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© Oxford Fajar Sdn. Bhd. (008974-T) 2012
CHAPTER 8 DIFFERENTIATION
Focus on Exam 8
1 (a) Let y = (x2 + 3)e-2x
dydx
= (x2 + 3)(-2e-2x) + e-2x(2x)
= 2e-2x(-x2 - 3 + x)
(b) Let u = x and y = sin3 x
= x12 y = sin3 u
dudx
= 12
x- 1
2 dydu
= 3 sin2 u (-cos u)
= 12 x
= -3 sin2 u cos u
Hence, dydx
= dydu
× dudx
= -3 sin2 u cos u × 12 x
= -3 sin2 x cos x
2 x
2 (a) Let y = ln (x3 e-3x)
dydx
= x3(-3e-3x) + e-3x(3x2)
x3e-3x
ddx
(x3e-3x)
= e-3x 3x2(-x + 1)
x3e-3x
= 3(-x + 1)x
Copy back x3e-3x.
(b) Let u = 5x
log5 u = x
ln uln 5
= x
ln u = x ln 5
1u
dudx
= ln 5
dudx
= u ln 5
Chap-08-FWS.indd 1 10/19/2012 10:37:43 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term2
dudx
= 5x ln 5
∴ ddx
(5x) = 5x ln 5
Let y = 5x
1 + 5x2
dydx
= (1 + 5x2)(5x ln 5) - (5x)(10x)(1 + 5x2)2
= 5x [ln 5 (1 + 5x2) - 10x](1 + 5x2)2
3 f (x) = e-2x sin 2x
f ′(x) = e-2x (2 cos 2x) + sin 2x (-2e-2x)
f ′(x) = 2e-2x cos 2x - 2 sin 2x e-2x
f ″(x) = 2e-2x - 2 sin 2x + cos 2x (- 4e-2x) - 2 sin 2x (-2e-2x) + e-2x (-4 cos 2x)
f ″(x) = -4e-2x sin 2x - 4 cos 2x e-2x + 4 sin 2x e-2x - 4e-2x cos 2x
f ″(x) = -8e-2x cos 2x
When x = π6
, f ″(x) = -8e-
π3 cos π
3
= -8e-
π3 1
2 = -4e
-π3
4 ey = x + 12x - 3
y = ln x + 12x - 3
= ln (x + 1) - ln (2x - 3)
dydx
= 1x + 1
- 22x - 3
At the x-axis, y = 0.
e0 = x + 12x - 3
1 = x + 12x - 3
2x - 3 = x + 1 x = 4
The gradient of the tangent at the point (4, 0) = 14 + 1
- 22(4) - 3
= - 15
Hence, the equation of the tangent at the point (4, 0) is
y - 0 = -15
(x - 4)
5y = -x + 4
Chap-08-FWS.indd 2 10/19/2012 10:37:43 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 3
5 x2 - xy + y2 = 7
When x = 3, 32 -3y + y2 = 7 y2 - 3y + 2 = 0 (y - 1)(y - 2) = 0 y = 1 or 2 x2 - xy + y2 = 7
Differentiating implicitly with respect to x,
2x - x dydx
+ y(-1) + 2y dydx
= 0
(-x + 2y) dydx
= -2x + y
dydx
= -2x + y-x + 2y
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 2y = ln (xy)
2dydx
= x
dydx
+ y(1)
xy
2xy dydx
= x dydx
+ y
(2xy - x) dydx
= y
dydx
= y
2xy - x
At the point P(e2, 1), dydx
= 12(e2)(1) - e2
= 1e2
Therefore, the gradient of the tangent is 1e2
.
Hence, the equation of the tangent at the point P(e2, 1) is
y - 1 = 1e2
(x - e2)
e2y - e2 = x - e2
e2y = x
7 x = e 4t = e2 t
dxdt
= 2 12 t e2 t
dxdt
= e2 t
t
Chap-08-FWS.indd 3 10/19/2012 10:37:43 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term4
y = e6t
y = (e6t)12
y = e3t
dydt
= 3e3t
x = e2 t
ln x = 2 t
(ln x)2 = 4t
t = 14
(ln x)2
dydx
=
dydtdxdt
= 3e3t
e2 t
t
= 3e3t t
e2 t e3t = e
34
(ln x)2
= 3(12 ln x)e34
(ln x)2
x e2 t = x
= 32
ln xx
e34
(ln x)2 t =
12
ln x
8 x = e2t - 2
dxdt
= 2e2t
dydx
=
dydtdxdt
= et + 12e2t
y = et + t
dydt
= et + 1
When t = ln 2, x = e2 ln 2 - 2 = eln 22
- 2 = 22 - 2 = 2 alogax = x
When t = ln 2, y = eln 2 + 2 = 2 + 2 = 4
Chap-08-FWS.indd 4 10/19/2012 10:37:43 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 5
When t = ln 2, dydx
= eln 2 + 12e2 ln 2
= 2 + 12(2)2
= 38
Hence, the equation of the tangent at the point where t = ln 2 is
y - 4 = 38
(x - 2)
8y - 32 = 3x - 6 8y = 3x + 26
9 x = - cos2 2θ
dxdθ
= -2 cos 2θ (-2 sin 2θ)
= 4 cos 2θ sin 2θ y = sin2 2θ
dydθ
= 2 sin 2θ (2 cos 2θ)
= 4 sin 2θ cos 2θ
∴ dydx
=
dydθdxdθ
= 4 sin 2θ cos 2θ4 cos 2θ sin 2θ
= 1
The gradient of the tangent is 1. Hence, the gradient of the normal is -1.
When θ = π8
, x = - cos2 π4
= - 12
2
= -12
and y = sin2 π4 = 1
22
= 12
Hence, the equation of the normal is
y - 12
= -13x - - 12
y - 12
= -x - 12
y = -x
Chap-08-FWS.indd 5 10/19/2012 10:37:44 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term6
10 y = e2x - 6x + 7
= (e2x - 6x + 7)12
dydx
= 12
(e2x - 6x + 7)- 1
2 (2e2x - 6)
= 2e2x - 62 e2x - 6x + 7
= 2e2x - 62y
= e2x - 3
y
y dydx
= e2x - 3
y d2ydx2
+ dydx
dydx
= 2e2x
y d2ydx2 + dy
dx2
= 2e2x [Shown]
11 y = ex ln x
dydx
= ex 1x + ln x ex
dydx
= ex 1x + y
x dydx
= ex + xy …
x d2ydx2
+ dydx
(1) = ex + x dydx
+ y(1)
x d2ydx2
+ (1 - x) dydx
- y = ex From ,
ex = x dydx
- xy
x d2ydx2
+ (1 - x) dydx
- y = x dydx
- xy
x d2ydx2
+ (1 - 2x) dydx
+ (x - 1) y = 0 [Shown]
12 y = cos xx
xy = cos x
x dydx
+ y(1) = -sin x
x dydx
+ y = -sin x
x d2ydx2
+ dydx
(1) + dydx
= -cos x
x d2ydx2
+ 2 dydx
= -xy
x d2ydx2 + 2
dydx
+ xy = 0 [Shown]
Chap-08-FWS.indd 6 10/19/2012 10:37:44 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 7
13 y = cos x
= cos12 x
dydx
= 12
(cos x)- 1
2 (-sin x)
= -sin x
2 cos x
= -sin x2y
2y dydx
= -sin x
2y d2ydx2
+ dydx
2 dydx = -cos x
2y d2ydx2
+ 2dydx
2
+ cos x = 0
2y d2ydx2
+ 2dydx
2
+ y2 = 0 [Shown]
14 y = e-2x sin x
dydx
= e-2x cos x - 2 sin x e-2x
dydx
= e-2x cos x - 2y e-2x sin x = y
d2ydx2 = -e-2x sin x - 2 cos x e-2x - 2
dydx
d2ydx2 = -y - 2dy
dx + 2y - 2
dydx
cos x e-2x =
dydx
+ 2y
d2ydx2 + 4
dydx
+ 5y = 0 [Shown]
15 y = ln (1 - cos x)
dydx
= sin x1 - cos x
…
d2ydx2
= (1 - cos x)(cos x) - sin x sin x
(1 - cos x)2
= cos x - cos2 x - sin2 x(1 - cos x)2
= cos x - (cos2 x + sin2 x)
(1 - cos x)2
= cos x - 1(1 - cos x)2
= - 1 - cos x(1 - cos x)2
e-2x sin x = y
Chap-08-FWS.indd 7 10/19/2012 10:37:44 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term8
= -11 - cos x
= 1
cos x - 1
But from , 1cos x - 1
= - 1sin x
dydx.
∴ d2y
dx2 = - 1sin x
dydx
sin x d2ydx2 = -dy
dx
sin x d2ydx2 + dy
dx = 0 [Shown]
16 y = esin x
dydx
= cos x esin x
dydx
= y cos x
d2ydx2 = y(-sin x) + cos x dy
dx
d2ydx2 = -y sin x + cos x dy
dx
d2ydx2 = -y sin x + 1
y dydx
dydx
cos x = 1y dydx
d2ydx2 = -y sin x + 1
y dy
dx2
y d2ydx2 = -y2 sin x + dy
dx2
y d2ydx2 = -y2 ln y + dy
dx2
y d2ydx2 + y2 ln y - dy
dx2
= 0 [Shown]
17 y = ln (sin x + cos x)
dydx
= cos x - sin xsin x + cos x
dydx
2
+ 1 = cos x - sin xsin x + cos x
2
+ 1
= (cos x - sin x)2 + (sin x + cos x)2
(sin x + cos x)2
= cos2 x - 2 sin x cos x + sin2 x + sin2 x + 2 sin x cos x + cos2 x
(sin x + cos x)2
esin x = y
sin x = ln y
Chap-08-FWS.indd 8 10/19/2012 10:37:44 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 9
= cos2 x + sin2 x + sin2 x + cos2 x(sin x + cos x)2
= 1 + 1(sin x + cos x)2
sin2 x + cos2 x = 1
= 2(sin x + cos x)2
[Shown]
d2ydx2 =
(sin x + cos x)(-sin x - cos x) - (cos x - sin x)(cos x - sin x)(sin x + cos x)2
= -sin2 x - 2 sin x cos x - cos2 x - (cos2 x - 2 sin x cos x + sin2 x)
(sin x + cos x)2
= -2 sin2 x - 2 cos2 x(sin x + cos x)2
= -2(sin2 x + cos2 x)
(sin x + cos x)2
= -2(1)
(sin x + cos x)2
= - 3dydx
2
+ 1 ∴ d
2ydx2 + dy
dx2 + 1 = 0 [Shown]
18 (a) y = x2
(x + 3)(x - 1)
= x2
x2 + 2x - 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.
limx → ±∞
y = limx → ±∞
x2
x2 + 2x - 3
= lim
x→± ∞
x2
x2
x2
x2 + 2x
x2 - 3
x2
= limx→± ∞
1
1 + 2x - 3
x2 = 1
1 + 0 + 0
= 1
Therefore, y = 1 is the horizontal asymptote.
Chap-08-FWS.indd 9 10/19/2012 10:37:44 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term10
(b) y = x2
x2 + 2x - 3
dydx
= (x2 + 2x - 3)(2x) - x2(2x + 2)
(x2 + 2x - 3)2
= 2x3 + 4x2 - 6x - 2x3- 2x2
(x2 + 2x - 3)2
= 2x2 - 6x(x2 + 2x - 3)2
d2ydx2 = (x
2 + 2x - 3)2(4x - 6) - (2x2 - 6x) 2(x2 + 2x - 3)(2x + 2)(x2 + 2x - 3)4
= 2(x2 + 2x - 3)[(x2 + 2x - 3)(2x - 3) - (2x2 - 6x)(2x + 2)](x2 + 2x - 3)4
= 2[(x2 + 2x - 3)(2x - 3) - (2x2 - 6x)(2x + 2)]
(x2 + 2x - 3)3
When dydx
= 0, 2x2 - 6x(x2 + 2x - 3)2
= 0
2x2 - 6x = 0
2x(x - 3) = 0
x = 0 or 3
When x = 0, y = 0 and
d2ydx2 =
2[(-3)(-3) - 0](-3)3
= -23 (<0)
Therefore, (0, 0) is a turning point and it is a local maximum point.
When x = 3, y = 96(2)
= 34
and
d2ydx2 =
2[(32 + 2 × 3 - 3)(2 × 3 - 3) - 0](32 + 2 × 3 - 3)3 = 6(>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 + 2x - 3 is as shown.
Chap-08-FWS.indd 10 10/19/2012 10:37:44 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 11
x
Y
−3 O
31
1
34
,� �
19 (a) y = 4(x - 3)2 - 1x - 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 - 1x - 3
= 0
4(x - 3)2 = 1x - 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 - 1x - 3
= 4(x - 3)2 - (x - 3)-1
dydx
= 8(x - 3)1(1) + (x - 3)-2(1)
= 8(x - 3) + 1(x - 3)2
d2ydx2 = 8 - 2(x - 3)-3(1)
= 8 - 2(x - 3)3
Chap-08-FWS.indd 11 10/19/2012 10:37:45 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term12
When dydx
= 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 = 452 - 32
- 1
52 - 3 = 1 + 2 = 3
d2ydx2 = 8 - 2
52 - 33
= 8 - (-16) = 24 (> 0)
Therefore, the turning point is 212
, 3 and it is a local minimum point.
(d) When d2ydx2
= 0,
8 - 2(x - 3)3
= 0
2(x - 3)3
= 8
(x - 3)3 = 14
x = 1
413
+ 3
x = 3.63
From (b), when x = 3.63, y = 0.
d3ydx3
= 6(x - 3)-4(1)
= 6(x - 3)4
When x = 3.63, d3ydx3
= 6(3.63 - 3)4
= 38.1 (i.e. ≠ 0)
Hence, (3.63, 0) is a point of inflexion.
Chap-08-FWS.indd 12 10/19/2012 10:37:45 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 13
(e) The graph of y = 4(x - 3)2 - 1x - 3
is as shown below.
x
y
3 3.63O
2 12
, 3� �
3613
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)
= 4x2 - x3
2y dydx
= 8x - 3x2
dydx
= 8x - 3x2
2y
dydx
= 8x - 3x2
2(± x 4 - x )
dydx
= x(8 - 3x)
± 2x 4 - x
dydx
= 8 - 3x± 2 4 - x
When dydx
= 0,
8 - 3x± 2 4 - x
= 0
8 - 3x = 0
x = 83
Chap-08-FWS.indd 13 10/19/2012 10:37:45 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term14
When x = 83
, y = ± 83
4 - 83
= ±3.08
Hence, 223
, 3.08 and 223
, -3.08 are turning points (whose tangents are horizontal).
When dydx
= ∞,
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 - e2x
1 + e2x
dydx
= (1 + e2x)(-2e2x) - (1 - e2x)(2e2x)
(1 + e2x)2
dydx
= -2e2x[1 + e2x + (1 - e2x)]
(1 + e2x)2
dydx
= -4e2x
(1 + e2x)2
Since e2x > 0 and (1 + e2x)2 > 0, thus dydx
= -4e2x
(1 + e2x)2 < 0 [Shown]
(b) y = 1 - e2x
1 + e2x
y + ye2x = 1- e2x
e2x(1 + y) = 1 - y
e2x = 1 - y1 + y
2x = ln 1 - y1 + y
x = 12
ln 1 - y1 + y [Shown]
Chap-08-FWS.indd 14 10/19/2012 10:37:46 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 15
dydx
= -4e2x
(1 + e2x)2
= -4e212 ln 1 - y
1 + y
3(1 + e)214 ln 2 1 - y
1 + y2
=
-41 - y1 + y
1 + 1 - y1 + y
2 alogax = x
= -41 - y
1 + y1 + y + 1 - y
1 + y 2
= -41 - y
1 + y4
(1 + y)2
= -(1 - y)(1 + y)
= y 2 - 1 [Shown]
d2ydx2
= 2ydydx
Since dydx
< 0, d2ydx2
< 0 if y > 0 and d2ydx2
> 0 if y < 0 [Shown]
(c) limx→∞ 1 - e2x
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 of y = 1 - e2x
1 + e2x is as shown below.
y
xO
1
−1
Chap-08-FWS.indd 15 10/19/2012 10:37:46 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term16
22 (a)
k cm
6 cm
(2k + 6) cm
Q
CRSD
x cm
BA
P
∆CQR and ∆CBS are similar triangles.
k cm
B
Q
CRS
x cm
[(2k + 6) − 6] cm
Thus, RCSC
= QR
BS
RC(2k + 6) - 6
= xk
RC2k
= xk
RC = 2x
Thus, DR = DC - RC
= 2k + 6 - 2x
(b) Area of PQRD,
L = DR × QR
L = (2k + 6 - 2x)(x)
L = (2k + 6)x - 2x2 [Shown]
(c) When L has a stationary value,
dLdx
= 0
2k + 6 - 4x = 0
4x = 2k + 6
x = 2k + 64
x = 2(k + 3)4
x = k + 32
d2Ldx2
= -4 (negative)
Thus, L has a maximum value.
Chap-08-FWS.indd 16 10/19/2012 10:37:47 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 17
Hence, the maximum value of L is
L = (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 ∆QMC, sin x = MCr
MC = r sin x
∴ AC = 2MC
= 2r sin x
r cmr c
m
O
MCA
B
x
x x
r cm
In ∆OMC, cos x = OMr
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 2x
L = 12
(2r 2 sin x + r 2 sin 2x)
L = r2
2 (2 sin x + sin 2x) [Shown]
dLdr
= r2
2 (2 cos x + 2 cos 2x)
Chap-08-FWS.indd 17 10/19/2012 10:37:48 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term18
When L has a stationary value,
dLdx
= 0
r2
2 (2 cos x + 2 cos 2x) = 0
cos x + cos 2x = 0 cos x + 2 cos2 x - 1 = 0 2 cos2 x + cos x - 1 = 0 (2 cos x - 1)(cos x + 1) = 0
cos x = 12
or cos x = -1
x = π3
x = π (not accepted)
d2Ldx2
= r2
2 (-2 sin x - 4 sin 2x)
When x = π3
,
d2Ldx2
= r 2
2 -2 sin π3
- 4 sin 2π3
= -2.60r 2 (< 0)
Hence, L is a maximum.
Lmax
= r 2
2 2 sin
π3
+ sin 2π3
= r2
2 2 × 3
2 + 3
2 = 3 3
4 r2 [Shown]
24 In ∆ORQ, cos a = ORr
OR = r cos a QM = MP = OR = r cos a
r cmr cm
M
R O
PQ
a a
In ∆ORQ, sin a = QRr
QR = r sin a Therefore, the perimeter of ORQP,
y = OR + RQ + QM + MP + PO y = r cos a + r sin a + r cos a + r cos a + r y = r + r sin a + 3r cos a y = r(1 + sin a + 3 cos a) [Shown]
Chap-08-FWS.indd 18 10/19/2012 10:37:48 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 19
dyda
= r(cos a - 3 sin a)
When y has a stationary value,
dydr
= 0
r(cos a - 3 sin a) = 0 cos a - 3 sin a = 0 cos a = 3 sin a
13
= sin acos a
tan a = 13
a = tan-1 13 rad [Shown]
d2yda2
= r(-sin a - 3 cos a)
Since sin a > 0 and cos a > 0, d2yda2
< 0.
Thus, y is a maximum.
ymax
= r(1 + sin a + 3 cos a)
ymax
= r1 + 1
10 + 3 × 3
10 y
max = r1 + 10
10 + 3 × 3 10
10 y
max = (1 + 10 )r [Shown]
√10
3
1
a
sin a = 1
10, cos a =
3
10
25 (a) dVdt
= Change in volume
Change in time
= 12
3
- 13
24
= - 7
192 m3 hour-1
(b) V = x3
dVdx
= 3x2
dxdV
= 1
3x2
dxdt
= dxdV
× dVdt
= 1
3x2 × - 7
192 = 1
3(0.7)2 × - 7
192 = -0.0248 m hour-1
∴ Rate of decrease = 0.0248 m hour-1
Chap-08-FWS.indd 19 10/19/2012 10:37:49 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term20
26 AB = x2 + 32
= (x2 + 9)12
d(AB)dx
= 12
(x2 + 9)-1
2 (2x)
= xx2 + 9
d(AB)
dt =
d(AB)dx
× dxdt
= xx2 + 9
× 2
= 442 + 9
× 2
= 1.6 units s-1
27 (a) (x + 2 − r ) cm
r cm
r cm
N
O
P
Q
x cm
R(x + 2) cm
∆RNO and ∆RQP are similar triangles.
Thus, NO QP
= NR QR
rx
= x + 2 - rx + 2
r(x + 2) = x(x + 2 - r) rx + 2r = x2 + 2x - rx 2rx + 2r = x2 + 2x r(2x + 2) = x2 + 2x
r = x2 + 2x
2x + 2 [Shown]
(b) r = x2 + 2x
2x + 2
drdx
= (2x + 2)(2x + 2) - (x2 + 2x)(2)
(2x + 2)2
drdx
= 4x2 + 8x + 4 - 2x2 - 4x(2x + 2)2
drdx
= 2x2 + 4x + 4(2x + 2)2
drdx
= 2(x2 + 2x + 2)
[2(x + 1)]2
drdx
= x2 + 2x + 22(x + 1)2
Chap-08-FWS.indd 20 10/19/2012 10:37:49 AM
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Fully Worked Solution 21
dxdt
= dxdr
× drdt
= 2(x + 1)2
x2 + 2x + 2 × (-0.4)
= 2(4 + 1)2
42 + 2(4) + 2 × (-0.4)
= 5026
× -0.4
= -0.769 cm s-1
(c) δrδx
≈ drdx
δr ≈ drdx
× δx
= x2 + 2x + 22(x + 1)2 × (4.002 - 4)
= 42 + 2(4) + 22(4 + 1)2 × 0.002
= 0.00104 cm
28 y = xex + 1
dydx
= x ex + 1 + ex + 1 …
= ex + 1(x + 1)
δy ≈ dydx
× δ x
= ex + 1(x + 1) d x x changes from 1 to 1.01. So, δ x = 1.01 - 1.
ynew
= yoriginal
+ δ y
1.01e2.01 = 1(e1 + 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 dydx
when x = 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 = cos x
x dydx
+ y(1) = -sin x
x dydx
+ y = -sin x
Chap-08-FWS.indd 21 10/19/2012 10:37:49 AM
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ACE AHEAD Mathematics (T) Second Term22
x d2ydx2
+ dydx
(1) + dydx
= -cos x
x d2ydx2
+ 2 dydx
= -xy
x d2ydx2 + 2
dydx
+ xy = 0 [Shown]
30 y = x ln (x + 1)
dydx
= x 1x + 1 + ln (x + 1)(1)
= xx + 1
+ ln (x + 1)
δ y = dydx
× δ x
= ( xx + 1
+ ln (x + 1)) δ x
ynew
= yoriginal
+ δ y x changes from 1 to 1.01. So, δ x = 1.01 - 1.
1.01 ln (1.01 + 1) = 1 ln (1 + 1) + 3 11 + 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 dydx
when x = 1.
1.01(ln 2.01) = 0.70508 ∴ ln 2.01 = 0.698
31 (a) f (t) = 4ekt - 14ekt + 1
f (0) = 4e0 - 14e0 + 1
= 35
(b) f ′(t) = (4ekt + 1)(4kekt) - (4ekt - 1)(4kekt)
(4ekt + 1)2
f ′(t) = (16ke2kt + 4kekt - 16ke2kt + 4kekt)
(4ekt + 1)2
f ′(t) = 8kekt
(4ekt + 1)2
Since k is a positive integer, f ′(t) > 0.
(c) LHS = k{1 - [f (t)]2}
= k {1 - 34ekt - 14ekt + 1
2} = k {(4ekt + 1)2 - (4ekt - 1)2
(4ekt + 1)2 }
Chap-08-FWS.indd 22 10/19/2012 10:37:50 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 23
= k {16e2kt + 8ekt + 1 - (16e2kt - 8ekt + 1)(4ekt + 1)2 }
= 16kekt
(4ekt + 1)2
= 23 8kekt
(4ekt + 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→∞
4ekt
ekt - 1
ekt
4ekt
ekt + 1ekt
= limt→∞
4 - 1ekt
4 + 1ekt
= 4 - 04 + 0
= 1
(e) lim t→∞
4ekt - 14ekt + 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 - 14ekt + 1 = 0
4ekt - 1 = 0
ekt = 14
kt = ln 14
t = 1k ln 1
4
Chap-08-FWS.indd 23 10/19/2012 10:37:50 AM
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ACE AHEAD Mathematics (T) Second Term24
t
f(t )
O
f(t ) =4e(kt) − 1
4e(kt) + 1
35
−1
1
41—1
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
dydx
= (1 + x2)(1) - x(2x)
(1 + x2)2
dydx
= 1 - x2
(1 + x2)2
dydx
= 1 - x2
xy
2
dydx
= (1 - x2) y2
x2
x2 dydx
= (1 - x2) y2 [Shown]
33 y = sin x - cos xsin x + cos x
(sin x + cos x)y = sin x - cos x
(sin x + cos x) dydx
+ y (cos x - sin x) = cos x + sin x
(sin x + cos x)dydx
- 1 + y(cos x - sin x) = 0
(sin x + cos x)dydx
- 1 - y(sin x - cos x) = 0
dydx
- 1 - ysin x - cos xsin x + cos x = 0
dydx
- 1 - y(y) = 0
Chap-08-FWS.indd 24 10/19/2012 10:37:50 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 25
dydx
- 1 - y2 = 0
d2ydx2
- 2ydydx
= 0
d2ydx2 = 2y
dydx
[Shown]
34 y = x3
x2 - 1
dydx
= (x2 - 1)(3x2) - x3(2x)
(x2 - 1)2
= 3x4 - 3x2 - 2x4
(x2 - 1)2
= x4 - 3x2
(x2 - 1)2
= x2(x2 - 3)(x2 - 1)2
d2ydx2 =
(x2 - 1)2(4x3 - 6x) - (x4 - 3x2)(2)(x2 - 1)1(2x)(x2 - 1)4
= (x2 - 1)2(2x)(2x2 - 3) - 4x(x4 - 3x2)(x2 - 1)
(x2 - 1)4
= 2x(x2 - 1)[(x2 - 1)(2x2 - 3) - 2(x4 - 3x2)]
(x2 - 1)4
= 2x[2x4 - 5x2 + 3 - 2x4 + 6x2](x2 - 1)3
= 2x(x2 + 3)(x2 - 1)3
d3ydx3 = (x
2 - 1)3(6x2 + 6) - (2x3 + 6x)(3)(x2 - 1)2(2x)(x2 - 1)6
= 6(x2 - 1)3(x2 + 1) - 6x(2x3 + 6x)(x2 - 1)2
(x2 - 1)6
= 6(x2 - 1)2[(x2 - 1)(x2 + 1) - x(2x3 - 6x)]
(x2 - 1)6
= 6(x4 - 1 - 2x4 - 6x2)(x2 - 1)4
= 6(-x4 - 6x2 - 1)(x2 - 1)4
When dydx
= 0
x2(x2 - 3) = 0
x = 0 or ± 3
When x = 0, y = 0 and d2ydx2 =
2(0)(02 + 3)(02 - 1)3
= 0
Chap-08-FWS.indd 25 10/19/2012 10:37:50 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
ACE AHEAD Mathematics (T) Second Term26
When x = 0, d3ydx3 =
6(-04 - 6(0)2 - 1)(02 - 1)4
= -6.
Since d3ydx3 ≠ 0, then (0, 0) is a point of reflextion.
When x = 3, y = 3 3
3 - 1
= - 3 3
2 and
d2ydx2 = 2 3(3 + 3)
(3 - 1)3
= 32 3.
Since d2ydx2 > 0, then 3, 3
2 3 is a minimum point.
When x = - 3, y = (- 3)3
3 - 1
= - 3 3
2 and
d2ydx2 =
2(- 3)(3 + 3)(3 - 1)3
= -32 3.
Since d2ydx2 < 0, then - 3, -3
2 3 is a maximum point.
When the denominator of y = x3
x2 - 1 is 0, x2 - 1 = 0 ⇒ x = ±1
Hence, x = -1 and x = 1 are asymptotes.
The graph of y = x3
x2 - 1 is as shown below.
y
√3 , 3√23 ��
√3 , 3√23 ��
1−1 Ox
−−
Chap-08-FWS.indd 26 10/19/2012 10:37:51 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 27
x3 = k (x2 - 1)
x3
x2 - 1 = k
By sketching the straight lines y = k on the above graph and as k varies, we obtain the following results.
Value of k Number of real roots
k > 32 3 3
k = 32 3 2
-32 3 < k < 3
2 3 1
k = -32 3 2
k < -32 3 3
35 y = xx2 - 1
dydx
= (x2 - 1)(1) - x(2x)
(x2 - 1)2
= -x2 - 1(x2 - 1)2
= -(x2 + 1)(x2 - 1)2
(that is <0)
Since dydx
<0 for all real values of x, then the gradient of the curve is always decreasing.
d2ydx2 = -(x2 - 1)2(2x) + (x2 + 1)(2)(x2 - 1)(2x)
(x2 - 1)4
= -2x(x2 - 1)[x2 - 1 - 2(x2 + 1)]
(x2 - 1)4
= -2x(-x2 - 3)(x2 - 1)3
= 2x(x2 + 3)(x2 - 1)3
When d2ydx2 = 0,
2x(x2 + 3)(x2 - 1)3
= 0
x = 0
When x = 0, y = 002 - 1
= 0
Chap-08-FWS.indd 27 10/19/2012 10:37:51 AM
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ACE AHEAD Mathematics (T) Second Term28
d3ydx3 = (x
2 - 1)3(6x2 + 6) - (2x)(x2 + 3)(3)(x2 - 1)2(2x)(x2 - 1)6
= 6(x2 - 1)3(x2 + 1) - (12x2)(x2 + 3)(x2 - 1)2
(x2 - 1)6
= 6(x2 - 1)2[(x2 - 1)(x2 + 1) - (2x2)(x2 + 3)](x2 - 1)6
= 6(x2 - 1)2(x4 - 1 - 2x4 - 6x2)(x2 - 1)6
= 6(-x4 - 1 - 6x2)(x2 - 1)4
When x = 0, d3ydx3 =
6[-04 - 1 - 6(0)2](02 - 1)4
= -6 (that is ≠ 0)
Since d2ydx2 = 0 and
d3ydx3 ≠ 0 when x = 0, then (0, 0) is the point of inflexion.
When the curve concaves upwards,
d2ydx2 > 0
2x(x2 + 3)(x2 - 1)3
> 0
2x(x2 + 3)
[(x + 1)(x - 1)]3 > 0
2x(x2 + 3)(x + 1)3(x - 1)3
> 0
−
+
+
++
+
x
x > 0
− − +(x − 1)3 > 0
+
−
−
−
+ + +x2 + 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 = xx2 - 1
is as shown below.
Ox
y
−1 1
Chap-08-FWS.indd 28 10/19/2012 10:37:52 AM
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
Fully Worked Solution 29
36 (a) x = t - 2t y = 2t + 1
t
dxdt
= 1 + 2t2
dydt
= 2 - 1t2
dydx
=
dydtdxdt
= 2 - 1
t2
1 + 2t2
= 2t2 - 1t2 + 2
t2 + 2 2
2t2 - 1 2t2 + 4 - 5
∴ dydx
= 2 - 5t2 + 2
[Shown]
Let m = dydx
m = 2 - 5t2 + 2
(m - 2) = - 5t2 + 2
(m - 2)(t2 + 2) = -5
mt2 + 2m - 2t2 - 4 = -5
(m - 2)t2 = -1 - 2m
t2 = -1 - 2mm - 2
t2 = 1 + 2m2 - m
t2 > 0
1 + 2m2 - m
> 0
+
−
+
−+
+
x
1 + 2m > 0
2 − m > 0
+−− −212
Hence, -12
< m < 2, that is, - 12
< dydx
< 2 [Shown]
The question states that t ≠ 0.So, we write t2 > 0 and not t2 ≥ 0.
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ACE AHEAD Mathematics (T) Second Term30
(b) dydx
= 13
2 - 5t2 + 2
= 13
2 - 13
= 5t2 + 2
53
= 5t2 + 2
t2 + 2 = 3
t2 = 1
t = ±1
When t = 1, x = 1 - 21
= -1
and
y = 2(1) + 11
= 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).
Chap-08-FWS.indd 30 10/19/2012 10:37:52 AM