chapter 8 fws

8/9/2019 Chapter 8 FWS

1/32

Oxford Fajar Sdn. Bhd. (008974-T) 2014

Fully Worked Solutions 1


CHAPTER 8 DIFFERENTIATION

Focus on Exam 8

1 (a) Lety=(x2+ 3)e2x

dy

dx=(x2+3)(2e2x) +e2x(2x)

= 2e2x

(x2

3+x)

(b) Let u = x and y=sin3 x

=x1

2 y=sin3u

du

dx=

1

2x

12

dy

du=3 sin2u (cos u)

=1

2 x = 3 sin2ucos u

Hence,dy

dx=

dy

du

du

dx

= 3 sin2ucos u 12 x

=3sin2 x cos x

2 x

2 (a) Lety=ln (x3e3x)

dy

dx=

x3(3e3x) +e3x(3x2)

x3e3x

d

dx(x3e3x)

=e3x3x2(x +1)

x3

e3x

=3(x+1)

x

Copy backx3e3x.

(b) Let u =5x

log5u =x

ln u

ln 5=x

ln u =x ln 5

1

u

du

dx=ln 5

du

dx= u ln 5


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ACE AHEADMathematics (T) Second Term2

du

dx= 5xln 5

d

dx(5x) =5xln 5

Lety =5x

1+

5x

2

dy

dx=

(1 + 5x2)(5x ln 5) (5x)(10x)

(1 + 5x2)2

=5x[ln 5(1+ 5x2) 10x]

(1 + 5x2)2

3 f(x) =e2xsin 2x

f (x)=e2x(2 cos 2x) +sin 2x(2e2x)

=2e2xcos 2x 2 sin 2xe2x

f (x)=2e2x 2 sin 2x+ cos 2x(4e2x) 2 sin 2x(2e2x) + e2x(4 cos 2x)

=4e2x

sin 2x 4 cos 2x

e

2x

+4 sin 2xe

2x

4e2x

cos 2x

=8e2xcos 2x

Whenx=

6, f(x)= 8e

3cos

3

= 8e

312 =4e

3

4 ey=x +1

2x 3

y=ln x +12x 3 =ln (x+1) ln (2x 3)

dy

dx=

1

x +1

2

2x 3

At thex-axis,y=0.

e0=x +1

2x 3

1 =x +1

2x 3 2x 3 =x+ 1

x=4

The gradient of the tangent at the point (4, 0) =1

4+1

2

2(4)3

=1

5

Hence, the equation of the tangent at the point (4, 0) is

y0 = 1

5(x 4)

5y=x+ 4


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5x2xy+y2=7

Whenx=3, 323y+y2=7

y23y +2 =0

(y1)(y2) =0

y=1 or 2

x2

xy+y2

=7

Differentiating implicitly with respect tox,

2xxdy

dx+y(1) +2y

dy

dx=0

(x+2y)dy

dx=2x+y

dy

dx=

y 2x

2y x

The gradient of the tangent at the point (3, 1) is

1 2(3)

2(1) 3 =5.

The gradient of the tangent at the point (3, 2) is2 2(3)

2(2) 3=4.

6 2y =ln (xy)

2dy

dx=

xdy

dx+y(1)

xy

2xydy

dx=x

dy

dx+y

(2xy x)dy

dx=y

dy

dx=

y

2xy x

At the pointP(e2, 1),dy

dx=

1

2(e2)(1) e2

=1

e2

Therefore, the gradient of the tangent is1

e2.

Hence, the equation of the tangent at the pointP(e2, 1) is

y1 =1

e2(xe2)

e2ye2=xe2

e2y=x

7 x=e 4t=e2 t

dx

dt=2 12 te2 t

dxdt= e2 t

t


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y= e6t

y=(e6t)12

y=e3t

dy

dt=3e3t

x=e2 t

lnx=2 t

(lnx)2=4t

t =1

4(lnx)2

dy

dx=

dy

dtdx

dt

= 3e3t

e2 t

t

=3e3t t

e2 t e3t=e

3

4(lnx)

2

=3(12lnx)e3

4(lnx)2

x e2 t=x

=3e

3

4 (lnx)2

lnx t=1

2 lnx 2x

8 x=e2t2

dx

dt=2e2t

y=et+t

dy

dt=et+1

dy

dx=

dy

dt

dxdt

=et+1

2e2t

When t=ln 2,x=e2 ln 2 2

=eln 22

2

=222

=2aloga x=x

When t=ln 2,y=eln 2+ ln 2

=2 +ln 2


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When t=ln 2,dy

dx=

eln 2+ 1

2e2 ln 2

=2 +1

2(2)2

=3

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

y(2 + ln 2) =3

8(x 2)

8y16 8 ln 2 =3x6

8y=3x+10 + 8 ln 2

9 x =cos22

dx

d=2 cos 2 (2 sin 2)

=4 cos 2sin 2

y

=sin2

2

dy

d= 2 sin 2(2 cos 2)

= 4 sin 2 cos 2

dy

dx=

dy

ddx

d

=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

= 122

=1

2

y=sin2

4 = 12

2

=1

2

Hence, the equation of the normal is

y1

2=1x 12

y1

2=x

1

2 y

= x


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10 y= e2x6x+7

=(e2x6x+7)1

2

dy

dx=

1

2(e2x6x+7)

1

2 (2e2x6)

=

2e2x6

2 e2x6x+7

=2e2x6

2y

=e2x3

y

ydy

dx=e2x3

yd2y

dx2+

dy

dx

dy

dx=2e2x

y

d2y

dx2 +dy

dx 2

=2e2x

[Shown]

11 y=exlnx

dy

dx= ex1x+ lnx ex

dy

dx= ex1x+ y

xdy

dx= ex+xy

xd2y

dx2+

dy

dx(1)=ex+x

dy

dx+y(1)

xd2y

dx2+(1 x)

dy

dxy =ex

From,

ex=xdy

dxxy

xd2y

dx2+(1 x)

dy

dxy =x

dy

dxxy

xd2y

dx2+(1 2x)

dy

dx+ (x1)y=0 [Shown]

12 y=cosx

x

xy=cosx

xdy

dx+y(1) =sinx

xdy

dx+y=sinx

xd2y

dx2+

dy

dx(1) +

dy

dx= cosx

xd2y

dx2+2

dy

dx= xy

x

d

2

ydx2+ 2dydx+xy=0 [Shown]


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13 y= cosx

=cos1

2x

dy

dx=

1

2(cosx)

1

2(sinx)

=

sinx

2 cosx

=sinx

2y

2ydy

dx= sinx

2yd2y

dx2+

dy

dx2 dydx= cosx

2yd2y

dx2+2dydx

2

+ cosx = 0

2y

d2ydx2

+2dydx2

+y2= 0 [Shown]

14 y=e2xsinx

dy

dx= e2xcosx 2 sinx e2x

dy

dx= e2xcosx 2y e2xsinx =y

d2y

dx2= e

2xsinx 2 cosx e2x2dy

dx

d2y

dx2= y 2dydx+ 2y 2

dy

dx

cosx e2x=dy

dx+2y

d2y

dx2+ 4

dy

dx+ 5y =0 [Shown]

15 y =ln(1 cosx)

dy

dx=

sinx

1 cosx

d2y

dx2=

(1 cosx)(cosx) sinx sinx

(1 cosx)2

=cosx cos2x sin2x

(1 cosx)2

=cosx (cos2x+ sin2x)

(1 cosx)2

=cosx 1

(1 cosx)2

=1 cosx

(1 cosx)2

e2xsinx =y


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=1

1 cosx

=1

cosx 1

But from,1

cosx 1

= 1

sinx

dy

dx.

d2y

dx2=

1

sinxdydx

sinxd2y

dx2=

dy

dx

sinxd2y

dx2+

dy

dx= 0 [Shown]

16 y=esinx

dy

dx= cosx

esinx

dy

dx=y cosx

d2y

dx2=y(sinx) + cosxdydx

d2y

dx2= ysinx+ cosxdydx

d2y

dx2= y sinx +

1

ydy

dx

dy

dx

cosx=1

ydy

dx

d2y

dx2= y sinx +

1

ydydx

2

yd2y

dx2= y2sinx +dydx

2

yd2y

dx2= y2lny +dydx

2

yd2y

dx2+y2lny dydx

2

=0 [Shown]

17 y =ln(sinx + cosx)

dy

dx=

cosx sinx

sinx

+ cosx

dydx2

+ 1 =cosx sinxsinx+ cosx2

+ 1

=(cosx sinx)2+ (sinx+ cosx)2

(sinx+cosx)2

=cos2x 2 sinx cosx+ sin2x + sin2x + 2 sinxcosx + cos2x

(sinx+cosx)2

esinx=y

sinx =lny


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=cos2x +sin2x+ sin2x+ cos2x

(sinx+cosx)2

=1 +1

(sinx+cosx)2sin2x +cos2x=1

=2

(sinx+cosx)2 [Shown]

d2y

dx2=

(sinx +cosx)(sinx cosx) (cosxsinx)(cosx sinx)

(sinx+cosx)2

=sin2x 2 sinxcosx cos2x (cos2x2 sinxcosx +sin2x)

(sinx+cosx)2

=2 sin2x 2 cos2x

(sinx+cosx)2

=2(sin2x + cos2x)

(sinx+cosx)2

= 2(1)(sinx+cosx)2

=dydx2

+1

d2y

dx2+ dydx

2

+1= 0 [Shown]

18 (a)y =x2

(x+3)(x1)

= x

2

x2+2x3

Asy ,the denominator ofx2

(x+3)(x1) 0

(x+3)(x1) 0

x 3 or 1

Therefore,x=3 andx=1are vertical asymptotes.

limx

y = limx

x2

x2+2x3

= lim

x

x

2

x2

x2

x2+

2x

x2

3

x2 = lim

x

1

1 +2

x

3

x2

=1

1 +0 +0

=1

Therefore,y=1 is the horizontal asymptote.


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(b) y =x2

x2+2x3

dy

dx=

(x2+2x3)(2x) x2(2x + 2)

(x2+2x3)2

=

2x3+4x26x2x3 2x2

(x2+2x3)2

=2x2 6x

(x2+2x3)2

d2y

dx2=

(x2+2x3)2(4x 6) (2x2 6x) 2(x2+ 2x 3)(2x + 2)

(x2+2x3)4

=2(x2+2x3)[(x2+2x3)(2x 3) (2x2 6x)(2x + 2)]

(x2+2x3)4

=

2[(x2+2x3)(2x3) (2x2 6x)(2x + 2)](x2+2x3)3

Whendy

dx= 0,

2x2 6x

(x2+2x3)2=0

2x2 6x =0

2x(x 3) =0

x = 0 or 3

Whenx=0,y=0 and

d2y

dx2=2[(3)(3) 0]

(3)3

=2

3(< 0)

Therefore, (0, 0) is a turning point and it is a local maximum point.

Whenx=3, y=9

6(2)

=3

4

d2

ydx2

=2[(32

+ 2 3 3)(2 3 3) 0](32+ 2 3 3)3

=6 (> 0)

Therefore, 3,34is a turning point and it is a local minimum point. (c) Wheny=0, x=0.

Hence, the graph ofy=x2

(x+3)(x1)

= x2

x2+2x3is as shown.


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x

Y

3 O

3

1

1

3

4,

19 (a)y =4(x 3)21

x3

x

=

3 is the vertical asymptote.

(b) Whenx=0, y=4(3)21

(3)

=361

3

Thus, the graph cuts they-axis at 0, 3613. Wheny=0, 4(x3)2

1

x3=0

4(x3)2=1

x3

(x3)3=1

4

x3 =1

413

x=1

413

+ 3

x= 3.63

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

(c) y= 4(x3)21

x3

=4(x3)2 (x3)1

dy

dx= 8(x3)1(1) + (x3)2(1)

= 8(x3) +1

(x3)2

d2y

dx2=8 2(x3)3(1)

=8

2

(x3)3


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Whendy

dx= 0,

8(x3) +1

(x3)2=0

8(x3) =1

(x3)2

(x3)3=18

x 3 =1

2

x =2 .5

Whenx=21

2,

y=452 32

1

5

2 3

= 1 + 2 = 3

d2y

dx2=8

2

52 33

=8 (16)

=24 (> 0)

Therefore, the turning point is 2.5, 3and it is a local minimum point. (d) When

d2ydx2

=0,

8 2

(x 3)3= 0

2

(x 3)3= 8

(x3)3=1

4

x=1

4

1

3

+3

x=3.63

From (b), whenx=3.63,y=0.

d3y

dx3=6(x3)4(1)

=6

(x 3)4

Whenx=3.63,d3y

dx3=

6

(3.633)4

=38.1 (i.e. 0)

Hence, (3.63, 0) is a point of inflexion.


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(e) The graph ofy=4(x 3)21

x 3is as shown below.

x

y

33.63O

2.5 , 3

361

3

20 (a) Thex-axisis the axis of symmetry.

(b) y 2=x2(4 x)

y20

x2(4 x) 0

Sincex20,x2(4 x) 0 only if 4 x0 i.e.x4.

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

(c) y2=x2(4 x)

=4x2x3

2ydy

dx=8x3x2

dy

dx=

8x3x2

2y

dy

dx=

8x3x2

2(x 4 x )

dydx= x(8 3x)2x 4 x

dy

dx=

8 3x

2 4 x

Whendy

dx=0,

8 3x

2 4 x=0

8 3x=0

x =83


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Whenx=8

3,y=

8

3 4

8

3

=3.08

Hence, 223, 3.08and 22

3,3.08are turning points (whose tangents are horizontal).

When dydx

=,

2 4 x =0

x=4

Whenx=4, y=4 4 4

=0

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

(d) The graph ofy 2=x2(4 x) is as shown below.

x

y

O 4

22

3, 3.08

22

3, 3.08

21 (a) y=1 e2x

1 +e2x

dy

dx=

(1 +e2x)(2e2x) (1 e2x)(2e2x)

(1 +e2x)2

dy

dx=

2e2x[1 +e2x+(1 e2x)]

(1 +e2x)2

dy

dx=

4e2x

(1 + e2x)2

Since e2x>0 and (1 +e2x)2>0, thusdy

dx=

4e2x

(1+ e2x)2


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dy

dx=

4e2x

(1 + e2x)2

=4e

212ln 1 y

1 +y

1 + e21

2 ln 1 y

1 +y2

=

41 y1 +y

1 + 1 y1 +y2 a

logax=x

=

41 y1 +y

1 +y+1 y

1 +y

2

=

41 y1 +y4

(1 +y)2

=(1 y)(1 +y)

=y 2 1 [Shown]

d2y

dx2=2ydy

dx

Sincedy

dx< 0,

d2y

dx2< 0 ify> 0 and

d2y

dx2> 0ify< 0. [Shown]

(c) limx 1 e

2x

1 +e2x=1 and limx 1 e2x

1 +e2x=1

(d) When y =0,1 e2x

1 +e2x=0

1 e2x=0 e2x=1

2x =ln 1

2x =0

x =0

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

Hence, the graph ofy=1 e2x

1 +e2xis as shown beside.

y

xO

1

1


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22 (a)

kcm

6 cm

(2k + 6) cm

Q

CRSD

x cm

BA

P

CQRand CBSare similar triangles.

k cm

B

Q

CRS

x cm

[(2k + 6) 6] cm

Thus, RCSC

=QRBS

RC

(2k +6) 6=

x

k

RC

2k=

x

k

RC=2x

Thus,DR =DC RC

=2k+6 2x

(b) Area ofPQRD,

L =DR QR

L =(2k +6 2x)(x)

L =(2k

+

6)x2x2 [Shown]

(c) WhenLhas a stationary value,

dL

dx=0

2k+6 4x=0

4x=2k+6

x=2k+6

4

x=2(k+3)

4

x=k+3

2

d2L

dx2=4 (negative)

Thus,Lhas a maximum value.


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Hence, the maximum value ofL

=(2k+6) k+32 2k+3

2 2

=2(k+3) k+32 2(k+3)2

4

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

2

=(k

+3)2

2

23 In OMC, sinx =MC

r MC =r sinx

AC =2MC

=2rsinx

rcmrcm

O

MCA

B

x

x x

rcm

In OMC, cosx =OM

r OM =r cosx

Area of ABC,

L =1

2AC BM

L =1

2AC (BO +OM)

L =1

2 (2r sinx)(r +r cosx)

L =r2

sinx + r2

sinx cosx

L =r2sinx +1

2r2(2 sinx cosx)

L =r2sinx +1

2r2sin 2x

L =1

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

L =r2

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

dLdr=r

2

2 (2 cosx + 2 cos 2x)


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WhenL has a stationary value,

dL

dx=0

r2

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

cosx +cos 2x=0

cosx+2 cos2

x 1 =0 2 cos2 x+cosx1 =0

(2 cosx1)(cosx +1) =0

cosx =1

2or cosx=1

x

=

3 x=(not accepted)

d2L

dx2=

r2

2(2 sinx4 sin 2x)

Whenx =

3,

d2L

dx2=

r 2

2 2 sin

34 sin

2

3 =2.60r 2(< 0)

Hence,Lis a maximum.

Lmax

=r 2

22 sin 3 +sin

2

3

=r2

22 32 + 32

=3 3

4r2 [Shown]

24 In ORQ,cos=OR

r

OR=rcos

QM =MP

=OR

=r cos

In ORQ, sin =QR

r

QR =r sin

Therefore, the perimeter of ORQP,

y=OR+RQ+QM+MP+PO

y=rcos +r sin +r cos +r cos+r

y=r +r sin + 3r cos

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

dy

da=r(cos 3 sin )

rcm

rcm

M

R O

PQ

a a


19/32



Whenyhas a stationary value,

dy

dr=0

r(cos3 sin) =0

cos 3 sin =0

cos =3 sin

1

3=

sin

cos

tan =1

3

=tan113rad [Shown]

d2y

d2=r(sin 3 cos )

Since sin >0 and cos >0,d2y

d2


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26 AB= x2+32

=(x2+9)1

2

d(AB)

dx=

1

2(x2+9)

1

2(2x)

=x

x2+9

d(AB)

dt=

d(AB)

dx

dx

dt

=x

x2+9 2

=4

42+9 2

=1.6 units s1

27 (a)(x+2 r)cm

rcm

rcm

N

O

P

Q

x cm

R(x+ 2) cm

RNOand RQPare similar triangles.

Thus,NO

QP=

NR

QR

r

x=

x+2r

x+2

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

rx+2r=x2+2xrx

2rx +2r =x2+2x

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

r =x2+2x

2x+2[Shown]

(b) r =x2+2x

2x+

2

dr

dx=

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

(2x+2)2

dr

dx=

4x2+ 8x+4 2x2 4x

(2x+2)2

dr

dx=

2x2+4x + 4

(2x+2)2

dr

dx=

2(x2+2x + 2)

[2(x+1)]2

dr

dx=x2+2x + 2

2(x+1)2


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dx

dt=

dx

dr

dr

dt

=2(x + 1)2

x2+2x + 2(0.4)

=2(4+ 1)2

42

+2(4)+ 2(0.4)

=50

260.4

=0.769 cm s1

28 y=xex +1

dy

dx=x ex+1+ex+1

=ex+1(x+1)

y

dy

dxx

=ex+1(x+1) x x changes from 1 to 1.01. So, x=1.01 1.

ynew

=yoriginal

+y

1.01e2.01=1(e1+1) +[e1+1(1 +1)](1.01 1)]

The value ofy

whenx= 1.01.

The value of

ywhenx=1.

The value ofdy

dx

whenx=1.

1.01e2.01=e2+2e2(0.01)

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

1.01

e2.01=7.46

29 y=cosx

x

xy=cosx

xdy

dx+y(1)=sinx

x dydx

+y =sinx

xd2y

dx2+

dy

dx(1) +

dy

dx=cosx

xd2y

dx2+2

dy

dx=xy

x

d2y

dx2

+

2

dy

dx

+xy=0 [Shown]


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30 y =x ln(x+1)

dy

dx=x 1x+1+ln (x+1)(1)

=x

x+1+ln (x+1)

y=dy

dxx

=( xx+1 +ln (x+1))x y

new=y

original+y x changes from 1 to 1.01. So, x= 1.01 1.

1.01 ln (1.01 +1) =1 ln (1 +1) + 11 +1+ln (1 +1)(1.01 1)

The value ofywhenx= 1.01.

The value ofywhenx= 1.

The value ofdy

dxwhenx= 1.

1.01(ln 2.01) =0.70508

ln 2.01 =0.698

31 (a) f(t) =4ekt1

4ekt+1

f(0) =4e01

4e0+1

=3

5 (b) f (t) =

(4ekt+1)(4kekt) (4ekt1)(4kekt)

(4ekt+1)2

f (t) =(16ke2kt+4kekt16ke2kt+4kekt)

(4ekt+1)2

f (t) =8kekt

(4ekt+1)2

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

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

=k{1 4ekt1

4ekt+1

2

} =k{(4e

kt+1)2(4ekt1)2

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

2kt+8ekt+1 (16e2kt8ekt+1)

(4ekt+1)2 } =

16kekt

(4ekt+1)2

=2 8kekt

(4ekt+1)2 =2f(t)

=RHS


23/32



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

kk[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 kand f(t) are both positive, f(t) 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. 1kln1

4, 0.

t

f(t)

O

f(t) = 4ekt 1

4ekt+ 1

35

1

1

4

1

1

k ln


24/32



32 y=x

1 +x2

dy

dx=

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

(1 +x2)2

dy

dx=

1 x2

(1 + x2)2

dy

dx=

1 x2

xy2

dy

dx=

(1 x2)y2

x2

x2dy

dx=(1 x2)y2 [Shown]

33 y =sinx cosx

sinx+ cosx (sinx+cosx)y=sinxcosx

(sinx+cosx)dy

dx+y(cosxsinx) =cosx+sinx

(sinx+cosx)dydx 1+y(cosxsinx) = 0

(sinx+cosx)dydx 1y(sinxcosx) = 0

dy

dx 1

y

sinxcosx

sinx+cosx= 0

dydx 1y(y) = 0

dy

dx 1 y2= 0

d2y

dx2 2y

dy

dx= 0

d2y

dx2= 2y

dy

dx [Shown]

34 y =x3

x2 1

dy

dx=

(x21)(3x2) x3(2x)

(x21)2

=3x43x22x4

(x21)2

=x43x2

(x21)2

=x2(x2

3)(x21)2


25/32



d2y

dx2=

(x21)2(4x36x) (x43x2)(2)(x21)1(2x)

(x21)4

=(x21)2(2x)(2x23) 4x(x43x2)(x21)

(x21)4

=2x(x21)[(x21)(2x23) 2(x43x2)]

(x2

1)4

=2x[2x45x2+3 2x4+6x2]

(x21)3

=2x(x2+3)

(x21)3

d3y

dx3=

(x21)3(6x2+6) (2x3+6x)(3)(x21)2(2x)

(x21)6

=6(x21)3(x2+1) 6x(2x3+6x)(x21)2

(x21)6

=

6(x21)2[(x21)(x2+1) x(2x3+6x)]

(x21)6

=6(x412x46x2)

(x21)4

=6(x46x2 1)

(x21)4

Whendy

dx=0

x2(x23) =0

x=0 or 3

Whenx=0,y=0 andd2y

dx2=

2(0)(02+3)

(021)3

= 0

Whenx=0,d3y

dx3=

6(04 6(0)21)

(021)4

= 6

Sinced3y

dx3 0, then (0, 0) is a point of reflextion.

Whenx= 3,y=( 3 )3

3 1

=3 3

2

and

d2y

dx2=

2 3(3 + 3)

(3 1)3

=3

2 3

Sinced2y

dx2> 0,then 3 ,3

2 3is a minimum point.


26/32



Whenx= 3,y=( 3)

3

3 1

=3 3

2 and

d2

ydx2

= 2( 3)(3 + 3)(3 1)3

= 3

2 3.

Sinced2y

dx2< 0,then 3, 32 3is a maximum point.

When the denominator ofy=x3

x2 1is 0,x21 =0 x=1

Hence,x=1 andx= 1 are asymptotes.

The graph ofy= x3

x2 1is as shown below.

y

3 ,3

2

3

3 , 3

2

3 11 O

x

x3= k(x2 1)

x3

x2 1=k

By sketching the straight linesy=kon the above graph and as kvaries, we obtain the

following results.

Value of k Number of real roots

k>3

2 3 3

k=3

2 3 2

3

2 3


27/32



35 y =x

x2 1

dy

dx=

(x21)(1) x(2x)

(x21)2

=x21

(x

2

1)

2

=(x2+1)

(x21)2(that is


28/32



When the curve concaves upwards,

d2y

dx2> 0

2x(x2+3)

(x21)3> 0

2x(x2

+3)[(x+1)(x1)]3

> 0

2x(x2+3)

(x+1)3(x1)3> 0

+

+

++

+

x

x >0

+(x1)3>0

+

+ + +

x2+3 >0

(x+1)3>0

01 + +1

Hence, the intervals for which the curve is concave upwards are 1 1.

The curvey =x

x2 1is as shown below.

Ox

y

1 1

36 (a) x =t2

t y =2t+

1

t

dx

dt=1 +2

t2

dy

dt=2 1

t2

dy

dx=

dy

dtdx

dt

=2

1

t2

1 +2

t2

=2t21

t2+2


29/32



t2+2

2

2t2 1 2t2+ 4

5

dy

dx=2

5

t2+2 [Shown]

Let m=dy

dx

m=2 5

t2+2

(m2) =5

t2+2

(m2)(t2+ 2) =5

mt2+2m 2t24 =5

(m 2)t2

=1 2m t2=

1 2m

m 2

t2=1 +2m

2m

t2> 0

1 +2m

2m>0

+

+

+

+

x

1 +2m>0

2 m>0

+ 21

2

Hence, 1

2


30/32



and

y=2(1) +1

1 =3

When t=1,x=(1) 2

(1)

=1

and

y=2(1) +1

(1)

=3

Hence, the coordinates of the required points are (1, 3) and (1,3).

37 x y

x y

2 3

2 3

sin cos

cos sin

q q

qq

qqd

ddd

d

d

d

dd

d

sin

cos

tan

y

x

y

x= =

=

q

q

q

q

q

3

2

3

2

When q = =

=

4

3

2 4

3

2

, tand

d

y

x

Gradient of tangent = -3

2

Gradient of normal =2

3

When q = =

=

=

4

24

2 1

2

2

, sinx

When q p p

43

4

3 1

2

3 2

2

, cosy


31/32



Equation of normal is

y x- = -( )3 22

2

32

6 9 2 4 2

6 9 2 4 4 2

6 4 5 2

y x

y x

y x

- = -( )- = -

= +

38 (a) x xy y

xy

xy

y

x

xx y x y

y

x

x y

2 24

2 1 2 0

2 2

2

+ + =

+ + ( ) + =

+( ) =

=

x y

y

d

d

d

d

d

d

d

d xx y

y

x

x y

x y

+

++

+

=

2

2

20

d

d [Shown]

(b) Atx-axis,y=0,

x2+ 0 + 0 =4

x=2

\(2, 0) and (2, 0)

(2, 0): Gradient =- ( )++ ( )

2 2 0

2 2 0

=- 2

(2, 0): Gradient = ( )

+ ( )

2 2 0

2 2 0

=-2

Aty-axis,x=0,

02+ 0(y) + y2=4

y=2

\(0, 2) and (0, 2)

(0, 2): Gradient =- ( )-

+ ( )

2 0 2

0 2 2

=-12

(0, 2): Gradient =- ( ) - -( )

+ -( )

2 0 2

0 2 2

=-12


32/32


(c) At stationary points,

d

d

y

x=0

- -

+

=

- - =

= -

2

20

2 0

2

x y

x y

x y

y x

Substitutingy=2xintox2+ xy+ y2=4

x2+ x(2x) + 4x2=4

3x2=4

x

x

2 4

3

2

3

=

=

y= -

2

2

3

= 4

3

Stationary points are2

3

4

3,-

and

2

3

4

3, .

(d)

2

2

y

xO2

2

3, 4

3

2

3, 4

3

2

chapter 8 fws

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