (a) (i) T = 20ABBBlβββ10
T 2 = 4001 lβββ10 2
l = T 2βββ40
dlβββdT
= 2Tβββ40
= Tβββ20
(ii) dT = 2.2 β 2.0 = 0.2
When T = 2.0,
dlβββdT
= 2βββ20
= 0.1
dl = dlβββdT
Γ dT
= (0.1) Γ (0.2) = 0.02
(b) drβββdt
= β0.5
When r = 10,
Volume = 4β3 Ο(10)3
= 4000βββββ3 Ο
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Additional Mathematics SPM Chapter 9
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Half of original volume = 2000βββββ3 Ο
4β3 Οr3 = 2000βββββ
3 Ο
r3 = 500 r = 3ABBB500 r = 7.937
Time taken = 10 β 7.937βββββββββ0.5
= 4.126 seconds
58. (a) x + x + y + y = 24 2x + 2y = 24 x + y = 12 y = 12 β x
Volume of 1 unit of cuboid = t Γ t Γ 1 = t2 cm3
Total surface area of the cuboid = 2t2 + 4(t Γ 1) = 2t2 + 4t
Number of cuboids = xy
βββββββ2t2 + 4t = y
x = 2t2 + 4t
Volume of total cuboids, V = yt2
= (12 β x)t2
= (12 β 2t2 β 4t)t2
= t2(12 β 2t2 β 4t) = 2t2(6 β t2 β 2t)
(b) V = 2t2(6 β t2 β 2t) = 12t2 β 2t4 β 4t3
dVβββdt
= 24t β 8t3 β 12t2
When dVβββ
dt = 0,
24t β 8t3 β 12t2 = 0 8t3 + 12t2 β 24t = 0 4t(2t2 + 3t β 6) = 0
t = β3 Β± ABBBBBBBBB32 β 4(2)(β6)βββββββββββββββββ2(2)
= β3 Β± ABB57βββββββββ4
= β2.637, 1.137
Since t . 0 for the length, then t = 1.137
d 2Vββββdt2 = 24 β 24t2 β 24t
When t = 1.137,
d 2Vββββdt2 = 24 β 24(1.137)2 β 24(1.137)
= 24(1 β 1.1372 β 1.137) = β34.31 , 0
Hence, the volume is a maximum when t = 1.137.
1. 1βu + 1βv = 1βββ12
v + uβββββuv = 1βββ12
12(v + u) = uv 12v + 12u = uv uv β 12v = 12u v(u β 12) = 12u
v = 12uββββββu β 12
dv
βββdu =
(u β 12) dβββdu
(12u) β 12u dβββdu
(u β 12)ββββββββββββββββββββββββββββββ
(u β 12)2
= (u β 12)(12) β 12u(1)βββββββββββββββββββ
(u β 12)2
= 12u β 144 β 12uββββββββββββββ(u β 12)2
= β 144ββββββββ(u β 12)2
2. 4xy = y + x 4xy β y = x y(4x β 1) = x y = xββββββ
4x β 1
dy
βββdx =
(4x β 1) dβββdx
(x) β x dβββdx
(4x β 1)βββββββββββββββββββββββββ
(4x β 1)2
= (4x β 1) β x(4)βββββββββββββ(4x β 1)2
= 4x β 1 β 4xββββββββββ(4x β 1)2
= β 1ββββββββ(4x β 1)2
3. y
βββx2 = (5 β 2x)4
y = x2(5 β 2x)4
dy
βββdx = x2 dβββ
dx(5 β 2x)4 + (5 β 2x)4 dβββ
dx(x2)
= x2 Β· 4(5 β 2x)3(β2) + (5 β 2x)4(2x) = β8x2(5 β 2x)3 + 2x(5 β 2x)4
= 2x(5 β 2x)3[β4x + (5 β 2x)] = 2x(5 β 2x)3(5 β 6x)
When dy
βββdx = 0,
2x(5 β 2x)3(5 β 6x) = 0x = 0, 5 β 2x = 0, 5 β 6x = 0 x = 5β
2 , x = 5β6
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4. Given the gradient of the normal is β1,therefore, the gradient of the tangent is 1.
y = aβx + bx
dy
βββdx = β aβββ
x2 + b .....................1
Substitute dy
βββdx = 1 for x = 1 into 1,
1 = β a + b β a + b = 1 ..............................2
Substitute x = 1, y = 7 into y = aβx + 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 β 5)2
dyβββdx = x dβββ
dx(x β 5)2 + (x β 5)2 dβββ
dx(x)
= x Β· 2(x β 5)(1) + (x β 5)2(1) = (x β 5)[2x + (x β 5)] = (x β 5)(3x β 5)
For stationary points, dy
βββdx = 0
(x β 5)(3x β 5) = 0 x = 5, 5β
3When x = 5,y = 5(5 β 5)2
= 0
When x = 5β3 ,
y = 5β3 1 5β
3 β 522
= 5β3 1β 10βββ
3 22
= 500βββ27
The stationary points are (5, 0) and ( 5β3 , 500ββββ
27).
dy
βββdx = (x β 5)(3x β 5)
d 2yββββdx2 = (x β 5)(3) + (3x β 5)(1)
= 3x β 15 + 3x β 5 = 6x β 20
For (5, 0),d 2y
ββββdx2 = 6(5) β 20
= 10 . 0Therefore, (5, 0) is a minimum point.
For 1 5β3 , 500ββββ
27 2,d 2y
ββββdx2 = 61 5β
3 2 β 20
= 10 β 20 = β10 , 0
Therefore, 1 5β3 , 500ββββ
27 2 is a maximum point.
When x = 0, y = 0(0 β 5)2
= 0When y = 0, x = 0, 5
0x
y
5
οΏ½β, βοΏ½
5β3
500β27
53
50027
6. (a) Area of rectangle ABCD = (2k)(k) = (2k2) cm2
Area of βABE = 1β2 (2k)(x)
= (kx) cm2
Area of βFEC = 1β2 (2x)(k β x)
= x(k β x) = (kx β x2) cm2
Area of βADF = 1β2 k(2k β 2x)
= (k2 β kx) cm2
Area of βAEF = 2k2 β [kx + (kx β x2) + (k2 β kx)] = 2k2 β kx β kx + x2 β k2 + kx = (k2 β kx + x2) cm2
(b) Let the area of βAEF be y y = k2 β kx + x2
dy
βββdx = βk + 2x
When dy
βββdx = 0,
βk + 2x = 0 x = 1β
2 k
d 2y
ββββdx2 = 2 . 0
Therefore, y is a minimum when x = 1β2 k.
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(c) Minimum area of βAEF
= k2 β k1 1β2 k2 + 1 1β
2 k22
= k2 β 1β2 k2 + 1β
4 k2
= 3β4 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 β x ......................1
Area of the rectangle, A = xy = x(7p β x) = 7px β x2
dAβββdx
= 7p β 2x
= 71 22βββ7 2 β 2x
= 22 β 2x
When dAβββdx
= 0,
22 β 2x = 0 x = 11
y = 7p β 11
= 71 22βββ7 2 β 11
= 11
d 2Aββββdx2 = β2 , 0
Therefore, A is a maximum when x = 11 and y = 11.
Maximum area of the rectangle = 11 Γ 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.
dVβββdt
= 30p cm3 sβ1
dAβββdt
= dAβββdV
Β· dVβββdt
= 1 dAβββdr
Γ drβββdV 2 dVβββ
dt
A = 4pr2 and V = 4β3 pr3
dAβββdr
= 8pr dVβββdr
= 4pr2
Therefore, dAβββdt
= 8pr Γ 1 1ββββ4Οr2 2 Γ 30p
= 60pβββr
When r = 4,dAβββdt
= 60pββββ4
= 15p cm2 sβ1
9. (a) Given dVβββdt
= 8 cm3 sβ1
Let the length of the side be x. V = x3
dVβββdx
= 3x2
dxβββdV
= 1ββββ3x2
dxβββdt
= dxβββdV
Β· dVβββdt
= 1 1ββββ3x2 2(8)
= 8ββββ3x2
When x = 2,
dxβββdt
= 8βββββ3(2)2
= 2β3 cm sβ1
(b) Total surface area, A = 6x2
dAβββdx
= 12x
dAβββdt
= dAβββdx
Β· dxβββdt
= 12x1 2β3 2
= 12(2)1 2β3 2
= 16 cm2 sβ1
10. Given dVβββdt
= k cm3 sβ1
Let h be the height of the water level.
Given dhβββdt
= 2 cm sβ1
(a) V = p(5)2h = 25ph dVβββ
dh = 25p
dVβββdt
= dVβββdh
Β· dhβββdt
k = (25p)(2) = 50p
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(b) Area of the curved surface, A = 2p(5)h A = 10ph
dAβββdh
= 10p
dAβββdt
= dAβββdh
Β· dhβββdt
= (10p)(2) = 20p cm2 sβ1
(c) Time taken = 1000pββββββ50p
= 20 seconds
11. y = 1βββx3
dy
βββdx = β 3βββ
x4
When x = 1,dy
βββdx = β 3βββ
14
= β32βββββ
0.993 = 23 1βββββ0.993 4
= 23 1βββ13 +
dyβββdx Β· dx4 Where dx = 0.99 β 1
= β0.01 = 2[1 + (β3)(β0.01)] = 2.06
12. (a) dy
βββdx = 3x2 β kx
Given equation of the normal is x β 7y + 13 = 0 7y = x + 13
y = 1β7 x + 13βββ
7 Gradient of the tangent at (1, 2) is β7.
Therefore, dy
βββdx = β7 when x = 1
β7 = 3(1)2 β k(1) β7 = 3 β k k = 10
(b) Equation of the tangent at (1, 2) is y β 2 = β7(x β 1) y = β7x + 7 + 2 y = β7x + 9
13. y = x2 + nx + 2
dy
βββdx = 2x + n
(3, β7) is a minimum point,
therefore, dy
βββdx = 0 when x = 3
0 = 2(3) + nn = β 6
14. (a) y = ax3 + bx2 + 9x
dy
βββdx = 3ax2 + 2bx + 9
For stationary points, dy
βββdx = 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 β 1, 6a = 6 a = 1
Substitute a = 1 into 2, 9(1) + 2b + 3 = 0 2b = β12 b = β6
(b) y = x3 β 6x2 + 9x When x = 1, y = 1 β 6 + 9 = 4 \ (1, 4)
When x = 3, y = 33 β 6(3)2 + 9(3) = 0 \ (3, 0)
The distance between the two stationary points
= ABBBBBBBBBBBB(3 β 1)2 + (0 β 4)2
= ABBBBB4 + 16
= 4.472 units
15. (a) BD2 = x2 + y2
BD = ABBBBBx2 + y2
Radius of the circle is 1β2
ABBBBBx2 + y2
Area of the shaded region, A = Area of the circle β Area of the rectangle
= p1 1β2
ABBBBBx2 + y2 22 β xy
= p1 1β4 2(x2 + y2) β xy
= 1β4 p(x2 + y2) β xy
(b) A = 1β4 px2 + 1β
4 py2 β xy
= 1β4 px2 + 1β
4 p(10)2 β x(10)
= 1β4 px2 + 25p β 10x
dAβββdx
= 1β2 px β 10
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Additional Mathematics SPM Chapter 9
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When dAβββdx
= 0,
1β2 px β 10 = 0
x = 20βββp
d 2Aββββdx2 = 1β
2 p . 0
Therefore, A is a minimum when x = 20βββp .
16. PQ = 2x β x2
Let PQ = s\ s = 2x β x2
dsβββdx
= 2 β 2x
When dsβββdx
= 0,
2 β 2x = 0 x = 1d 2sβββdx2 = β2 , 0
\ s is a maximum when x = 1.
Maximum distance of PQ = 2(1) β 12
= 1 unit
17. y = 2x + 1ββββββx β 4
dy
βββdx =
(x β 4)(2) β (2x + 1)(1)ββββββββββββββββββββ
(x β 4)2
= 2x β 8 β 2x β 1βββββββββββββ(x β 4)2
= β9βββββββ(x β 4)2
dy
βββdx = β9(x β 4)β2
d 2y
ββββdx2 = 18(x β 4)β3
= 18βββββββ(x β 4)3
\ d 2y
ββββdx2 +
dyβββdx = 18βββββββ
(x β 4)3 + β9βββββββ(x β 4)2
= 18 β 9(x β 4)βββββββββββ(x β 4)3
= 18 β 9x + 36βββββββββββ(x β 4)3
= 54 β 9xβββββββ(x β 4)3
d 2yββββdx2 +
dyβββdx = 0
54 β 9xβββββββ(x β 4)3 = 0
54 β 9x = 0 9x = 54 x = 6
18. (a) f (x) = 5ββββββ1 β 4x
= 5(1 β 4x)β1
f β²(x) = β5(1 β 4x)β2(β4) = 20(1 β 4x)β2
f β²β²(x) = β 40(1 β 4x)β3(β4) = 160(1 β 4x)β3
f β²β²(0) = 160(1 β 0)β3
= 160
(b) (i) y = x(x2 β 12) = x3 β 12x
dy
βββdx = 3x2 β 12
dxβββdy
= 1ββββdy
βββdx
= 1ββββββββ3x2 β 12
(ii) dy
βββdx = 0
3x2 β 12 = 0 3x2 = 12 x2 = 4 x = Β±2
When x = 2, y = 2(22 β 12) = β16
When x = β2, y = β2[(β2)2 β 12] = 16 \y = Β±16
19. (a) y = 4x2 β 8x + 1
dy
βββdx = 8x β 8
d 2y
ββββdx2 = 8
When d 2y
ββββdx2 =
dyβββdx
8 = 8x β 8 8x = 16 x = 2 y = 4(2)2 β 8(2) + 1 = 1
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Additional Mathematics SPM Chapter 9
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(b) dy
βββdx = 0
8x β 8 = 0 x = 1
y = 4(1)2 β 8(1) + 1 = β3 Therefore, the stationary point is (1, β3).
d 2y
ββββdx2 = 8 . 0
\ (1, β3) is a minimum point.
When x = 0, y = 1
When y = 0, 4x2 β 8x + 1 = 0
x = 8 Β± 64 β 4(4)(1)ABBBBBBBBBββββββββββββββββ
2(4)
= 8 Β± ABB48ββββββββ8
= 8 Β± 4AB3βββββββ8
= 1 Β± AB3βββ2
1 β βx
y
023
1
(1, β3)
οΏ½οΏ½1 + β2
3οΏ½οΏ½
20. Given dxβββdt
= β0.01 cm sβ1
x cm2x cm
16 cm
Total surface area,A = 2(2x2) + 2(16x) + 2(2x)(16) = 4x2 + 32x + 64x = 4x2 + 96x
dAβββdx
= 8x + 96
dAβββdt
= dAβββdx
Β· dxβββdt
= (8x + 96)(β0.01)
When volume = 162, (2x)(x)16 = 162
x2 = 162ββββ32
= 81βββ16
x = 9β4 x . 0
\ dAβββdt
= 18 Γ 9β4 + 962(β0.01)
= β1.14 cm2 sβ1
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.
lβββ12 = rβββ
10
r = 10βββ12 l
= 5β6 l
V = 1β3 pr2l
= 1β3 p1 5β
6 l22l
= 25ββββ108 pl3
dVβββdl
= 25ββββ108 Γ 3pl2
= 25βββ36 pl2
V = p102h = 100phdVβββdh
= 100p
dhβββdt
= 0.2 cm sβ1
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Additional Mathematics SPM Chapter 9
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dlβββdt
= dlβββdh
Β· dhβββdt
= 1 dlβββdV
Β· dVβββdh 2(0.2)
= 1 36βββββ25pl2 2(100p)(0.2)
= 36βββββββ25p(6)2 (100p)(0.2)
= 0.8 cm sβ1
Therefore, the water level in the cone is decreasing at a constant rate of 0.8 cm sβ1.
22. (a) Given dhβββdt
= 10βββt m sβ1
dVβββdt
= β 20βββt p cm3 sβ1
= β 20βββt p Γ 10β6 m3 sβ1
dVβββdh
= dVβββdt
Γ dtβββdh
= 1β 20βββt p2(10β6)1 tβββ10 2
= β2 Γ 10β6 p
(b) dVβββdt
= dVβββdh
Γ dhβββdt
= (β2 Γ 10β6 p)(2) = β 4 Γ 10β6 p m3 sβ1
= β 4 Γ 10β6 p Γ 106 cm3 sβ1
= β 4p cm3 sβ1
23. (a) p = 3x + 2, dpβββdt
= 6 units per second
dpβββdx
= 3
dxβββdt
= dxβββdp
Γ dpβββdt
= 1 1β3 2(6)
= 2 units sβ1
(b) y = β 5βββp2
= β 5ββββββββ(3x + 2)2
= β5(3x + 2)β2
dy
βββdx
= 10(3x + 2)β3(3)
= 30ββββββββ(3x + 2)3
(c) dx = 1.01 β 1 = 0.01
dy = dy
βββdx
Β· dx
= 3 30ββββββββ(3x + 2)3 4 dx
= 30βββββββββ
[3(1) + 2]3 Γ (0.01)
= 30βββ53 Γ 0.01
= 0.0024
24. y = 3x2 β 4x + 3
dy
βββdx
= 6x β 4
At point (1, 2),dy
βββdx
= 6(1) β 4 = 2
(a) dy = 2.01 β 2 = 0.01
dx = dxβββdy
Β· dy
= 1 1β2 2(0.01)
= 0.005
(b) dxβββdt
= dxβββdy
Β· dy
βββdt
= 1 1β2 2(0.4)
= 0.2 unit sβ1
25. (a) Given dxβββdt
= 0.1 cm sβ1
Radius, r = 1β2 x
drβββdx
= 1β2
drβββdt
= drβββdx
Β· dxβββdt
= 1 1β2 2(0.1)
= 0.05 cm sβ1
(b) Area of the metal, A = x2 β pr2
= x2 β p1 1β2 x2
2
= x2 β 1β4 px2
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Additional Mathematics SPM Chapter 9
Β© Penerbitan Pelangi Sdn. Bhd.
dAβββdx
= 2x β 1β2 px
dAβββdt
= 1 dAβββdx 21 dxβββ
dt 2 = 12x β 1β
2 px2(0.1)
= 32(2) β 1β2 p(2)4(0.1)
= (4 β p)(0.1) = 0.08584 cm2 sβ1
(c) dx = 2.1 β 2 = 0.1
dA = dAβββdx
Β· dx
= 12x β 1β2 px2dx
= 32(2) β 1β2 p(2)4(0.1)
= (4 β p)(0.1) = 0.08584 cm2