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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