chapter 8 differentiation.pdf

7/27/2019 Chapter 8 Differentiation.pdf

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O xford Fajar Sdn. Bhd. (008974-T) 2012

CHAPTER 8 DIFFERENTIATION

Focus on Exam 8

1 (a) Let y = ( x2 + 3)e - 2 x

d yd x

= ( x2 + 3)(- 2e - 2 x) + e- 2 x(2 x)

= 2e- 2 x

(- x 2

- 3 + x )(b) Let u = x and y = sin3 x

= x12 y = sin3 u

d ud x

= 12

x- 1

2 d yd u

= 3 sin 2 u (- cos u)

= 12 x

= - 3 sin 2 u cos u

Hence,d yd x

=d yd u

d ud x

= - 3 sin 2 u cos u 12 x

= - 3 sin2 x cos x 2 x

2 (a) Let y = ln ( x3 e- 3 x)

d yd x

= x3(- 3e - 3 x) + e- 3 x(3 x2)

x3e- 3 x

d d x

( x3e- 3 x)

= e- 3 x 3 x2(- x + 1)

x3e- 3 x

= 3(- x + 1)

x

Copy back x3e- 3 x.

(b) Let u = 5 x

log5 u = x

ln uln 5

= x

ln u = x ln 5

1u

d ud x

= ln 5

d ud x

= u ln 5


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Fully Worked Solution 3

5 x2 - xy + y2 = 7When x = 3, 3 2 - 3 y + y2 = 7

y2 - 3 y + 2 = 0( y - 1)( y - 2) = 0

y = 1 or 2

x2 - xy + y2 = 7Differentiating implicitly with respect to x,

2 x - x d yd x

+ y(- 1) + 2 y d yd x

= 0

(- x + 2 y)d yd x

= - 2 x + y

d yd x

= - 2 x + y- x + 2 y

The gradient of the tangent at the point (3, 1) is- 2(3) + 1- 3 + 2(1) = 5.

The gradient of the tangent at the point (3, 2) is- 2(3) + 2- 3 + 2(2)

= - 4.

6 2 y = ln ( xy)

2d yd x

= x

d yd x

+ y(1)

xy

2 xyd yd x

= xd yd x

+ y

(2 xy - x) d yd x

= y

d yd x

= y2 xy - x

At the point P (e 2, 1),d yd x

= 12(e 2)(1) - e2

= 1e2

Therefore, the gradient of the tangent is 1

e2.

Hence, the equation of the tangent at the point P (e 2, 1) is

y - 1 = 1e2

( x - e2)

e2 y - e2 = x - e2

e2 y = x

7 x = e 4t = e2 t

d xd t

= 2 12 t

e2 t

d xd t = e2 t

t


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When t = ln 2,d yd x

= eln 2 + 12e 2 ln 2

= 2 + 12(2)2

= 3

8Hence, the equation of the tangent at the point where t = ln 2 is

y - 4 = 38

( x - 2)

8 y - 32 = 3 x - 6 8 y = 3 x + 26

9 x = - cos 2 2

d xd

= - 2 cos 2 (- 2 sin 2 )

= 4 cos 2 sin 2

y = sin 2 2

d yd

= 2 sin 2 (2 cos 2 )

= 4 sin 2 cos 2

d yd x

=

d yd d xd

= 4 sin 2 cos 2

4 cos 2 sin 2

= 1

The gradient of the tangent is 1. Hence, the gradient of the normal is - 1.

When = 8

, x = - cos2 4

= - 12

2

= - 12

and y = sin 2

4

= 12

2

= 12

Hence, the equation of the normal is

y - 12

= - 13 x - - 12y - 1

2= - x - 1

2 y = - x


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ACE AHEAD Mathematics (T) Second Term6

10 y = e2 x - 6 x + 7

= (e2 x - 6 x + 7)12

d yd x

= 12

(e2 x - 6 x + 7)- 1

2 (2e 2 x - 6)

= 2e 2 x - 6

2 e 2 x - 6 x + 7

= 2e2 x - 62 y

= e2 x - 3 y

yd yd x

= e2 x - 3

yd 2 yd x2

+ d yd x

d yd x

= 2e2 x

y d 2 yd x 2 +

d yd x

2

= 2e2 x

[Shown]

11 y = e x ln x

d yd x

= e x 1

x+ ln x e x

d yd x

= e x 1 x

+ y

xd yd x

= e x + xy

x

d 2 y

d x2 + d y

d x (1) = e x

+ xd y

d x + y(1)

xd 2 yd x2

+ (1 - x)d yd x

- y = e x From ,

e x = xd yd x

- xy

xd 2 yd x2

+ (1 - x)d yd x

- y = xd yd x

- xy

x d 2 yd x 2

+ (1 - 2 x )d yd x

+ ( x - 1) y = 0 [Shown]

12 y = cos x x

xy = cos x

xd yd x

+ y(1) = - sin x

xd yd x

+ y = - sin x

xd 2 yd x2

+ d yd x

(1) + d yd x

= - cos x

xd 2 yd x2

+ 2d yd x

= - xy

x d2

yd x 2 + 2 d yd x + xy = 0 [Shown]


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13 y = cos x

= cos12 x

d yd x

= 12

(cos x)- 1

2 (- sin x)

=- sin x

2 cos x

= - sin x2 y

2 yd yd x

= - sin x

2 yd 2 yd x2

+ d yd x

2d yd x

= - cos x

2 yd 2 yd x2

+ 2d yd x

2

+ cos x = 0

2 y d2 y

d x 2 + 2 d y

d x

2 + y2 = 0 [Shown]

14 y = e- 2 x sin x

d yd x

= e- 2 x cos x - 2 sin x e- 2 x

d yd x

= e- 2 x cos x - 2 y e- 2 x sin x = y

d 2 y

d x2 = - e- 2 x sin x - 2 cos x e- 2 x - 2

d y

d x

d 2 yd x2

= - y - 2d yd x

+ 2 y - 2d yd x

cos x e- 2 x =

d yd x

+ 2 y

d 2 yd x 2

+ 4 d yd x

+ 5 y = 0 [Shown]

15 y = ln (1 - cos x)

d yd x

= sin x1 - cos x

d 2 yd x2

= (1 - cos x)(cos x) - sin x sin x

(1 - cos x)2

= cos x - cos2 x - sin 2 x

(1 - cos x)2

= cos x - (cos 2 x + sin 2 x)

(1 - cos x)2

= cos x - 1(1 - cos x)2

= - 1 - cos x(1 - cos x)2

e- 2 x sin x = y


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= cos2 x + sin 2 x + sin 2 x + cos 2 x

(sin x + cos x)2

= 1 + 1(sin x + cos x)2

sin 2 x + cos2 x = 1

= 2

(sin x + cos x )2 [Shown]

d 2 yd x2

= (sin x + cos x)(- sin x - cos x) - ( cos x - sin x)(cos x - sin x)

(sin x + cos x)2

= - sin 2 x - 2 sin x cos x - cos 2 x - ( cos 2 x - 2 sin x cos x + sin 2 x)

(sin x + cos x)2

= - 2 sin2 x - 2 cos 2 x

(sin x + cos x)2

= - 2(sin 2 x + cos 2 x)

(sin x + cos x)2

= - 2(1)(sin x + cos x)2

= - 3d yd x2

+ 1

d 2 yd x 2

+d yd x

2 + 1 = 0 [Shown]

18 (a) y = x2

( x + 3)( x - 1)

= x2

x2 + 2 x - 3

As y , the denominator of x2

( x + 3)( x - 1) 0

( x + 3)( x - 1) 0 x - 3 or 1

Therefore, x = - 3 and x = 1 are vertical asymptotes.

lim x

y = lim x

x2

x2 + 2 x - 3

= lim

x

x2

x2 x2

x2 + 2 x

x2 - 3

x2

= lim x

1

1 + 2 x

- 3 x2

= 11 + 0 + 0

= 1

Therefore, y = 1 is the horizontal asymptote.


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(b) y = x2

x2 + 2 x - 3

d yd x

=( x2 + 2 x - 3)(2 x) - x2(2 x + 2)

( x2 + 2 x - 3)2

=2 x3 + 4 x2 - 6 x - 2 x3- 2 x2

( x2 + 2 x - 3)2

= 2 x2 - 6 x

( x2 + 2 x - 3)2

d 2 yd x2

=( x2 + 2 x - 3)2(4 x - 6 ) - (2 x2 - 6 x) 2( x2 + 2 x - 3)( 2 x + 2)

( x2 + 2 x - 3)4

=2( x2 + 2 x - 3)[( x2 + 2 x - 3)(2 x - 3 ) - (2 x2 - 6 x)(2 x + 2)]

( x2 + 2 x - 3)4

= 2[( x2

+ 2 x - 3)(2 x - 3) - (2 x2

- 6 x)(2 x + 2)]( x2 + 2 x - 3)3

Whend yd x

= 0, 2 x2 - 6 x

( x2 + 2 x - 3)2 = 0

2 x2 - 6 x = 02 x( x - 3) = 0

x = 0 or 3

When x = 0, y = 0 and

d 2

yd x2

= 2[( -3 )(-3 ) - 0](- 3)3

= - 23

(0)

Therefore, 3, 34

is a turning point and it is a local minimum point .

(c) When y = 0, x = 0.

Hence, the graph of y = x2

( x + 3)( x - 1)

= x2

x2 + 2 x - 3 is as shown.


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x

Y

3 O

31

1

34

,

19 (a) y = 4( x - 3 )2 - 1 x - 3

x = 3 is the vertical asymptote.

(b) When x = 0, y = 4(- 3)2 - 1(- 3)

= 36 13

.

Thus, the graph cuts the y-axis at 0, 36 13

.

When y = 0, 4( x - 3)2 - 1 x - 3

= 0

4( x - 3)2 = 1 x - 3

( x - 3)3 = 14

x - 3 = 1

413

x = 1

413

+ 3

x = 3.63

Thus, the graph cuts the x-axis at ( 3.63, 0 ).

(c) y = 4 ( x - 3)2 - 1 x - 3

= 4( x - 3)2 - ( x - 3) - 1

d yd x

= 8( x - 3)1(1) + ( x - 3)- 2(1)

= 8( x - 3) + 1( x - 3)2

d 2 yd x2

= 8 - 2( x - 3) - 3(1)

= 8 -2

( x - 3)3


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Whend yd x

= 0,

8( x - 3) + 1( x - 3)2

= 0

8( x - 3) = - 1( x - 3)2

( x - 3)3 = - 18

x - 3 = - 12

x = 2 12

When x = 212

,

y = 4 52

- 32

- 15

2 - 3

= 1 + 2= 3

d 2 yd x2

= 8 - 252

- 33

= 8 - (- 16) = 24 ( > 0)

Therefore, the turning point is 2 12

, 3 and it is a local minimum point .

(d) When d 2 y

d x2 = 0,

8 - 2( x - 3)3

= 0

2( x - 3)3

= 8

( x - 3)3 = 14

x = 1

41

3

+ 3

x = 3.63

From (b), when x = 3.63, y = 0.

d 3 yd x3

= 6( x - 3) - 4(1)

= 6( x - 3)4

When x = 3.63,d 3 yd x3

= 6(3.63 - 3)4

= 38.1 (i.e. 0)Hence, (3.63, 0) is a point of inflexion .


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(e) The graph of y = 4( x - 3 )2 - 1 x - 3

is as shown below.

x

y

33.63O

2 12

,3

361

3

20 (a) The x -axis is the axis of symmetry.

(b) y 2 = x2(4 - x) y2 0 x2(4 - x) 0

Since x2 0, x2(4 - x) 0 only if 4 - x 0 i.e. x 4.

Hence, the set of values of x where the graph does not exist is { x : x > 4}.

(c) y2 = x2(4 - x) = 4 x2 - x3

2 yd yd x

= 8 x - 3 x2

d yd x

= 8 x - 3 x2

2 y

d yd x

= 8 x - 3 x2

2( x 4 - x)

d yd x = x(8 - 3 x) 2 x 4 - x

d yd x

= 8 - 3 x 2 4 - x

Whend yd x

= 0,

8 - 3 x 2 4 - x

= 0

8 - 3 x = 0

x = 83


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When x = 83

, y = 83

4 - 83

= 3.08

Hence, 2 23

, 3.08 and 2 23

, - 3.08 are turning points (whose tangents are horizontal).

When d yd x

= ,

2 4 - x = 0 x = 4

When x = 4, y = 4 4 - 4 = 0

Hence, (4, 0) is also a turning point where tangent is vertical.

(d) The graph of y 2 = x2(4 - x) is as shown below.

x

y

O 4

2 23

, 3.08

2 23

, 3.08

21 (a) y = 1 - e2 x

1 + e2 x

d yd x

= (1 + e2 x)(- 2e 2 x) - (1 - e2 x)(2e 2 x)

(1 + e2 x)2

d yd x

= - 2e 2 x[1 + e2 x + (1 - e2 x)]

(1 + e2 x)2

d yd x

= - 4e2 x

(1 + e2 x)2

Since e 2 x > 0 and (1 + e2 x)2 > 0, thusd yd x

= - 4e2 x

(1 + e2 x )2 < 0 [Shown]

(b) y = 1 - e2 x

1 + e2 x

y + ye2 x = 1- e2 x

e2 x(1 + y) = 1 - y

e2 x = 1 - y1 + y

2 x = ln1 - y1 + y

x = 12

ln1 - y1 + y

[Shown]


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d yd x

= - 4e2 x

(1 + e2 x)2

= - 4e2

12 ln

1 - y1 + y

3(1 + e)2 14 ln 2 1 - y1 + y2

= - 4

1 - y1 + y

1 +1 - y1 + y

2 aloga x = x

= - 4

1 - y1 + y

1 + y + 1 - y

1 + y

2

= - 4

1 - y1 + y4

(1 + y)2

= - (1 - y)(1 + y)

= y 2 - 1 [Shown]

d 2 yd x2 = 2 y

d yd x

Sinced yd x

< 0,d 2 yd x 2

< 0 if y > 0 and d 2 yd x 2

> 0 if y < 0 [Shown]

(c) lim x

1 - e2 x

1 + e2 x = - 1 and lim

x -

1 - e2 x

1 + e2 x = 1

(d) When y = 0, 1 - e2 x

1 + e2 x = 0

1 - e2 x = 0e2 x = 12 x = ln 12 x = 0

x = 0

Thus, (0, 0) is a point of inflexion.

Hence, the graph of y = 1 - e2 x

1 + e2 xis as shown below.

y

x O

1

1


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22 (a)

k cm

6 cm

(2k + 6) cm

Q

C R S D

x cm

B A

P

CQR and CBS are similar triangles.

k cm

B

Q

C R S

x cm

[(2k + 6) 6] cm

Thus, RC SC

= QR BS

RC (2k + 6) - 6

= xk

RC 2k

= xk

RC = 2 x

Thus, DR = DC - RC = 2k + 6 - 2 x

(b) Area of PQRD ,L = DR QRL = (2k + 6 - 2 x)( x)L = (2k + 6) x - 2 x 2 [Shown]

(c) When L has a stationary value,

d Ld x

= 0

2k + 6 - 4 x = 04 x = 2k + 6

x = 2k + 64

x = 2(k + 3)4

x = k + 32

d 2 L

dx2 = - 4 (negative)

Thus, L has a maximum value.


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Hence, the maximum value of L is

L = (2k + 6) k + 32

- 2 k + 32

2

= 2(k + 3) k + 32

-2(k + 3)2

4

= (k + 3)2 - (k + 3)2

2

= (k + 3) 2

2

23 In QMC , sin x = MC r

MC = r sin x AC = 2 MC = 2r sin x

r c m r c m

O

M C A

B

x

x x

r cm

In OMC , cos x = OM

r OM = r cos x

Area of ABC ,

L = 12

AC BM

L = 12

AC ( BO + OM )

L = 12

(2r sin x) (r + r cos x)

L = r 2

sin x + r 2

sin x cos x

L = r 2 sin x + 12

r 2(2 sin x cos x)

L = r 2 sin x + 12

r 2 sin 2 x

L = 12

(2r 2 sin x + r 2 sin 2 x)

L = r 2

2(2sin x + sin 2 x ) [Shown]

dLdr = r 2

2 (2 cos x + 2 cos 2 x)


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When L has a stationary value,

d Ld x

= 0

r 2

2(2 cos x + 2 cos 2 x) = 0

cos x + cos 2 x = 0

cos x + 2 cos2

x - 1 = 02 cos 2 x + cos x - 1 = 0

(2 cos x - 1)(cos x + 1) = 0

cos x = 12

or cos x = - 1

x = 3

x = (not accepted)

d 2 L

d x2 = r

2

2(- 2 sin x - 4 sin 2 x)

When x =

3

,

d 2 L

d x2 = r

2

2- 2 sin

3 - 4 sin 2

3

= - 2.60 r 2 (< 0)

Hence, L is a maximum.

Lmax = r 2

2 2 sin

3

+ sin 23

= r 2

2 2 3

2+ 3

2

= 3 34

r 2 [Shown]

24 In ORQ, cos a = ORr

OR = r cos a QM = MP

= OR = r cos a

r c mr c m

M

R O

P Q

a a

In ORQ, sin a = QRr

QR = r sin a

Therefore, the perimeter of ORQP ,

y = OR + RQ + QM + MP + POy = r cos a + r sin a + r cos a + r cos a + r

y = r + r sina

+ 3r cos a

y = r (1 + sin a + 3 cos a ) [Shown]


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d yd a

= r (cos a - 3 sin a )

When y has a stationary value,

d yd r

= 0

r (cosa

- 3 sin a

) = 0cos a - 3 sin a = 0cos a = 3 sin a

13

= sin a

cos a

tan a = 13

a = tan - 1 13

rad [Shown]

d 2 yd a 2

= r (- sin a - 3 cos a )

Since sin a > 0 and cos a > 0,d 2 yd a 2

< 0.

Thus, y is a maximum.

ymax = r (1 + sin a + 3 cos a )

ymax = r 1 +110

+ 3 310

ymax = r 1 +10

10 + 3

3 1010

ymax = (1 + 10 )r [Shown]

10

3

1

a

sin a = 110

, cos a = 310

25 (a) d V d t

= Change in volume

Change in time

= 12

3

- 13

24

= - 7

192 m 3 hour - 1

(b) V = x3

d V

d x = 3 x2

d xd V

= 1

3 x2

d xd t

= d xd V

d V d t

= 1

3 x2 -

7192

= 13(0.7) 2

-7

192 = - 0.0248 m hour - 1

Rate of decrease = 0.0248 m hour - 1


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26 AB = x2 + 32

= ( x2 + 9)12

d ( AB)d x

= 12

( x2 + 9)- 1

2 (2 x)

= x x2 + 9

d( AB)d t

= d( AB)

d x d x

d t

= x x2 + 9

2

= 442 + 9

2

= 1.6 units s - 1

27 (a)(x + 2 r ) cm

r cmr cm

N

O

P

Q

x cm

R (x + 2) cm

RNO and RQP are similar triangles.

Thus, NO QP

= NR QR

r x

= x + 2 - r x + 2

r ( x + 2) = x( x + 2 - r ) rx + 2r = x2 + 2 x - rx 2rx + 2r = x2 + 2 x

r (2 x + 2) = x2 + 2 x

r = x 2 + 2 x

2 x + 2[Shown]

(b) r = x2 + 2 x

2 x + 2 d r

d x =

(2 x + 2)(2 x + 2) - ( x2 + 2 x)(2)(2 x + 2)2

d r d x

= 4 x2 + 8 x + 4 - 2 x2 - 4 x

(2 x + 2)2

d r d x

= 2 x2 + 4 x + 4

(2 x + 2)2

d r d x

= 2( x2 + 2 x + 2)

[2( x + 1)]2

d r d x =

x2 + 2 x + 22( x + 1)2


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d xd t

= d xd r

d r d t

= 2( x + 1)2

x2 + 2 x + 2 (-0.4)

= 2(4 + 1)2

42 + 2(4) + 2 (-0.4)

= 5026

-0.4

= - 0.769 cm s - 1

(c) r x

d r d x

r d r d x

x

= x2 + 2 x + 2

2( x + 1)2 (4.002 - 4)

= 42 + 2(4) + 22(4 + 1)2

0.002

= 0.00104 cm

28 y = xe x + 1

d yd x

= x e x + 1 + e x + 1

= e x + 1( x + 1)

y d yd x x

= e x + 1( x + 1) d x x changes from 1 to 1.01. So, x = 1.01 - 1.

ynew = yoriginal + y

1.01e 2.01 = 1(e 1 + 1) + [e1 + 1(1 + 1)](1.01 - 1)

The value of y when x = 1.01.

The value of y when x = 1.

The value of d yd x

when x = 1.

1.01e 2.01 = e2 + 2e2(0.01)

e2.01 = 7.3891 + 2(7.3891)(0.01)

1.01 e2.01 = 7.46

29 y = cos x x

xy = cos x

xd yd x

+ y(1) = - sin x

xd yd x + y = - sin x


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xd 2 yd x2

+ d yd x

(1) + d yd x

= - cos x

xd 2 yd x2

+ 2 d yd x

= - xy

x d 2 y

d x 2 + 2

d y

d x

+ xy = 0 [Shown]

30 y = x ln ( x + 1)

d yd x

= x 1 x + 1

+ ln ( x + 1)(1)

= x x + 1

+ ln ( x + 1)

y = d yd x

x

= ( x x + 1 + ln ( x + 1)) x ynew = yoriginal + y x changes from 1 to 1.01. So, x = 1.01 - 1.

1.01 ln (1.01 + 1) = 1 ln (1 + 1) + 311 + 1 + ln (1 + 1) (1.01 - 1)The value of y when x = 1.01.

The value of y when x = 1.

The value of d yd x

when x = 1.

1.01(ln 2.01) = 0.70508 ln 2.01 = 0.698

31 (a) f (t ) = 4ekt - 1

4ekt + 1

f (0) = 4e0 - 1

4e0 + 1

= 35

(b) f (t ) = (4e kt + 1)(4 k ekt ) - (4e kt - 1)(4 k ekt )

(4e kt + 1)2

f (t ) = (16 k e2kt + 4k ekt - 16k e2kt + 4k ekt )

(4e kt + 1)2

f (t ) = 8k ekt

(4e kt + 1)2

Since k is a positive integer, f (t ) > 0.

(c) LHS = k {1 - [f ( t )]2}

= k {1 - 34e kt - 14e kt + 12}

= k {(4ekt + 1)

2 - (4e

kt - 1)

2

(4e kt + 1)2 }


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= k {16e 2kt + 8ekt + 1 - (16e 2kt - 8ekt + 1)(4e kt + 1)2 } = 16k e

kt

(4e kt + 1)2

= 23 8k ekt (4e

kt

+ 1)2

= 2f (t ) = RHS

k {1 - [f( t )]2} = 2f (t ) k - k [f (t )]2 = 2f (t ) - 2k [f( t )] f (t ) = 2f (t ) - k [f (t )] f (t ) = f (t )

f (t ) = - k [ f (t )] f (t )Since k and f (t ) are both positive, f (t ) < 0.

(d) limt

f (t ) = limt

4ekt

- 14ekt + 1

= limt

4e kt

ekt - 1

ekt

4e kt

ekt +

1ekt

= limt

4 - 1

ekt

4 + 1ekt

= 4 - 04 + 0 = 1

(e) limt

4ekt - 1

4e kt + 1 = lim

t

4 - 1ekt

4 + 1ekt

= 4 - 04 + 0

= 1

When t = 0, f (0) = 4(1) - 14(1) + 1 = 35

Therefore, the graph of f( t ) intersects the f( t )-axis at the point 0, 35

.

When f (t ) = 0, 4ekt - 1

4e kt + 1 = 0

4ekt - 1 = 0

ekt = 14

kt = ln 14

t = 1k

ln 14


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t

f(t )

O

f(t ) =4e (kt) 1

4e (kt) + 1

35

1

1

411

k ln

Key Point:

If f (t ) = - k [f ( t )] f (t ), since k > 0 and f (t ) > 0, f (t ) > 0 only when f (t ) = 0. Therefore, the point of

inflexion is on the t -axis, i.e. 1k

ln 14

, 0 .

32 y = x1 + x2

d yd x

= (1 + x2)(1) - x(2 x)

(1 + x2)2

d yd x

= 1 - x2

(1 + x2)2

d yd x =

1 - x2

x y

2

d yd x

= (1 - x2) y2

x2

x 2 d yd x

= (1 - x 2) y2 [Shown]

33 y = sin x - cos xsin x + cos x

(sin x + cos x) y = sin x - cos x

(sin x + cos x)d yd x

+ y (cos x - sin x) = cos x + sin x


- 1 + y (cos x - sin x) = 0


- 1 - y(sin x - cos x) = 0

d yd x

- 1 - y sin x - cos xsin x + cos x

= 0

d yd x - 1 - y( y) = 0


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When x = 0,d 3 yd x3

= 6(- 04 - 6 (0) 2 - 1)

(02 - 1)4

= -6.

Sinced 3 yd x3

0, then (0, 0) is a point of reflextion .

When x = 3, y = 33

3 - 1

= - 3 32

and

d 2 yd x2

=2 3(3 + 3)

(3 - 1)3

= 32

3.

Since d 2

yd x2 > 0, then 3 , 32 3 is a minimum point .

When x = - 3, y = (- 3)

3

3 - 1

= - 3 32

and

d 2 yd x2

=2(- 3 )(3 + 3)

(3 - 1)3

= -32 3.

Sinced 2 yd x2

< 0, then - 3, - 32

3 is a maximum point .

When the denominator of y = x3

x2 - 1is 0, x2 - 1 = 0 x = 1

Hence, x = - 1 and x = 1 are asymptotes.

The graph of y = x3

x2 - 1is as shown below.

y

3 ,3

23

3 ,3

23

11 O x


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d 3 yd x3

=( x2 - 1)3(6 x2 + 6) - (2 x)( x2 + 3)(3)( x2 - 1)2(2 x)

( x2 - 1)6

=6( x2 - 1)3( x2 + 1) - (12 x2)( x2 + 3)( x2 - 1)2

( x2 - 1)6

=6( x2 - 1)2[( x2 - 1)( x2 + 1) - (2 x2)( x2 + 3)]

( x2

- 1)6

=6( x2 - 1)2( x4 - 1 - 2 x4 - 6 x2)

( x2 - 1)6

=6(- x4 - 1 - 6 x2)

( x2 - 1)4

When x = 0, d 3 yd x3

= 6[- 04 - 1 - 6(0) 2]

(02 - 1)4

= - 6 (that is 0)

Sinced 2 y

d x2 = 0 and

d 3 y

d x3 0 when x = 0, then (0, 0) is the point of inflexion.

When the curve concaves upwards,

d 2 yd x2

> 0

2 x( x2 + 3)( x2 - 1)3

> 0

2 x( x2 + 3)

[( x + 1)( x - 1)] 3 > 0

2 x( x2 + 3)

( x + 1)3( x - 1)3 > 0

+

+

+ +

+

x

x > 0

+(x 1)3 > 0

+

+ + +

x 2 + 3 > 0

(x + 1)3 > 0

0 1 + +1

Hence, the intervals for which the curve concaves upwards are - 1 < x < 0 or x > 1.

The curve y = x

x2

- 1is as shown below.

O x

y

1 1


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36 (a) x = t - 2t y = 2t + 1

t

d xd t

= 1 + 2t 2

d yd t

= 2 - 1t 2

d yd x =

d y

d t d xd t

= 2 - 1

t 2

1 + 2t 2

= 2t 2 - 1

t 2 + 2

t 2

+ 2

2

2t 2

- 12t 2 + 4 - 5

d yd x

= 2 - 5t 2 + 2

[Shown]

Let m = d yd x

m = 2 - 5t 2 + 2

(m - 2) = - 5t 2 + 2

(m - 2)(t 2 + 2) = - 5 mt 2 + 2m - 2 t 2 - 4 = - 5 (m - 2) t 2 = - 1 - 2m

t 2 = - 1 - 2mm - 2

t 2 = 1 + 2m2 - m

t 2 > 0

1 + 2m2 - m

> 0

+

+

+

+

x

1 + 2 m > 0

2 m > 0

+ 212

Hence, -12 < m < 2, that is, -

12 0 and not t 2 0.


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chapter 8 differentiation.pdf

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