1. No category

MA1102R
Calculus
AY 2008/2009 Sem 2
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Sean Lim Wei Xinq, Terry Lau Shue Chien
MA1102R Calculus
AY 2008/2009 Sem 2
Question 1
(a) Check that
lim
x→∞
x2 + 2x + 3
sin
4x + 5
6
7x
= lim
y→0
1 + 2y + 3y 2
sin
4y + 5y 2
6y
7
Then by L’Hopital’s rule, we have
(2 + 6y) sin
lim
6y
7
+ 76 cos
6y
7
(1 + 2y + 3y 2 )
=
4 + 10y
y→0
6/7
3
=
4
14
(b) Applying L’Hopital’s rule twice, we have
lim
x→0
(c) Let f (x) :=
1
1
1
2 x +3 x +4 x
3
1
1
− x
x e −1
ex − x − 1
x→0 x(ex − 1)
ex − 1
= lim x
x→0 xe + ex − 1
ex
= lim x
x→0 xe + 2ex
1
1
= lim
=
x→0 x + 2
2
= lim
x
and consider
1
1
1
2x + 3x + 4x
lim ln f (x) = lim x ln
x→∞
x→∞
3
1 1 1
ln 2 x +33x +4 x
= lim
1
x→∞
!
x
Then, by L’Hopital’s rule we have
3
lim
1
1
1
2 x +3 x +4 x
·
1
3
1
1
1
2 x ln 2 · − x12 + 3 x ln 3 · − x12 + 4 x ln 4 · − x12
x→∞
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− x12
1
= lim
x→∞
1
1
2 x ln 2 + 3 x ln 3 + 4 x ln 4
1
1
1
2x + 3x + 4x
ln 2 + ln 3 + ln 4
ln 24
=
=
3
3
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Calculus
AY 2008/2009 Sem 2
Then,
ln 24
3
lim f (n) = lim f (x) = exp
n→∞
x→∞
= 241/3 = 2(3)1/3
Question 2
dx
x
(a) Let u = ln x. Then du =
⇒ eu du = dx. Then we have
Z
1
3
1
Z
u
ue4u du
x (u)(e du) =
0
0
Using Integration by Parts gives us
1
4
(b) Let u =
√
x. Then du =
dx
√
2 x
1
Z
Z 1
1
4u
4u 1
e du
u d(e ) =
(ue )|0 −
4
0
1 4 1 4u 1
=
e − (e )|0
4
4
4
3e + 1
=
16
4u
0
⇒ 2udu = dx. Then we have
1
Z
−1
tan
√
( x) dx =
Z
1
2u tan−1 u du
0
0
Using Integration by Parts gives us
1
Z
−1
2u tan
0
Z
1
tan−1 u d(u2 )
0
Z 1
u2
2
−1
1
= (u tan u)|0 −
du
2
0 1+u
Z 1
π
u2
= −
du
2
4
0 1+u
u du =
Here, let u = tan v. Then du = sec2 vdv. Hence we have
π
−
4
Z
0
1
u2
π
du = −
1 + u2
4
Z
π
−
4
Z
π
4
π
4
π
4
π
2
Z
=
=
=
=
=
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π
4
0
π
4
tan2 v
(sec2 vdv)
sec2 v
tan2 v dv
0
−
π
4
(sec2 v − 1) dv
0
π
− (tan v − v)|04
π
π
− tan( ) +
4
4
−1
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MA1102R
Calculus
AY 2008/2009 Sem 2
(c) We have
Z
4
1
x2 + 4x + 4
dx =
x2 (x2 + 4)
4
(x + 2)2
dx
2 2
1 x (x + 4)
Z 4
2 2 1
dx
=
1+
x
x2 + 4
1
Z
Now let x = 2 tan u. Then dx = 2 sec2 udu. This gives us
Z
1
4
2
1+
x
2
1
dx =
2
x +4
Z
tan−1 2 tan−1 12
tan−1
1
2
Z
1
2
Z
1
=
2
Z
=
=
1
1+
tan u
2
1
(2 sec2 udu)
4 sec2 udu
2
(1 + cot u)2 du
tan−1 12
tan−1 2
(1 + 2 cot u + cot2 u) du
tan−1 12
tan−1 2
(csc2 u + 2 cot u) du
tan−1 12
1
−1 2
= (− cot u + 2 ln | sin u|)|tan
−1 1
tan
2
2
1 1
1
−1
= (− + 2 + 2(ln | sin(tan 2)| − ln | sin(tan−1 )|))
2 2
2
3
1
= + ln | sin(tan−1 2)| − ln | sin(tan−1 )|
4
2
3
2
1
= + ln √
− ln √
4
5
5
3
= + ln 2
4
Question 3
(a) Consider f (x) :=
1
(x2 +1)1/x
=
1
x2 +1
1
x
. Then
ln
ln lim f (x) = lim
x→∞
x→∞
= lim
1
x2 +1
x
(x2 + 1) · − (x22x
+1)2
1
2x
= lim − 2
=0
x→∞ x + 1
x→∞
by L’Hopital’s Rule. Then we have
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Calculus
AY 2008/2009 Sem 2
lim f (n) = lim f (x) = e0 = 1
n→∞
x→∞
Hence, we have
lim (−1)n f (n) 6= 0
n→∞
which shows that the series diverges by the Divergence Test.
(b) Using the Limit Comparison Test and comparing it with
√
lim
n→∞
√
n+1− n−1
n
1
n3/2
= lim
n→∞
1
,
n3/2
we have
√
√ √
n( n + 1 − n − 1)
√
√
√
√
n+1+ n−1
√
= lim n ( n + 1 − n − 1) · √
n→∞
n+1+ n−1
√ n + 1 − (n − 1)
√
= lim n √
n→∞
n+1+ n−1
√
2 n
√
= lim √
n→∞
n+1+ n−1
2
2
q
= lim q
=1
=
n→∞
1+1
1+ 1 + 1− 1
√
n
n
Then we have
∞
X
1
3/2
n
n=1
converges if and only if
∞
X
n=1
(c) Note that
p
(2k − 1)(2k + 1) =
√
n+1−
n
√
n−1
converges
p
√
4k 2 − 1 < 4k 2 = 2k
Observe that
p
√
√
3 · ( 3 · 5) · · · ( (2n − 3)(2n − 1)) · 2n − 1
4 · 6 · 8 · · · (2n + 2)
√
√
3 · 4 · 6 · · · · · (2n − 2) · 2n − 1
<
4 · 6 · 8 · · · (2n + 2)
√
√
3 · 4 · 6 · · · · · (2n − 2) · 2n
<
4 · 6 · 8 · · · (2n + 2)
√
3
=√
2n(2n + 2)
√
3
<√
2n2n
r
3 1
=
8 n 32
1 · 3 · 5 · · · (2n − 1)
=
4 · 6 · 8 · · · (2n + 2)
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√
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Calculus
AY 2008/2009 Sem 2
This is dominated by the p-series where p = 32 , which converges. Therefore
∞
X
1 · 3 · 5 · · · (2n − 1)
n=1
4 · 6 · 8 · · · (2n + 2)
converges.
Question 4
First, we want to find the point of intersection of the two curves.
√
1
1
x = x − 1 ⇒ x2 − x + 1 = 0 ⇒ x = 2
2
4
So the region of integration will be [0,2].
By observations and visualization of the rotation of the shaded area through the given diagram,
the total surface area of solid generated is in fact√the surface of the cone,S1 generated by y = 21 x
and surface of the parabola,S2 generated by y = x − 1.
Using the formula
Z b
p
A(S) = 2π
f (x) 1 + [f 0 (x)]2 dx
a
we evaluate A(S1 ) and A(S2 ).
Z
2
r
1
A(S1 ) = 2π
1 + dx
4
√ 0Z 2
5
=
π
x dx
2
0
√ 2 2
5
x
=
π
2
2 0
√
= 5π
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1
x
2
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MA1102R
Calculus
2√
Z
A(S2 ) = 2π
AY 2008/2009 Sem 2
s
x−1
1+
1
Z
2
1
dx
4(x − 1)
r
1
x − 1 + dx
4
1
Z 2r
3
x − dx
= 2π
4
1
" 3 #2
3 2
2
x−
= 2π
3
4
! 1
√
4
5 5−1
= π
3
8
√
5 5−1
=
π
6
= 2π
Therefore
√
√
6 5+3 3+1
A(S) = A(S1 ) + A(S2 ) =
π
6
Question 5
(a) We have
x+y =1⇒y =1−x
Hence, substituting this into the equation of the curve gives us
1 − x = px2 + qx
px2 + (q + 1)x − 1 = 0
Since the line is tangent to the curve, the discriminant is 0, i.e.
(q + 1)2 − 4(p)(−1) = 0
4p + (q + 1)2 = 0
(b) The curve cuts the x-axis at 0 and − pq , and so the area, A(p, q) is given by
− pq
Z
A(p, q) =
(px2 + qx) dx
0
=
px3 qx2
+
3
2
− q
p
0
q3
= 2
6p
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Calculus
AY 2008/2009 Sem 2
2
Now p = − (q+1)
4 , and so
A(q) =
q3
=
2 2
6 − (q+1)
4
8q 3
3(q + 1)4
Then we have
3(q + 1)4 (24q 2 ) − 8q 3 (12(q + 1)3 ))
9(q + 1)8
24q 2 (3(q + 1) − 4q)
=
9(q + 1)5
2
8q (3(q + 1) − 4q)
=
3(q + 1)5
8q 2 (3 − q)
=
3(q + 1)5
A0 (q) =
Now if A0 (q) = 0, we have q = 0 or q = 3, but q > 0 is given. Then, check that A0 (q) > 0 if q < 3
and A0 (q) < 0 if q > 3. So q = 3 is a maximum point. Then, we have p = −4.
Question 6
(a) First, note that we can f is one to one is important for us so that g = f −1 is in fact well-defined.
We might want to simplify the second part of the integration. Let x = f (y), and thus we have
dx
0
dy = f (y). Note that also x = d ⇒ y = b and x = c ⇒ y = a. Therefore we have
Z
d
Z
d
g(x) dx =
c
g(g −1 (y)) dx
c
Z
b
=
y
a
Z
b
=
dx
dy
dy
yf 0 (y) dy
a
We perform by parts on the first part, then we can evaluate the given integral
Z
b
Z
d
f (x) dx +
a
[xf (x)]ba
g(x) dx
c
Z b
Z
b
xf (x) dx +
yf 0 (y) dy
a
a
Z b
Z b
0
= bf (b) − af (a) −
xf (x) dx +
xf 0 (x) dx
=
−
0
a
a
= bd − ac
(b) Observe that
∞
X
n=0
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xn =
1
1−x
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MA1102R
Calculus
AY 2008/2009 Sem 2
for 0 < x < 1. Differentiating both sides gives us
∞
X
nxn−1 =
n=0
1
(1 − x)2
(1)
Differentiating it again and rearranging the terms (which is possible because the series is convergent) yields
∞
X
n(n − 1)xn−2 =
2
(1 − x)3
(n2 xn−2 − nxn−2 ) =
2
(1 − x)3
n=0
∞
X
n=0
By letting x = 1/3, we have
2
∞
X
n
n
27
9
−
=
3n 3n
4
n=0
∞ 2
X
n
3
n
− n =
n
3
3
4
n=0
By (1), we have
n−1
∞
X
1
1
n
=
3
(1 − 1/3)2
n=0
∞
X
n
3
=
n
3
4
n=0
So we have
∞
X
n2
n=0
3n
=
3 3
3
+ =
4 4
2
Question 7
(a) By Mean Value Theorem for [a, a + δ] for a δ > 0, there exists c ∈ (a, a + δ) such that
f 0 (c) =
f (a + δ) − f (a)
δ
Taking limits on both sides gives us
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AY 2008/2009 Sem 2
f (a + δ) − f (a)
δ→0
δ
lim f 0 (c) = lim
δ→0
Now δ → 0 implies c → a+ , and so we have
lim f 0 (x) = f 0 (a)
x→a+
Similarly, by Mean Value Theorem for [a − δ, a] for a δ > 0, there exists c ∈ (a − δ, a) such that
f 0 (c) =
f (a) − f (a − δ)
δ
Taking limits on both sides gives us
f (a) − f (a − δ)
δ→0
δ
lim f 0 (c) = lim
δ→0
Now δ → 0 implies c → a− , and so we have
lim f 0 (x) = f 0 (a)
x→a−
Then note that
lim f 0 (x) = lim f 0 (x) = L = f 0 (a)
x→a−
x→a+
and this completes the proof.
(b) By the Extreme Value Theorem, there exists c1 , c2 ∈ [a, b] such that c1 is the global min and c2 is
the global max, i.e.
f (c1 ) ≤ f (x) ≤ f (c2 )
for all x ∈ [a, b]. Since f is non-constant, we have f (c1 ) 6= f (c2 ). Now let c = f (c1 ) and d = f (c2 ).
Then the range of f is the closed interval [c, d], as desired.
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