chapter 8 fws
Post on 01-Jun-2018
280 Views
Preview:
TRANSCRIPT
-
8/9/2019 Chapter 8 FWS
1/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 1
Oxford Fajar Sdn. Bhd. (008974-T) 2014
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
-
8/9/2019 Chapter 8 FWS
2/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
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
-
8/9/2019 Chapter 8 FWS
3/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 3
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
-
8/9/2019 Chapter 8 FWS
4/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term4
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
-
8/9/2019 Chapter 8 FWS
5/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 5
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
-
8/9/2019 Chapter 8 FWS
6/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term6
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]
-
8/9/2019 Chapter 8 FWS
7/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 7
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
-
8/9/2019 Chapter 8 FWS
8/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term8
=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
-
8/9/2019 Chapter 8 FWS
9/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 9
=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.
-
8/9/2019 Chapter 8 FWS
10/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term10
(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.
-
8/9/2019 Chapter 8 FWS
11/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 11
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
-
8/9/2019 Chapter 8 FWS
12/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term12
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.
-
8/9/2019 Chapter 8 FWS
13/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 13
(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
-
8/9/2019 Chapter 8 FWS
14/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term14
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
-
8/9/2019 Chapter 8 FWS
15/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 15
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
-
8/9/2019 Chapter 8 FWS
16/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term16
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.
-
8/9/2019 Chapter 8 FWS
17/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 17
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)
-
8/9/2019 Chapter 8 FWS
18/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term18
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
-
8/9/2019 Chapter 8 FWS
19/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 19
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
-
8/9/2019 Chapter 8 FWS
20/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term20
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
-
8/9/2019 Chapter 8 FWS
21/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 21
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]
-
8/9/2019 Chapter 8 FWS
22/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term22
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
-
8/9/2019 Chapter 8 FWS
23/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 23
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
-
8/9/2019 Chapter 8 FWS
24/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term24
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
-
8/9/2019 Chapter 8 FWS
25/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 25
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.
-
8/9/2019 Chapter 8 FWS
26/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term26
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
-
8/9/2019 Chapter 8 FWS
27/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 27
35 y =x
x2 1
dy
dx=
(x21)(1) x(2x)
(x21)2
=x21
(x
2
1)
2
=(x2+1)
(x21)2(that is
-
8/9/2019 Chapter 8 FWS
28/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term28
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
-
8/9/2019 Chapter 8 FWS
29/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 29
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
-
8/9/2019 Chapter 8 FWS
30/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
ACE AHEADMathematics (T) Second Term30
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
-
8/9/2019 Chapter 8 FWS
31/32
Oxford Fajar Sdn. Bhd. (008974-T) 2014
Fully Worked Solutions 31
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
-
8/9/2019 Chapter 8 FWS
32/32
ACE AHEADMathematics (T) Second Term32
(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
top related