Remainder & Factor Theorems
Exercise 8A
6 (f) cos [pic] = (cos [pic] = ([pic]
(g) sin [pic] = (sin [pic] = ([pic]
7 (c) [pic] = [pic] = ([pic]
(d) [pic] = [pic] = [pic] × [pic]
= [pic] = [pic]
9 (e) Basic (, ( = [pic], 0 < ( < 2(
( = [pic], [pic]
( = [pic], ( ( [pic], ( + [pic], 2( ( [pic] = [pic], [pic], [pic], [pic]
10 (c) sin [pic] = (sin [pic]
(d) tan [pic] = tan [pic]
15 tan A = ([pic] tan B = 2
(a) sin A = [pic] (c) sin B = ([pic]
(b) cos A = ([pic] (d) cos B = ([pic]
17 Given: cos 160( = (q
(a) cos 20( = (cos 160( = q
(b) sin 160( = sin 20( = [pic] = [pic]
(c) tan ((20() = (tan 20( = ([pic]
(d) cos 70( = sin 20( = [pic]
Exercise 8B
4 (b) y = 4 + 2sin 2x, 0 ( x ( ( 6 g(x) = p sin x + q, p > 0
Since (p ( p sin x ( p
(p + q ( p sin x + q ( p + q
Given: p + q = 10 ----- (1)
And (p + q = (4 ----- (2)
(1) + (2): 2q = 6 ( q = 3
From (1): p = 3 + 4 = 7
(p = 7 and q = 3.
7 (b) y = 3sin x ( 1, 0 ( x ( 2( (c) y = 3 + 2cos 2x, 0 ( x ( 2(
The range is 1 ( y ( 5.
The range is (4 ( y ( 2.
9 y = sin 2x, and y = cos 3x, 0( ( x ( 360(
From the sketch, for sin 2x = cos 3x,
there are 6 solutions.
11 (i) For y = p cos ax, a = 3 and p = 2.
For y = q sin x + b, q = 4.
p and q refer to the amplitudes of the curves, while a refers to the number of cycles
within 360(.
(ii) b = (1, and it moves the original curve down by 1.
Exercise 8C
3 y = [pic] and y = sin x, 0 ( x ( ( 4 y = | 2sin x | and y = [pic], 0 ( x ( 2(
For [pic] and y = sin x, from the sketch, From the sketch, for
there are 2 solutions. (a) 5( sin x = 3x,
| 2sin x | = [pic], there are 4 solutions.
(b) 5( sin x = 3x,
2sin x = [pic], there are 2 solutions.
Exercise 8D
4 sec x = 2, 0( < x < 180(
( cos x = [pic]
(i) sin x = [pic]
(ii) cosec x = [pic] = [pic] = [pic]
(iii) cot x = [pic] = [pic] = [pic]
6 cot x = 2.5, x is acute
( tan x = [pic]
(i) sin x = [pic] = [pic]
(ii) sin [pic] = cos x = [pic] = [pic]
(iii) sec [pic] = [pic] = [pic] = [pic]
Exercise 8E
4 (e) tan 3x ( 3sin 30 = 0(, 0( < x < 360(
tan 3x = [pic] 0( < 3x < 1080(
Basic ( = 56.31(
3x = 56.31(, 180( + 56.31(, 360( + 56.31(, 540( + 56.31(, 720( + 56.31(,
900( + 56.31(
x = 18.8(, 78.8(, 138.8(, 198.8(, 258.8(, 318.8(
5 (e) tan (2x ( 78() = (1.57, 0( ( x ( 360(
Basic ( = 57.51( (78( ( 2x ( 78( ( 642(
2x ( 78( = 180( ( 57.51(, 360( ( 57.51(, 540( ( 57.51(, (57.51(
x = 100.2(, 190.2(, 280.2(, 10.2(
5 (f) sin ([pic]x ( 27() = 0.6, 0( ( x ( 360(
Basic ( = 36.87( (27( ( [pic]x ( 27( ( 153(
[pic]x ( 27( = 36.87(, 180( ( 36.87(
x = 127.7(, 340.3(
6 (c) [pic] = 3, (( < x < (
7sin x + 6 = 3 ( 3sin x
10sin x = (3
sin x = (0.3
Basic ( = 0.3047
x = (0.3047, ((( ( 0.3047)
x = (0.305, (2.84
7 (c) tan [pic] = (1.48, 0 ( x ( 2(
Basic ( = 0.9766
x ( [pic] = ( ( 0.9766, 2( ( 0.9766
x = 2.95, 6.09
7 (d) 2sin (2x + 0.4) = (0.83, 0 ( x ( 2(
sin (2x + 0.4) = (0.415
Basic ( = 0.4279
2x + 0.4 = ( + 0.4279, 2( ( 0.4279, 2( + 0.4279, 4( ( 0.4279
x = 1.58, 2.73, 4.73, 5.87
10 (c) sin2 x = [pic]sin x, 0 ( x ( 2(
9sin2 x ( sin x = 0
sin x (9sin x ( 1) = 0
sin x = 0 or sin x = [pic]
Basic ( = 0 Basic ( = 0.1113
x = 0, (, 2( x = 0.1113, ( ( 0.1113
x = 0, (, 2(, 0.1113, 3.03
10 (d) sin 2x tan 2x = sin 2x, 0 ( x ( 2(
sin 2x tan 2x ( sin 2x = 0
sin 2x (tan 2x ( 1) = 0
sin 2x = 0 or tan 2x = 1
Basic ( = 0 Basic ( = [pic]
2x = 0, (, 2(, 3(, 4( 2x = [pic], ( + [pic], 2( + [pic], 3( + [pic]
x = 0, [pic], (, [pic], 2(, [pic], [pic], [pic], [pic]
11 (c) sin x = tan x, 0( ( x ( 360(
sin x = [pic]
sin x cos x = sin x
sin x cos x ( sin x = 0
sin x (cos x ( 1) = 0
sin x = 0 or cos x = 1
Basic ( = 0( Basic ( = 0(
x = 0(, 180(, 360(, x = 0(, 360(
(x = 0(, 180(, 360(
11 (d) 4cos x = 2cot x, 0( ( x ( 360(
2cos x = [pic]
2cos x sin x = cos x
2cos x sin x ( cos x = 0
cos x (2sin x ( 1) = 0
cos x = 0 or sin x = [pic]
Basic ( = 90( Basic ( = 30(
x = 90(, 270(, x = 30(, 150(
(x = 30, 90(, 150(, 270(
Exercise 9A
3 (d) [pic] = 7 + [pic], 0( ( x ( 360(
3sec2 x = 7 + 4tan x
3(tan2 x + 1) ( 7 ( 4tan x
3tan2 x ( 4tan x ( 4 = 0
(tan x ( 2)(3tan x + 2) = 0
tan x = 2 or tan x = ([pic]
Basic ( = 63.43( Basic ( = 33.69(
x = 63.43(, 180( + 63.43(, x = 180( ( 33.69(, 360( ( 33.69(
(x = 63.4(, 243.4(, 146.3(, 326.3(
4 (c) 4 + sin x tan x = 4cos x, 0 ( x ( 2(
4 + [pic] = 4cos x
4cos x + sin2 x = 4cos2 x
4cos x + (1 ( cos2 x) = 4cos2 x
5cos2 x ( 4cos x ( 1 = 0
(cos x ( 1)(5cos x + 1) = 0
cos x = 1 or cos x = ([pic]
Basic ( = 0 Basic ( = 1.369
x = 0, 2( x = ( ( 1.369, ( + 1.369
x = 0, 2(, 1.77, 4.51
4 (d) [pic] ( 14cosec x + 5 = 0, 0 ( x ( 2(
[pic] ( 14cosec x + 5 = 0
8cosec2 x ( 14cosec x + 5 = 0
(2cosec x ( 1)(4cosec x ( 5) = 0
cosec x = [pic] or cosec x = [pic]
sin x = 2 or sin x = [pic]
(rejected as (1 ( sinx ( 1) Basic ( = 0.9273
x = 0.9273, ( ( 0.9273
x = 0, 927, 2.21
Exercise 9B
2 (e) Prove: [pic] = sec x
LHS = (1 + [pic]) ÷ (1 + cos x)
= [pic] × [pic] = [pic] = sec x = RHS
2 (f) Prove: [pic] = (sec2 x
LHS = [pic]
= ([pic] ÷ sin2 x
= ([pic] × [pic] = ([pic] = (sec2 x = RHS
3 (g) Prove: [pic] = sin x cos x
LHS = 1 ÷ ([pic] + [pic])
= 1 ÷ [pic]
= 1 ÷ [pic] = sin x cos x = RHS
3 (h) Prove: [pic] + [pic] = 2cosec x
LHS = [pic]
= [pic]
= [pic]
= [pic] = [pic] = 2cosec x = RHS
3 (q) Prove: cos2 x + cot2 x cos2 x = cot2 x
LHS = cos2 x (1 + cot2 x)
= cos2 x cosec2 x
= cos2 x × [pic] = cot2 x = RHS
3 (r) Prove: [pic] = sin2 x ( cos2 x
LHS = (1 ( [pic]) ÷ (1 + [pic])
= [pic] ÷ [pic]
= [pic] ÷ [pic]
= [pic] × sin2 x = sin2 x ( cos2 x = RHS
3 (s) Prove: sec4 x ( tan4 x = [pic]
LHS = (sec2 x)2 ( (tan2 x)2
= (sec2 x ( tan2 x)(sec2 x + tan2 x)
= (1)([pic] + [pic]) = [pic] = RHS
Exercise 9C
7 (i) sin (A + B)
= sin A cos B + cos A sin B
= (([pic])(([pic]) + (([pic])(([pic]) = [pic]
(ii) cos (A + B)
= cos A cos B ( sin A sin B
= (([pic])(([pic]) ( (([pic])(([pic]) = ([pic]
(iii) tan (A ( B) = [pic]
= ([pic] ( [pic]) ÷ [1 + ([pic])([pic])] = [pic]
9 Given: sin (A ( B) = [pic], sin A cos B = [pic], A & B are acute angles.
(i) sin (A ( B) = sin A cos B ( cos A sin B
[pic] = [pic] ( cos A sin B
cos A sin B = [pic] ( [pic] = [pic]
(ii) sin (A + B) = sin A cos B + cos A sin B
=[pic] + [pic] = [pic]
(iii) 3tan A cot B = 3([pic])([pic])
= 3([pic]) ÷ [pic] = 5[pic]
11 (i) sin (A + B)
= sin A cos B + cos A sin B
= ([pic])([pic]) + ([pic])(b)
= [pic]
(ii) cos (A ( B)
= cos A cos B + sin A sin B
= ([pic])([pic]) + ([pic])(b) = [pic]
(iii) tan (A ( B) = [pic]
= (a ( [pic]) ÷ [1 + (a)( [pic])]
= [pic] ÷ [pic] = [pic]
12 (i) sin (A + B)
= sin A cos B + cos A sin B
= (x)(y) + (([pic])(([pic])
= xy + [pic]
(ii) cos (A ( B)
= cos A cos B + sin A sin B
= (([pic])(y) + (x)(([pic]) = (y[pic] ( x[pic]
(iii) tan 2A = [pic]
= ([pic] ÷ [1 ( ([pic])2]
= ([pic] ÷ [pic]
= ([pic] = [pic]
(iv) sin 2B = 2sin B cos B
= 2(([pic])(y) = (2y[pic]
(v) cos 2B = 2cos2 B ( 1 = 2y2 ( 1
Exercise 9D
6 (i) sin 2A
= 2sin A cos A
= 2([pic])(([pic]) = ([pic]
(ii) cos A = 2cos2 [pic]A ( 1
2cos2 [pic]A = ([pic] + 1 = [pic] [90( ( A ( 180(, so 45( ( [pic]A ( 90(]
cos2 [pic]A = [pic]
cos [pic]A = ([pic] = [pic] ([pic]45( ( [pic]A ( 90()
(iii) cos 2B = 2cos2 B ( 1
= 2(([pic])2 ( 1 = ([pic]
9 (i) cos 2x = 2cos2 x ( 1
= 2(([pic])2 ( 1 = ([pic]
(ii) tan 2x = [pic]
= [pic] = 1[pic]
(iii) From (ii), [pic] = [pic]
sin 2x = [pic](([pic]) = ([pic]
sin 4x = 2sin 2x cos 2x
= 2[([pic]] [([pic]] = [pic]
10 (c) 4cos 2x ( 3sin x + 1 = 0, 0( ( x ( 360(
4(1 ( 2sin2 x) ( 3sin x + 1 = 0
8sin2 x + 3sin x ( 5 = 0
(8sin x ( 5)(sin x + 1) = 0
sin x = [pic] or sin x = (1
Basic ( = 38.68( Basic ( = 90(
x = 38.68(, 180( ( 38.68(, x = 180( + 90(, 360( ( 90(
(x = 38.7(, 141.3(, 270(
10 (d) 5tan 2x + 7tan x = 0, 0( ( x ( 360(
[pic] + 7tan x = 0
10tan x + 7tan x(1 ( tan2 x) = 0
7tan3 x ( 17tan x = 0
tan x(7tan2 x ( 17) = 0
tan x = 0 or tan x = ([pic]
Basic ( = 0( Basic ( = 57.31(
x = 0(, 180(, 360(, x = 57.31(, 180( ( 57.31(, 180( + 57.31(, 360( ( 57.31(
(x = 0(, 180(, 360(, 57.3(, 233.7(, 237.3(, 302.7(
11 (c) 5cos 2x + 11sin x ( 8 = 0, 0 ( x ( 2(
5(1 ( 2sin2 x) + 11sin x ( 8 = 0
10sin2 x ( 11sin x + 3 = 0
(2sin x ( 1)(5sin x ( 3) = 0
sin x = [pic] or sin x = [pic]
Basic ( = [pic] Basic ( = 0.6435
x = [pic], ( ( [pic], x = 0.6435, ( ( 0.6435
(x = [pic], [pic](, 0.644, 2.50
11 (d) 3tan 2x ( 8tan x = 0, 0 ( x ( 2(
[pic] ( 8tan x = 0
6tan x ( 8tan x(1 ( tan2 x) = 0
8tan3 x ( 2tan x = 0
2tan x(4tan2 x ( 1) = 0
tan x = 0 or tan x = ([pic]
Basic ( = 0 Basic ( = 0.4636
x = 0, (, 2(, x = 0.4636, ( ( 0.4636, ( + 0.4636, 2( ( 0.4636
(x = 0, (, 2(, 0.464, 2.68, 3.61, 5.82
12 (i) tan2 A = ([pic])2 = [pic]
(ii) sin 2A = 2sin A cos A
= 2([pic])(x) = 2x[pic]
(iii) cos 4A = 1 ( 2sin2 2A
= 1 ( 2(2x[pic])2 = 1 ( 8x2 + 8x4
(iv) cos A = 1 ( 2sin2 [pic]A [0( ( A ( 90(, so 0( ( [pic]A ( 45(]
2sin2 [pic]A = 1 ( x
sin2 [pic]A = [pic]
sin [pic]A = ([pic] = [pic] ([pic]0( ( [pic]A ( 45()
Exercise 9E
1 (i) Prove: sin 3A = 3sin A ( 4sin3 A
LHS = sin (A + 2A)
= sin A cos 2A + cos A sin 2A
= sin A (1 ( 2sin2 A) + cos A (2sin A cos A)
= sin A ( 2sin3 A + 2sin A cos2 A
= sin A ( 2sin3 A + 2sin A (1 ( sin2 A)
= sin A ( 2sin3 A + 2sin A ( 2sin3 A
= 3sin A ( 4sin3 A = RHS
1 (j) Prove: cosec 2A ( tan A = cot 2A
LHS = [pic] ( [pic]
= [pic] ( [pic]
= [pic] = [pic] = cot 2A = RHS
1 (k) Prove: (tan A + cot A) sin 2A = 2
LHS = ([pic] + [pic]) (2sin A cos A)
= [pic](2sin A cos A)
= [pic](2sin A cos A) = 2 = RHS
2 (h) Prove: [pic] + [pic] = 2sec 2A
LHS = [pic]
= [pic]
= [pic]
= 2sec 2 A ÷ (1 ( [pic])
= [pic] ÷ [pic]
= [pic] = [pic] = 2sec 2A = RHS
2 (i) Prove: [pic] = 1 ( tan2 A
LHS = 2tan A cot 2A
= 2([pic])([pic])
= ([pic])([pic])
= [pic] = 1 ( tan2 A = RHS
2 (j) Prove: tan [pic] = tan A + sec A
LHS = (tan [pic] + tan [pic]) ÷ (1 ( tan [pic] × tan [pic])
= (1 + [pic]) ÷ (1 ( [pic])
= [pic] ÷ [pic]
= [pic] × [pic]
= [pic]
= [pic] = tan A + sec A = RHS
2 (k) Prove: [pic] = cosec 2A
Simplifying:
tan [pic] and tan [pic]
= [pic] = [pic]
= [pic] = [pic]
LHS = [[pic] + [pic]] ÷ [[pic]([pic]]
= [pic] ÷ [pic]
= [pic]
= [pic]
= [pic]sec2 A cot A
= [pic]([pic])
= [pic] = [pic] = cosec 2A = RHS
3 Prove: tan 3A = [pic]
LHS = tan (A + 2A)
= [pic]
= (tan A + [pic]) ÷ [1 ( [pic]]
= [pic] ÷ [pic]
= [pic] ÷ [pic]
= [pic] = RHS
Exercise 9F
5 (c) 3cos ( + 4sin ( = 2, 0( < ( < 360(
Let 3cos ( + 4sin ( = Rcos (( ( ()
R = [pic] = 5
( = tan(1 ([pic]) = 53.13(
So, 3cos ( + 4sin ( = 2 becomes
5cos (( ( 53.13() = 2
cos (( ( 53.13() = 0.4
Basic ( = 66.42(
( ( 53.13( = 66.42(, 360( ( 66.42(
(( = 119.6(, 346.7(
5 (d) sin ( ( [pic]cos ( = 0.8, 0( < ( < 360(
Let sin ( ( [pic]cos ( = Rsin (( ( ()
R = [pic] = [pic]
( = tan(1 ([pic]) = 54.74(
So, sin ( ( [pic]cos ( = 0.8 becomes
[pic]sin (( ( 54.74() = 0.8
sin (( ( 54.74() = [pic]
Basic ( = 27.51(
( ( 54.74( = 27.51(, 180( ( 27.51(
(( = 82.3(, 207.2(
6 (c) 4cos x ( 3sin x = 1.9, 0 < x < 2(
Let 4cos x ( 3sin x = Rcos (x + ()
R = [pic] = 5
( = tan(1 ([pic]) = 0.6435
So, 4cos x ( 3sin x = 1.9 becomes
5cos (x + 0.6435) = 1.9
cos (x + 0.6435) = [pic]
Basic ( = 1.181
x + 0.6435 = 1.181, 2( ( 1.181
(x = 0.538, 4.46
6 (d) 5sin x + 2cos x = 1.4, 0 < x < 2(
Let 4sin x + 2cos x = Rsin (x + ()
R = [pic] = [pic]
( = tan(1 ([pic]) = 0.3805
So, 5sin x + 2cos x = 1.4 becomes
[pic]sin (x + 0.3805) = 1.4
sin (x + 0.3805) = [pic]
Basic ( = 0.2630
x + 0.3805 = ( ( 0.2630, 2( + 0.2630
(x = 2.50, 6.17
7 (b) 3cos 2x ( 2sin 2x = [pic], 0( ( x ( 360(
Let 3cos 2x ( 2sin 2x = Rcos (2x + ()
R = [pic] = [pic]
( = tan(1 ([pic]) = 33.69(
So, 3cos 2x ( 2sin 2x = [pic] becomes
[pic]cos (2x + 33.69() = [pic]
cos (2x + 33.69() = [pic]
Basic ( = 66.91(
2x + 33.69( = 66.91(, 360( ( 66.91(, 360( + 66.91(, 720( ( 66.91(
(x = 16.6(, 129.7(, 196.6(, 309.7(
7 (d) 7sin 3x ( 6cos 3x = 3.8, 0( ( x ( 360(
Let 7sin 3x ( 6cos 3x = Rsin (3x ( ()
R = [pic] = [pic]
( = tan(1 ([pic]) = 40.60(
So, 7sin 3x ( 6cos 3x = 3.8 becomes
[pic]sin (3x ( 40.60() = 3.8
sin (3x ( 40.60() = [pic]
Basic ( = 24.34(
3x ( 40.60( = 24.34(, 180( ( 24.34(, 360( + 24.34(, 540( ( 24.34(,
720( + 24.34(, 900( ( 24.34(
(x = 21.6(, 65.4(, 141.6(, 185.4(, 261.6(, 305.4(
8 y = 12cos x ( 5sin x
(i) Let 12cos x ( 5sin x = Rcos (x + ()
R = [pic] = 13
( = tan(1 ([pic]) = 22.62(
So, y = 13cos (x + 22.6()
(ii) Since (1 ( cos (x + 22.6() ( 1, then
(13 ( 13cos (x + 22.6() ( 13
Max. vertical displacement = 13
When cos (x + 22.6() = 1 ( x = 360( ( 0( ( 22.6( = 337.4(
Min. vertical displacement = (13
When cos (x + 22.6() = (1 ( x = 180( ( 22.6( = 157.4(
9 Let 5sin x ( 12cos x = Rsin (x ( ()
R = [pic] = 13
( = tan(1 ([pic]) = 67.38(
So, 5sin x ( 12cos x = 13sin (x ( 67.4()
(i) 5sin x ( 12cos x = 7, 0( ( x ( 360(
( 13sin (x ( 67.38() = 7
sin (x ( 67.38() = [pic]
Basic ( = 32.58(
x ( 67.38( = 32.58(, 180( ( 32.58(
(x = 100.0(, 214.8(
(ii) Since (13 ( 13sin (x ( 67.38() ( 13, then
0 ( 132 cos2 (x + 22.6() ( 132
Max. value = 132 = 169
Min. value = 0
12 (i) In (OAD, sin ( = [pic] ( OD = 0.6sin (
In (CDE, cos ( = [pic] ( DE = 1.4cos (
OE = OD + DE = 1.4cos ( + 0.6sin ( (Shown)
(ii) Let 1.4cos ( + 0.6sin ( = Rcos (( ( ()
R = [pic] = 1.523
( = tan(1 ([pic]) = 23.20(
So, OE = 1.52 cos (( ( 23.2()
(iii) Since (1 ( cos (( ( 23.20() ( 1, then
(1.523 ( 1.523cos (( ( 23.20() ( 1.523
Max. value = 1.52
When cos (( ( 23.20() = 1 ( x = 0( + 23.20( = 23.2(
(iv) OE = 1.1
( 1.523cos (( ( 23.20() = 1.1
cos (( ( 23.20() = [pic]
Basic ( = 43.76(
( ( 23.20( = 43.76(
(( = 67.0(
Exercise 11A
8 y = 3x2 ( 5x ( 1
[pic] = 6x ( 5
At y-axis, x = 0, so [pic] = (5
10 y = [pic] + [pic] = ax(2 + bx(1
[pic] = (2ax(3 ( bx(2 = ([pic] ( [pic]
Given at ((1, 5), [pic] = 4 ( ([pic] ( [pic] = 4
2a ( b = 4 ----- (1)
At ((1, 5): 5 = [pic] + [pic] ( a ( b = 5 ----- (2)
(1) ( (2): a = (1
From (2): b = (1 ( 5 = (6
(a = (1 and b = (6.
11 y = ax ( [pic] = ax ( bx(1
[pic] = a + bx(2 = a + [pic]
Given at ([pic], ([pic]), [pic] = 3 ( a + [pic] = 3
a + 4b = 3 ----- (1)
At ([pic], ([pic]): ([pic] = a([pic]) ( [pic] ( a ( 4b = (13 ----- (2)
(1) + (2): 2a = (10 ( a = (5
From (1): b = (3 + 5) ÷ 4 = 2
(a = (5 and b = 2.
12 pV = 3600 ( p = [pic] = 3600V (1
[pic] = (3600V (2 = ([pic]
When p = 40, 40V = 3600 ( V = 90
So, [pic] = ([pic] = ([pic]
Exercise 11B
5 y = 2t + 5 + [pic] = 2t + 5 + 4(5t ( 9)(1
[pic] = 2 ( 4(5t ( 9)(2(5) = 2 ( [pic]
When t = 2, [pic] = 2 ( [pic] = (18
7 y = [pic] = 24(3x ( 5)(2
[pic] = ( 24(2)(3x ( 5)(3(3) = ([pic]
At (2, 24), [pic] = ([pic] = (144
8 (c) Let y = [pic] = [pic]
[pic] = [pic][pic][1 + ((2x(2)]
= [pic][pic][pic]
8 (d) Let y = [pic] = 6[pic]
[pic] = 6(([pic])[pic] (4x + 7)
= ([pic]
9 y = [pic] = [pic]
[pic] = [pic][pic](2x ( 4) = [pic]
When [pic] = 0, [pic] = 0 ( x ( 2 = 0 ( x = 2
When x = 2, y = [pic] = 2
So, required coordinates = (2, 2)
Exercise 11C
5 y = (2x ( 3)3(x + 5)5
Let u = (2x ( 3)3 and v = (x + 5)5
Then [pic] = 3(2x ( 3)2(2), [pic] = 5(x + 5)4(1)
Using Product rule, [pic] = u[pic] + v[pic]
= 5(2x ( 3)3(x + 5)4 + 6(2x ( 3)2(x + 5)5
= (2x ( 3)2(x + 5)4 [5(2x ( 3) + 6(x + 5)]
= (2x ( 3)2(x + 5)4 (16x + 15)
When [pic] = 0, (2x ( 3)2(x + 5)4 (16x + 15) = 0
x = 1[pic], (5 or ([pic]
6 (d) Let y = (3x ( 1)[pic] = (3x ( 1)[pic]
Let u = 3x ( 1 and v = [pic]
Then [pic] = 3, [pic] = [pic][pic](4x)
Using Product rule, [pic] = u[pic] + v[pic]
= [pic] + 3[pic]
= [pic] = [pic]
6 (e) Let y = 2x[pic] = 2x[pic]
Let u = 2x and v = [pic]
Then [pic] = 2, [pic] = [pic][pic](3)
Using Product rule, [pic] = u[pic] + v[pic]
= 2x[[pic][pic]] + 2[pic]
= [pic][15x + 2(3x ( 1)]
= (21x ( 2)[pic]
8 y = (3x + 2)(2 ( x) (1
Let u = 3x + 2 and v = (2 ( x) (1
Then [pic] = 3, [pic] = ((2 ( x) (2((1)
Using Product rule, [pic] = u[pic] + v[pic]
= [pic] + [pic]
= [pic] = [pic]
When [pic] = 8, [pic] = 8 ( (2 ( x)2 = 1
2 ( x = (1
x = 3 or 1
9 y = (x + 4)[pic] = (x + 4)[pic]
Let u = x + 4 and v = [pic]
Then [pic] = 1, [pic] = [pic][pic]((1)
Using Product rule, [pic] = u[pic] + v[pic]
= [pic] + [pic]
= [pic] = [pic]
When [pic] = 0, [pic] = 0
x = 4[pic]
10 (d) Let y = [pic] (2ax + 3b) = [pic](2ax + 3b)
Let u = 2ax + 3b and v = [pic]
Then [pic] = 2a, [pic] = [pic][pic](1)
Using Product rule, [pic] = u[pic] + v[pic]
= [pic] + 2a[pic]
= [pic] = [pic]
Exercise 11D
6 y = [pic]
Let u = 2x ( 1 and v = x2 + 3
Then [pic] = 2, [pic] = 2x
Using Quotient rule, [pic] = [pic]
= [pic]
= [pic] = [pic]
At x-axis, y = 0, ( x = [pic]
When x = [pic], [pic] = [pic] = [pic]
7 y = [pic]
Let u = x ( 2 and v = x2 + 5
Then [pic] = 1, [pic] = 2x
Using Quotient rule, [pic] = [pic]
= [pic]
= [pic]
For y > 0, ( [pic] > 0 ( x ( 2 > 0 (since x2 + 5 > 0)
(x > 2
For [pic] > 0, ( [pic] > 0
( 5 + 4x ( x2 > 0 [since (x2 + 5)2 > 0]
( x2 ( 4x ( 5 < 0
( (x + 1)(x ( 5) < 0
( (1 < x < 5
For both y > 0 and [pic] > 0, 2 < x < 5
9 2x + 9y = 3 ( x = [pic] ----- (1)
xy + y + 2 = 0 ----- (2)
Subs (1) into (2): y([pic]) + y + 2 = 0
( 3y ( 9y2 + 2y + 4 = 0
( 9y2 ( 5y ( 4 = 0
( (9y + 4)(y ( 1) = 0
( y = ([pic] or 1
When y = ([pic] in (1): x = [pic]
When y = 1 in (1): x = (3
For the curve, y(x + 1) = (2 ( y = ([pic] = (2(x + 1) (1
[pic] = 2(x + 1)(2 = [pic]
When x = [pic], [pic] = [pic]
When x = (3, [pic] = [pic]
10 (c) Let y = [pic]
Let u = ax2 + bx + c and v = ax2 ( bx ( c
Then [pic] = 2ax + b, [pic] = 2ax ( b
Using Quotient rule, [pic] = [pic]
= [pic]
= [pic]
= [pic] = [pic]
10 (d) Let y = [pic] = [pic] = [pic]
Let u = ax ( b and v = cx + b
Then [pic] = a, [pic] = c
Using Quotient rule, [pic] = [pic]
= [pic]
= [pic]
= [pic] = [pic]
Exercise 11E
5 y = [pic]
Let u = x ( 2 and v = 2x + 1
Then [pic] = 1, [pic] = 2
Using Quotient rule, [pic] = [pic]
= [pic] = [pic]
On the x-axis, y = 0, so [pic] = 0 ( x = 2
Grad. of tangent = [pic] = [pic] ( grad. of normal = (5
Equation of normal is y ( 0 = (5(x ( 2)
y = 10 ( 5x
7 For y = 2x2 ( 7x + 3 with tangent y = x + 2, then [pic] = 1
So, [pic] = 4x ( 7, ( 4x ( 7 = 1 ( x = 2
When x = 2 on the curve, y = (3.
Required tangent is y + 3 = (1)(x ( 2) ( y = x ( 5
On y-axis for tangent, x = 0 ( y = (5
Required coordinates = (0, (5)
8 y = ax3 + bx2 + c, [pic] = 3ax2 + 2bx
At (0, 5), c = 5
At ((1, 0), (a + b + c = 0
a = b + 5 ----- (1)
Given [pic] = 0 when x = 1, then 3a + 2b = 0 ----- (2)
Subs. (1) into (2): 3(b + 5) + 2b = 0 ( b = (3
From (1): a = 2
(a = 2 ,b = (3 and c = 5.
10 y = (x ( 2)3, [pic] = 3(x ( 2)2
At (3, 1), [pic] = 3
Equation of tangent is y ( 1 = 3(x ( 3)
y = 3x ( 8 ----- (1)
Subs. (1) into curve: (x ( 2)3 = 3x ( 8
x3 ( 6x2 + 9x = 0
x(x2 ( 6x + 9) = 0
x(x ( 3)2 = 0
x = 0 or 3
When x = 0 in (1): y = (8
Required coordinates = (0, (8)
11 y = x2 ( x ( 1, [pic] = 2x ( 1
At A(2, 1), [pic] = 3
Equation of tangent is y ( 1 = 3(x ( 2)
y = 3x ( 5
If grad. of normal = 3 ( grad. of tangent = ([pic]
( 2x ( 1 = ([pic] ( x = [pic]
When x = [pic] in the curve: y = ([pic])2 ( [pic] ( 1 = (1[pic]
Required coordinates = ([pic], (1[pic])
Exercise 11F
8 V = [pic]h2(h + 4) = [pic]h3 + [pic]h2, [pic] = 2 cm/s, h = 5 cm.
[pic] = h2 + [pic]h
For rate of change, [pic] = [pic] × [pic]
= [(5)2 + [pic](5)] (2) = 76[pic] cm3/s
10 2xy = 195 ( y = [pic], [pic] = [pic] units/s, x = 15 units.
[pic] = ([pic]
For rate of change, [pic] = [pic] × [pic]
= (([pic]) ([pic]) = ([pic] unit/s
12 V = [pic], [pic] = 5 cm/s, r = 60 cm.
[pic] = [pic]
For rate of change, [pic] = [pic] × [pic]
= ([pic]) (5) = 200πh cm3/s
13 V = 6πx2 ( [pic]πx3, [pic] = 20 cm3/s
(i) [pic] = 12πx ( πx2, x = 4 cm
For rate of change, [pic] = [pic] × [pic]
20 = [12π(4) ( π(4)2] × [pic]
( [pic] = [pic] cm/s
(ii) When x = 5 cm, [pic] = [pic] × [pic]
20 = [12π(5) ( π(5)2] × [pic]
( [pic] = [pic] < [pic]
So [pic] will not increase when x = 5.
14 V = 0.05[(3x + 2)3 ( 8]
(i) [pic] = 0.05[3(3x + 2)2(3)] = 0.45(3x + 2)2
(ii) [pic] = 0.081 m3/s, V = 0.95
When V = 0.95, 0.05[(3x + 2)3 ( 8] = 0.95
(3x + 2)3 ( 8 = 19
3x + 2 = 3 ( x = [pic]
For rate of change, [pic] = [pic] × [pic]
0.081 = 0.45[3([pic]) + 2]2 × [pic]
( [pic] = 0.02 m/s
Exercise 12A
4 (e) Let f(x) = [pic] = [pic]x + [pic]x (1
f ((x) = [pic] ( [pic]x (2 = [pic]
f (((x) = [pic]x (3 = [pic]
4 (f) Let f(x) = [pic]
Let u = ax + b and v = cx ( d
Then [pic] = a, [pic] = c
Using Quotient rule, f ((x) = [pic]
= [pic] = ([pic]
Using f ((x) = ((ad + bc)(cx ( d)(2
f (((x) = 2(ad + bc)(cx ( d)(3(c) = [pic]
5 y = x2 + 2x + 3, [pic] = 2x + 2, [pic] = 2
Prove: ([pic])2 + ([pic])3 = 4y
L.H.S. = ([pic])2 + ([pic])3 = (2x + 2)2 + 23
= 4x2 + 8x + 12
= 4(x2 + 2x + 3) = 4y = R.H.S.
6 xy ( 3 = 2x2 ( y = 2x + 3x (1, [pic] = 2 ( 3x (2, [pic] = 6x (3 = [pic]
Prove: x2[pic] + x[pic] = y
L.H.S. = x2[pic] + x[pic] = x2([pic]) + x(2 ( [pic])
= [pic] + 2x ( [pic]
= 2x + [pic] = y = R.H.S.
Exercise 12B
2 (f) f(x) = x3 + 3x2 ( 24x + 11
f ((x) = 3x2 + 6x ( 24
For decreasing function, f ((x) < 0
3(x2 + 2x ( 8) < 0
( (x + 4)(x ( 2) < 0
( (4 < x < 2
(required set of values of x = {x : (4 < x < 2}
3 y = [pic], x > 1[pic]
Let u = x2 and v = 2x ( 3
Then [pic] = 2x, [pic] = 2
Using Quotient rule, [pic] = [pic]
= [pic] = [pic]
For increasing function, [pic] > 0
[pic] > 0
( 2x(x ( 3) > 0
( x < 0 or x > 3
But given, x > 1[pic], then x > 3.
4 P(x) = 2x3 ( 21x2 + 60x ( 5
P ((x) = 6x2 ( 42x + 60
For decreasing profit, P ((x) < 0
6(x2 ( 7x + 10) < 0
( (x ( 2)(x ( 5) < 0
( 2 < x < 5
Exercise 12C
4 (d) y = x + [pic] = x + (x ( 1)(1
[pic] = 1 ( (x ( 1)(2 = 1 ( [pic]
[pic] = 2(x ( 1)(3 = [pic]
For stationary points, [pic] = 0 ( (x ( 1)2 = 1
( x ( 1 = ±1
( x = 0 or 2
When x = 0, y = (1, and [pic] < 0.
When x = 2, y = 3, and [pic] > 0.
((0, (1) is a maximum point and (2, 3) is a minimum point.
6 y = [pic]
(i) Let u = 3x + 2 and v = 2x ( 1
Then [pic] = 3, [pic] = 2
Using Quotient rule, [pic] = [pic]
= [pic] = [pic]
(ii) For stationary points to exist, [pic] = 0
( [pic] = 0 which is not possible for all x values.
(the stationary points does not exist.
7 y = [pic]
(i) Let u = x2 ( 2x + 1 and v = x ( 3
Then [pic] = 2x ( 2, [pic] = 1
Using Quotient rule, [pic] = [pic]
= [pic]
= [pic] = [pic]
(ii) For stationary points, [pic] = 0
( (x ( 1)(x ( 5) = 0 ( x = 1 or 5
When x = 1, y = 0, and using ‘Sign-test’,
|x |1( |1 |1+ |
|[pic] | |__ | |
| |/ | |\ |
((1, 0) is a maximum point.
When x = 5, y = 8, and using ‘Sign-test’,
|x |5( |5 |5+ |
|[pic] |\ |__ |/ |
((5, 8) is a minimum point.
12 y = ax4 + bx3 + 5, [pic] = 4ax3 + 3bx2
(i) At (2, 85) on the curve: 85 = a(2)4 + b(2)3 + 5
16a + 8b = 80
2a + b = 10 ( b = 10 ( 2a ----- (1)
At min. pt. ((1, 4): 0 = 4a((1)3 + 3b((1)2
4a ( 3b = 0 ----- (2)
Subs. (1) into (2): 4a ( 3(10 ( 2a) = 0
10a = 30 ( a = 3
From (1): b = 10 ( 2(3) = 4
(a = 3 and b = 4.
(ii) For stationary points, [pic] = 0
( 12x3 + 12x2 = 0 ( x = 0 or (1
Since [pic] = 12x3 + 12x2, so [pic] = 36x2 + 24x
When x = 0, y = 5, and [pic] = 0.
Using ‘Sign-test’,
|x |0( |0 |0+ |
|[pic] | |__ |/ |
| |/ | | |
((0, 5) is a point of inflexion.
13 y = ax + [pic], [pic] = a ( [pic]
(i) At ([pic], 12): 12 = [pic]a + 4b ( a + 8b = 24 ----- (1)
and 0 = a ( 16b ----- (2)
Solving (1) & (2): 24b = 24 ( b = 1
From (2): a = 16(1) = 16
(a = 16 and b = 1.
(ii) For stationary points, [pic] = 0
( 16 ( [pic] = 0 ( x = [pic]
Since [pic] = 16 ( [pic], so [pic] = [pic] > 0 (min. value)
When x = [pic], y = 12.
(([pic], 12) is a minimum point.
(iii) For increasing function, [pic] > 0
( 16 ( [pic] > 0
( 8x3 ( 1 > 0
( x3 > [pic] ( x > [pic]
15 y = px3 + 3x2 + 36x + q, [pic] = 3px2 + 6x + 36
(i) At (2, 51): 0 = 3p(2)2 + 6(2) + 36
12p = (48 ( p = (4 ----- (1)
and 51 = (4(2)3 + 3(2)2 + 36(2) + q
( q = (1
(p = (4 and q = (1.
(ii) For stationary points, [pic] = 0
( (12x2 + 6x + 36 = 0
( 2x2 ( x ( 6 = 0
( (2x + 3)(x ( 2) = 0 ( x = (1[pic] or 2
When x = (1[pic], y = 34[pic].
So, the other stationary point = ((1[pic], 34[pic]).
(iii) Since [pic] = (12x2 + 6x + 36, so [pic] = (24x + 6
When x = 2, [pic] < 0.
When x = (1[pic], [pic] > 0.
((2, 51) is a maximum point and ((1[pic], 34[pic]) is a minimum point.
(iv) For increasing function, [pic] > 0
( (12x2 + 6x + 36 > 0
( 2x2 ( x ( 6 < 0
( (2x + 3)(x ( 2) < 0 ( (1[pic] < x < 2
Exercise 12D
7 4x + 3y = 800 ( y = [pic]
For area, A = 3xy = 3x([pic]) = 800x ( 4x2
[pic] = 800 ( 8x, so for stationary values, [pic] = 0
8(100 ( x) = 0 ( x = 100
[pic] = (8 < 0 (max. value)
When x = 100, y = [pic] = 133[pic]
(x = 100 m and y = 133[pic] m.
9 x + y = 15 ( y = 15 ( x
For volume, V = 40([pic]) + 60([pic]) = [pic] + 80π(15 ( x)3
[pic] = 160πx2 ( 240π(15 ( x)2,
For stationary values, [pic] = 0
160πx2 ( 240π(15 ( x)2 = 0
( 2x2 ( 3(225 ( 30x + x2) = 0
( x2 ( 90x + 675 = 0
( x = [pic]
= 8.2577 or 81.74 (rejected as x + y = 15)
[pic] = 320πx + 480π(15 ( x),
When x = 8.2577, y = 15 ( 8.2577 = 6.7423, [pic] > 0 (min. value)
(x = 8.26 cm and y = 6.74 cm.
10 Given: y = 1.5x, and 30 = xyh ( h = [pic] = [pic]
For total surface area, A = 2(xh) + 3(xy) + 4(yh)
= [pic] + 4.5x2 + [pic]
= 4.5x2 + [pic] (Shown)
[pic] = 9x ( [pic], so for stationary values, [pic] = 0
9x ( [pic] = 0
x3 = [pic] ( x = 2.6099
[pic] = 9 + [pic]
When x = 2.6099, y = 1.5(2.6099) = 3.9149,
h = [pic] = 2.9362, [pic] > 0 (min. value)
(x = 2.61 cm, y = 3.91 cm and h = 2.94 cm.
11 Given: πr2h = 600 ( h = [pic]
For total surface area, A = 2πrh + 2πr2
= [pic] + 2πr2
[pic] = ([pic] + 4πr, so for stationary values, [pic] = 0
([pic] + 4πr = 0
r3 = [pic] ( r = 4.5708
[pic] = [pic] + 4π
When x = 4.5708, h = [pic] = 9.1416, [pic] > 0 (min. value)
(x = 4.57 cm and h = 9.14 cm.
14 V = [pic]πr2h, r + h = a ( h = a ( r
So, V = [pic]πr2(a ( r) = [pic]πar2 ( [pic]πr3
[pic] = [pic]πar ( πr2
For stationary values, [pic] = 0 ( [pic]πar ( πr2 = 0
( πr([pic]a ( r) = 0
( r = [pic]a or 0 (rejected as r > 0)
[pic] = [pic]πa ( 2πr
When r = [pic]a, h = a ( [pic]a = [pic]a, [pic] < 0 (max. value)
(r = [pic]a and h = [pic]a.
Exercise 13A
3 (e) Let y = cosec 2x = [pic]
[pic] = [pic] = [pic]
= (2([pic])([pic])
= (2cot 2x cosec 2x
4 (g) Let y = [pic]
[pic] = [pic]
= [pic]
= [pic] = [pic]
5 (d) y = sin2 3x2 ( cos [pic]
[pic] = 2sin 3x2 (cos 3x2)(6x) ( [(sin [pic]] (([pic])
= 12x sin 3x2 cos 3x2 ( [pic]sin [pic]
5 (g) y = [pic] = 4cos (3x ( π)
[pic] = (4sin (3x ( π) (3) = (12sin (3x ( π)
5 (h) y = [pic] = 2 [cos (4x + [pic])] (3
[pic] = (6[cos (4x + [pic])] (4 [(sin (4x + [pic]) (4)]
= 24sin (4x + [pic]) ([pic])
= 24sin (4x + [pic]) sec4 (4x + [pic])
6 (d) Let y = tan3 (px + q)
[pic] = 3tan2 (px + q) sec2 (px + q) (p)
= 3p tan2 (px + q) sec2 (px + q)
7 y = 3x sin 2x
[pic] = 3x (2cos 2x) + 3sin 2x = 6x cos 2x + 3sin 2x
When x = 1.2, [pic] = 6 (1.2) cos 2.4 + 3sin 2.4 = (3.28
10 Given: [pic] = [pic]
L.H.S. = [pic]
= [pic]
= [pic] = [pic]
(h = 8 and k = 6.
Exercise 13B
3 f(x) = ln 5x, x > 0
f ((x) = [pic] = [pic] ( f ((1) = 1
4 (d) Let y = ln ([pic] ( 1)2 = 2 ln ([pic] ( 1)
[pic] = [pic]([pic]) = [pic]
5 (d) y = [pic]
[pic] = [pic] = [pic] = [pic]
6 y = x ln 2x
[pic] = x([pic]) + ln 2x = 1 + ln 2x
When x = [pic], [pic] = 1 + ln [pic] = 0.307
[pic] = [pic] = [pic]
When x = [pic], [pic] = 4
7 (f) ey = 2x3 + 7x ( y = ln (2x3 + 7x)
[pic] = [pic]
7 (h) ey = sec x ( y = ln (sec x)
= ln [pic] = ln 1 ( ln (cos x) = (ln (cos x)
[pic] = [pic] = tan x
Exercise 13C
4 (d) y = ex + [pic] = ex + e(x
[pic] = ex ( e(x = ex ( [pic]
5 (g) Let y = x2[pic]
[pic] = x2 ((2x[pic]) + 2x[pic] = 2x[pic](1 ( x2)
5 (h) Let y = [pic] = [pic]
[pic] = [pic] = [pic] ( x
= [pic] (or [pic] ( [pic])
6 y = e5x ( ecos x
[pic] = 5e5x ( ((sin x)ecos x = 5e5x + sin x ecos x
When x = 0, [pic] = 5 + 0 = 5
7 f(x) = [pic]
f ((x) = [pic]
= [pic] = [pic]
8 f(x) = [pic]
f ((x) = [pic]
= [pic] = [pic]
Exercise 13D
5 y = ln (9 ( 2x), [pic] = [pic]
From 2y ( x + 7 = 0, grad. = [pic]
( [pic] = (2
( [pic] = (2
( 9 ( 2x = 1 ( x = 4
Subs. x = 4 into y = ln (9 ( 2x), y = 0
Required coordinates = (4, 0)
9 y = x + 2cos x, 0 < x < (
[pic] = 1 ( 2sin x, [pic] = (2cos x
For stationary value, [pic] = 0 ( sin x = [pic]
x = [pic] or [pic]
When x = [pic], [pic] < 0 (max.) and y = [pic] + 2cos [pic] = [pic] + [pic]
When x = [pic], [pic] > 0 (min.)
(max. value of y = [pic] + [pic]
12 y = 2tan (3x + [pic]), [pic] = (0.25 rad./s
[pic] = 6sec2 (3x + [pic])
When x = [pic], [pic] = 6sec2 ([pic] + [pic]) = 12
For rate of change, [pic] = [pic] × [pic]
= 12((0.25) = (3 units/s
13 y = 2sin3 x ( 3sin x, 0 ( x ( (
[pic] = 6sin2 x cos x ( 3cos x
= 6cos x(1 ( cos2 x) ( 3cos x
= 3cos x ( 6cos3 x
For stationary value, [pic] = 0 ( 3cos x(1 ( 2cos2 x) = 0
cos x = 0 or cos x = ([pic]
x = [pic], [pic] or [pic]
[pic] = (3sin x + 18cos2 x sin x = 3sin x(6cos x ( 1)
When x = [pic], [pic] < 0 (max.) and y = 2sin3 [pic] ( 3sin [pic] = (1
When x = [pic], [pic] > 0 (min.) and y = 2sin3 [pic] ( 3sin [pic] = ([pic]
When x = [pic], [pic] > 0 (min.) and y = 2sin3 [pic] ( 3sin [pic] = ([pic]
(the points are min ([pic], ([pic]), max ([pic], (1) and min ([pic], ([pic]).
14 y = ln ([pic]) = ln (5 ( 7x) ( ln (8 + x)
[pic] = [pic] ( [pic] = [pic]
= [pic]
For stationary value, [pic] = 0 ( [pic] = 0
( no solution.
Hence, the curve has no stationary point for all real values of x. (Shown)
18 P = 8ln [pic], E = sin 0.5t, t = [pic] s
So, P = 8ln [pic] = 8ln (stn 0.5t) ( 8ln 10(12
[pic] = [pic]
When t = [pic], [pic] = [pic] = 4 Watts/s
21 y = 2x + 3sec2 2x, x = [pic]
= 2x + 3(1 + tan2 2x) = 2x + 3 + 3tan2 2x
[pic] = 2 + 12tan 2x sec2 2x
When x = [pic], [pic] = 2 + 12tan [pic] sec2 [pic]
= 2 + 12([pic])(4) = 2 + 48[pic]
Required grad. of normal = ( [pic] ( [pic]
= ( [pic]
= ( [pic] = [pic]
Exercise 14A
4 (f) [pic] = [pic]
= [pic] + 4x + [pic] + c = [pic] + 4x + [pic] + c
8 [pic] = k[pic] = [pic]
y = [pic] = [pic] + c = [pic] + c
At (1, 4), 4 = [pic] + c ----- (1)
At (8, 16), 16 = [pic] + c ----- (2)
(2) ( (1): [pic] = 12 ( k = [pic]
From (1): c = 4 ( [pic]([pic]) = [pic]
(y = [pic]([pic]) + [pic] = [pic] (or 5y = [pic] + 16)
9 [pic] = [pic] = 1 ( 4x(2
y = [pic] = x + 4x(1 + c
At (2, 7), 7 = 2 + 2 + c ( c = 3
(y = x + [pic] + 3
10 [pic] = kx + 3
(i) Since grad. of normal = ([pic] ( grad. of tangent = 15
( kx + 3 = 15
( k(3) = 12 ( k = 4
(ii) y = [pic] = 2x2 + 3x + c
At (3, 19), 19 = 18 + 9 + c ( c = (8
(y = 2x2 + 3x ( 8
(iii) For turning point, [pic] = 0 ( x = ([pic]
When x = ([pic], y = 2(([pic])2 + 3(([pic]) ( 8 = ([pic]
Required coordinates = (([pic], (9[pic])
Exercise 14B
1 (h) [pic] = [pic][pic]
= [pic][pic] + c
= [pic] + c
4 [pic] = [pic] = (3t + 6)[pic]
m = [pic]
= [pic] + c = [pic] + c
When t = 1, m = 7000, 7000 = [pic] + c
c = 6994
(m = [pic] + 6994
5 (b) [pic] = [pic]
= [pic] + [pic] + c
= [pic] + [pic]
The integral will be valid if 3x + 4 ( 0 ( x ( (1[pic]
Exercise 14C
5 (f) [pic] = [pic] + c
= sin ((x ( [pic]) + c
6 (c) [pic] = [pic]
= [pic]tan (px + q) ( x + c
7 [pic] = tan2 x
y = [pic] = [pic]
= tan x ( x + c
When x = [pic], y = [pic], [pic] = tan [pic] ( [pic] + c
c = [pic] ( 1
(y = tan x ( x + [pic] ( 1
9 f ((x) = 2cos2 (5x ( [pic]) + sin2 x ( x
= [cos 2(5x ( [pic]) + 1] + [pic](1 ( cos 2x) ( x
= cos 2(5x ( [pic]) ( [pic] + [pic] ( x
f(x) = [pic]
= [pic]sin (10x ( [pic]) ( [pic] + [pic] ( [pic] + c
Exercise 14D
3 (d) [pic] = [pic]
= ([pic] + [pic] + c = ([pic] + [pic] + c
4 (d) [pic] = [pic]
= ([pic] + [pic]ln x + c
= [pic]ln x ( [pic] + c
4 (e) [pic] = [pic]
= 9x ( 24ln x ( [pic] + c
5 (d) [pic] = [pic]
= [pic]
= ex + 6x ( [pic] + c
Exercise 14E
4 y = [pic]
Let u = 4x and v = (3x ( 2x2)[pic]
Then [pic] = 4, [pic] = [pic](3x ( 2x2)[pic](3 ( 4x)
= [pic]
Using Quotient rule, [pic] = [pic]
= [4[pic] ( [pic]] ( ([pic])2
= [pic] ( [pic]
= [pic] = [pic]
[pic] = [pic][pic]
= [pic]([pic]) + c = [pic] + c
6 y = e2x(5x ( 4)
Using Product rule, [pic] = 2e2x(5x ( 4) + 5e2x
= 10x e2x ( 3e2x
From above, [pic] = e2x(5x ( 4) + A
[pic][pic] ( [pic] = e2x(5x ( 4) + A
[pic][pic] ( [pic] = 5x e2x ( 4e2x + A
[pic][pic] = 5x e2x ( [pic] + A
([pic] = [pic](5x e2x ( [pic] + A)
= [pic] ( [pic] + c (c = [pic])
7 y = e3x(sin 3x + cos 3x)
Using Product rule, [pic] = 3e3x(sin 3x + cos 3x) + e3x(3cos 3x ( 3sin 3x)
= 6e3x cos 3x (Shown)
From above, 6[pic] = e3x(sin 3x + cos 3x) + A
[pic] = [pic]e3x(sin 3x + cos 3x) + c (c = [pic])
11 y = 3sin x cos x
Using Product rule, [pic] = 3cos x (cos x) + 3sin x ((sin x)
= 3(cos2 x ( sin2 x)
= 3[cos2 x ( (1 ( cos2 x)]
= 6cos2 x ( 3
From above, [pic] = 3sin x cos x + A
[pic][pic] ( [pic] = 3sin x cos x + A
[pic][pic] ( 3x = 3sin x cos x + A
[pic] = [pic](3x + 3sin x cos x + A)
= [pic] + c (c = [pic])
14 (i) [pic][ln (x2 + 3)] = [pic]
(ii) Let [pic] = [pic] + [pic]
×(2x + 5)(x2 + 3): 18 ( 15x = A(x2 + 3) + (Bx + C)(2x + 5)
Let x = ([pic]: 18 ( 15(([pic]) = A([pic] + 3) + 0
A = 6
Equating coeff. of x2: 0 = A + 2B ( B = (3
Equating constants: 18 = 3A + 5C ( C = 0
([pic] = [pic] ( [pic]
[pic] = [pic][pic]
= [pic][pic]
= [pic] ( [pic]
= [pic] ( [pic][pic]
= [pic]ln (2x + 5) ( [pic]ln (x2 + 3) + c
= ln (2x + 5) ( [pic]ln (x2 + 3) + c
-----------------------
x
y
(
[pic]
(2
(
(1
(4
5
x
y
(
(
3
70(
[pic]
1
20(
q
[pic]
(
4
0
(2
y
x
y = 2sin 2x
2
y = 4 + 2sin 2x
6
2
(2
y
0
x
2(
(
1
5
y = 2cos 2x
y = 3 + 2cos 2x
3
(3
y
0
x
2(
(
(4
2
y = 3sin x ( 1
y = 3sin x
180(
0
y
x
y = sin 2x
2
90(
360(
270(
3
(1
(3
1
5
y = cos 3x
0
y
x
y = 2sin x
2
2(
(2
(
y = [pic]
y = | 2sin x |
y
0
x
(
[pic]
1
y = | cos [pic]x |
y = sin x
1
2
[pic]
x
y
x
5
[pic]
2
x
y
x
x
y
A
(15
(
(8
17
x
y
B
(3
(
(4
5
1
[pic]
a
x
y
A
1
[pic]
b
x
y
B
([pic]
1
x
y
A
(
x
(8
17
x
y
A
(
15
(3
5
x
y
B
(
4
(1
[pic]
x
(
2
1
[pic]
x
x
y
A
................
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