MA1102R
Calculus
AY 2013/2014 Sem 2
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
MA1102R Calculus
AY 2013/2014 Sem 2
Version 1: November 19, 2014
Written by
Sherman Yuen
Audited by
Henry Morco
Contributors
–
Question 1
(a) (i) Since x2 and cos x − 1 are both equal to 0 when x = 0, we may use the L’Hˆopital’s Rule. So,
x2
2x
= lim
x→0 cos x − 1
x→0 − sin x
if the latter limit exists. Similarly, since 2x and − sin x are both equal to 0 when x = 0, we may
apply the L’Hˆ
opital’s Rule again.
lim
2
2
2x
= lim
=
= −2
x→0 − cos x
x→0 − sin x
−1
lim
Therefore we have
x2
= −2
x→0 cos x − 1
lim
(ii) We may use the substitution y = 1/x.
2
1
3 y + 5 sin 2y
3 + 5y 2 sin 2y
(3x2 + 5) sin x2
lim
= lim
= lim
x→∞
5x + 3
5y + 3y 2
y→0+
y→0+
5 y1 + 3
Since both the numerator and denominator equal to 0 when y = 0, we may use the L’Hˆopital’s
Rule. So,
3 + 5y 2 sin 2y
10y sin 2y + (3 + 5y 2 )(2 cos 2y)
0 + (3 + 0)(2)
6
lim
=
lim
=
=
2
+
+
5y + 3y
5 + 6y
5+0
5
y→0
y→0
Therefore we have
(3x2 + 5) sin x2
6
=
x→∞
5x + 3
5
lim
(iii)
lim
x→0
1 + 2 x + 3x
3
1
x
ln
= lim e
1
1+2x +3x x
3
x→0
1
= lim e x
ln
1+2x +3x
3
x→0
=e
limx→0
1
x
ln
1+2x +3x
3
The limits are the same as
e is a continuous function everywhere. We may now apply L’Hˆopital’s
x
x
Rule because ln 1+2 3+3 and x are both equal to 0 when x = 0. We have
2x ln 2+3x ln 3
√
1
1 + 2 x + 3x
2x ln 2 + 3x ln 3
ln 2 + ln 3
1
3
3
lim ln
= lim
= lim
=
=
ln
6
=
ln
6
x +3x 1+2
x→0 x
x→0
x→0 1 + 2x + 3x
3
3
3
3
Therefore,
lim
x→0
1 + 2 x + 3x
3
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1
x
=e
limx→0
1
x
ln
Page 1 of 8
1+2x +3x
3
= eln
√
3
6
=
√
3
6
NUS Mathematics Society
MA1102R
Calculus
AY 2013/2014 Sem 2
(b)
2
2
x − 2x
x − 6
x − 2x − 3x − 6 x2 − 5x − 6 (x + 1)(x − 6) =
= |x + 1| x + 2 − 3 = x+2 =
x + 2
x+2
x+2
Whenever |x + 1| < 12 , that is, − 21 < x + 1 <
1
2
13
(i) − 32 < x < − 12 ⇒ − 15
⇒
2 <x−6<− 2
(ii) − 32 < x < − 21 ⇒
1
2
<x+2<
3
2
⇒
1
2
or − 32 < x < − 12 , we must have:
13
2
< |x − 6| <
< |x + 2| <
3
2
⇒
15
2
1
2
⇒ |x − 6| <
15
2
< |x + 2|
Let > 0 be given. Choose δ = min{ 12 , 15
}. Then
2
x − 2x
x − 6
< |x + 1| 15/2 = |x + 1| × 15 < × 15 = 0 < |x + 1| < δ ⇒ − 3 = |x + 1| x+2
x + 2
1/2
15
Question 2
(a) Let g(x) = |f (x)| and let g 0 (0) = L. We have
g(x) − g(0)
g(x) − 0
g(x)
= lim
= lim
x→0
x→0 x − 0
x→0 x
x−0
L = lim
By taking limits for the positive side, we get
g(x) g(x)
|g(x)|
g(x) L = lim
= lim
= lim = lim
= |L|
x x→0+ x x→0+ x
x→0+ |x|
x→0+
By taking limits for the negative side, we get
g(x) g(x)
|g(x)|
g(x) L = lim
= lim
= lim − = − lim
= −|L|
x x→0− x
x→0− −|x|
x→0−
x→0− x
So we have −|L| = |L|, which means L = 0. So g 0 (0) = 0. Now we want to show that f 0 (0) = 0.
Let > 0 be given. Then there exists a δ > 0 such that
g(x)
− 0 < 0 < |x − 0| < δ ⇒ x
|f (x)| <
⇒ x f (x) <
⇒ x f (x) − 0
⇒ − 0 < x−0
f (x) − f (0)
⇒ − 0 < x−0
This shows that f 0 (0) = 0.
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Calculus
AY 2013/2014 Sem 2
(b) Since f is continuous at 0, we have
f (0) = lim f (x) = lim
x→0
x→0
f (x)
f (x)
× x = lim
× lim x = 2 × 0 = 0
x→0
x→0
x
x
By using the definition of the derivative,
f (x) − f (0)
f (x) − 0
f (x)
= lim
= lim
=2
x→0
x→0
x→0 x
x−0
x
f 0 (0) = lim
(c)
Z
5x
2
Z
5x
2
sin(t )dt =
cos x
Z
0
sin(t )dt +
0
2
Z
Z
2
cos x
sin(t )dt −
sin(t )dt =
cos x
5x
0
sin(t2 )dt
0
By using the Fundamental Theorem of Calculus, we have
Z 5x
Z 5x
Z cos x
d
d
d
2
2
sin(t )dt =
sin(t )dt −
sin(t2 )dt = 5 sin((5x)2 ) − sin(cos2 x)(− sin x)
dx cos x
dx 0
dx 0
So,
d
dx
Z
5x
sin(t2 )dt = 5 sin(25x2 ) + sin x sin(cos2 x)
cos x
Question 3
(a) Since f is differentiable at 1, the function must be continuous at 1.
πx π f (1) = lim f (x) = lim x cos
= lim 1 cos
=0
2
2
x→1−
x→1−
x→1−
So we must have f (1) = 0. Since 0 = f (1) = a(1)2 + b = a + b, we have b = −a.
The limit of the difference quotient must also exist at 1.
f 0 (1) = lim
x→1+
ax2 + b − 0
ax2 + b
f (x) − f (1)
= lim
= lim
x−1
x−1
x→1+
x→1+ x − 1
We may substitute b = −a into the equation to obtain
f 0 (1) = lim
x→1+
ax2 − a
a(x + 1)(x − 1)
= lim
= lim a(x + 1) = 2a
+
x−1
x−1
x→1
x→1+
By looking at the left-hand limit, we get
πx
x cos πx
−
0
x
cos
f (x) − f (1)
2
2
f (1) = lim
= lim
= lim
x−1
x−1
x−1
x→1−
x→1−
x→1−
0
We will use L’Hˆ
opital’s Rule since both the numerator and denominator are equal to 0
π
πx
π π
π cos πx
+
x(−
sin
π
0
2
2
2 )
f (1) = lim
= cos
− sin
=−
−
1
2
2
2
2
x→1
So we obtain 2a = − π2 , giving us a = − π4 , b = π4 .
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MA1102R
Calculus
AY 2013/2014 Sem 2
(b) We have f (2) = (2 − 2)g(2) = 0. So,
f 0 (x) = lim
x→2
(x − 2)g(x) − 0
f (x) − f (2)
= lim
= lim g(x) = g(2) = 1
x→2
x→2
x−2
x−2
since g is continuous at 2.
(c)
Z
π
4
√
ln( 1 + tan x)dx =
Z
0
π
4
0
r
Z π r
4
π
π
ln
1 + tan ( − t) (−1)dt =
1 + tan ( − t) dt
ln
4
4
0
We use the trigonometric addition formula, tan(A − B) =
1 + tan (
tan(A)−tan(B)
1+tan(A) tan(B) ,
to get
tan π4 − tan t
π
1 − tan t
1 + tan t + 1 − tan t
2
=1+
− t) = 1 +
=
=
π
4
1 + tan 4 tan t
1 + 1 · tan t
1 + tan t
1 + tan t
This gives
π
4
Z
I:=
0
π
4
Z
=
0
π
4
Z
=
0
π
4
Z
=
0
π
4
Z
=
√
ln( 1 + tan x)dx
r
π
ln
1 + tan ( − t) dt
4
!
r
2
ln
dt
1 + tan t
!
√
2
ln √
dt
1 + tan t
√
√
ln 2 − ln( 1 + tan t)dt
0
π
4
Z
=
√
Z
ln 2dt −
0
=
√
ln( 1 + tan t)dt
0
π
4
Z
π
4
√
ln 2dt − I
0
where we have made the remarkable observation that the last integral is also exactly equal to I!
Hence we must have
π
Z π √
4
4
1
π ln 2
2I =
ln 2dt =
t ln 2 =
2
8
0
0
π
R
√
ln 2
Hence 04 ln( 1 + tan x)dx = I = π 16
.
Question 4
(a) (i)
Z
1
00
Z
x(1 − x)f (x)dx =
0
1
00
Z
xf (x)dx −
0
1
x2 f 00 (x)dx
0
For the first integral, we can integrate by parts to get
Z 1
Z 1
0 1
00
xf (x)dx = xf (x) 0 −
f 0 (x)dx = (1 · f 0 (1) − 0 · f 0 (0)) − (f (1) − f (0)) = f 0 (1)
0
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Calculus
AY 2013/2014 Sem 2
For the second integral, we may integrate by parts repeatedly on the given integral
Z 1
f (x)dx
1=
0
Z 1
1
xf 0 (x)dx
= [xf (x)]0 −
0
!
1 Z 1
1 2 0
1 2 00
= (1 · f (1) − 0 · f (0)) −
x f (x) −
x f (x)dx
2
0 2
0
Z 1
1
1 2 00
= (0 − 0) − ( f 0 (1) − 0 · f 0 (0)) −
x f (x)dx
2
0 2
Z 1
0
f (1)
1 2 00
x f (x)dx −
=
2
0 2
R1
So we have 0 x2 f 00 (x)dx = 2 + f 0 (1). Therefore,
Z
1
x(1 − x)f 00 (x)dx =
0
Z
1
xf 00 (x)dx −
0
Z
1
x2 f 00 (x)dx = f 0 (1) − (2 + f 0 (1)) = −2
0
(ii) It suffices to find a c ∈ [0, 1] such that f 00 (c) = −12. Define g(x) = −6x2 + 6x − f (x). Note
that g(x) is twice differentiable such that its second derivative is continuous. Let G(x) be an
anti-derivative of g(x). Then, by the Fundamental Theorem of Calculus,
Z
1
G(1) − G(0) =
g(x)dx
0
Z
1
−6x2 + 6x − f (x)dx
0
Z 1
3
2 1
= −2x + 3x |0 −
f (x)dx
=
0
= (−2 + 3) − 1
=0
This means that G(0) = G(1). By the Mean Value Theorem, there exists d ∈ (0, 1) such that
g(d) = G0 (d) =
G(1) − G(0)
=0
1−0
Now, by direct calculations, we also have g(0) = g(1) = 0. So by Mean Value Theorem again,
there exists a ∈ (0, d) and b ∈ (d, 1) such that
g 0 (a) =
g(d) − g(0)
= 0,
d−0
g 0 (b) =
g(1) − g(d)
=0
1−d
And by the Mean Value Theorem one more time, there exists c ∈ (a, b) such that
g 00 (c) =
g 0 (b) − g 0 (a)
=0
b−a
Now, since g(x) = −6x2 + 6x − f (x), we have g 0 (x) = −12x + 6 − f 0 (x) and g 00 (x) = −12 − f 00 (x).
So this gives
0 = g 00 (c) = −12 − f 00 (c)
⇒
f 00 (c) = −12
This completes the proof.
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Calculus
AY 2013/2014 Sem 2
(b) Let F (x) be an anti-derivative of f (x) and G(x) be an anti-derivative of F (x) (so G0 (x) = F (x)
and F 0 (x) = f (x)). Integrating by parts, we get
Z
1
0=
0
1 Z
xf (x)dx = xF (x) −
1
0
0
1
F (x)dx = 1F (1) − 0F (0) − G(x) = F (1) − (G(1) − G(0))
0
This gives G(1) − G(0) = F (1). By the Mean Value Theorem, there exists c ∈ (0, 1) such that
F (c) = G0 (c) =
F (1)
G(1) − G(0)
=
= F (1)
1−0
1
Also, we have
1
Z
0=
0
1
f (x)dx = F (x) = F (1) − F (0)
0
So altogether we obtain F (0) = F (c) = F (1) for some c ∈ (0, 1). By Rolle’s Theorem, there exists
a ∈ (0, c) and b ∈ (c, 1) such that F 0 (a) = 0, F 0 (b) = 0. So f (a) = f (b) = 0.
(c)
Z
Volume of solid =
4
√
4
Z
2
π( x) dx =
0
0
4
1 2 πxdx = πx = 8π
2
0
Question 5
(a)
dy
dv
=v+x
dx
dx
So, by substituting this into the equation, we get
dy
+ x2 + y 2
dx dv
2
+ x2 + (vx)2
= (x − x(vx)) v + x
dx
dv
2
= x (1 − v) v + x
+ x2 + v 2 x2
dx
dv
= x2 (1 − v) v + x
+ 1 + v2
dx
0 = (x2 − xy)
But x 6= 0. So we may divide x2 from both sides.
dv
+ 1 + v2
0 = (1 − v) v + x
dx
dv
dv
= v − v2 + x
− vx
+ 1 + v2
dx
dx
dv
dv
=v+x
− vx
+1
dx
dx
dv
= x(1 − v)
+1+v
dx
This gives
1
=
x
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v−1
v+1
dv
=
dx
1−
Page 6 of 8
2
v+1
dv
dx
NUS Mathematics Society
MA1102R
Calculus
AY 2013/2014 Sem 2
Integrating both sides with respect to x, we get
Z
Z 2
1
1−
dv
dx =
x
v+1
ln x = v − 2 ln (v + 1) + C
y
y
ln x = − 2 ln
+1 +C
x
x
Substituting x = 1, y = 0, we get 0 = 0 − 2 ln 1 + C = C. So
y
y
ln x = − 2 ln
+1
x
x
(b) An integrating factor of the first order linear differential equation is
e
R
−(1+ x3 )dx
= e−x−3 ln x = e−x eln x
−3
= x−3 e−x
So we have
3
dy
− (1 + )y = x + 2
dx
x
dy
3
x−3 e−x
− x−3 e−x (1 + )y = x−3 e−x (x + 2)
dx
x
d
−3 −x
x e y = x−3 e−x (x + 2)
dx
Integrating both sides with respect to x, we get
Z
x−3 e−x y = x−3 e−x (x + 2)dx
To integrate the RHS, we make the observation that x−3 e−x (x + 2) =
Quotient Rule, looks like some form of the derivative of
e−x
.
x2
e−x (x2 +2x)
(x2 )2
which, by using
Indeed,
d e−x
−x2 (e−x ) − 2x(e−x )
=
= −x−3 e−x (x + 2)
dx x2
(x2 )2
Therefore, we have
x
−3 −x
e
Z
y=−
−x−3 e−x (x + 2)dx
e−x
+C
x2
y = −x + Cx3 ex
x−3 e−x y = −
where C is an arbitrary constant. Substitute the initial value and we get C = 1. Therefore we have
y = x3 e x − x
END OF SOLUTIONS
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MA1102R
Calculus
AY 2013/2014 Sem 2
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