MA1102R
Calculus
AY 2009/2010 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
solutions prepared by Boyan,Tay Jun Jie
MA1102R Calculus
AY 2009/2010 Sem 1
Question 1
(a) By L’Hˆopital’s Rule,
limπ
x→ 4
sec2 x − 2 tan x
2 sec2 x tan x − 2 sec2 x
= limπ
1 + cos 4x
−4 sin 4x
x→ 4
2 sec4 x + 4 tan2 x sec2 x − 4 tan x sec2 x
−16 cos 4x
x→ 4
8+8−8
=
−16
1
=−
2
= limπ
(b) −x ≤ x cos √1x ≤ x
Therefore, we have
1
lim −x ≤ lim x cos √ ≤ lim x
x→0
x→0
x x→0
1
0 ≤ lim x cos √ ≤ 0
x→0
x
By Squeeze Theorem,
1
lim x cos √ = 0
x
x→0
Question 2
(a) Given > 0, let δ = 3 , such that |x − 2| < δ implies,
2
2
3x − 3x − 6 3x − x − 4
= 3 |x − 2| < 3δ = − 2 = x+1
x+1
(b)
√
√
dy
((ex + 1) x2 + 2)0 (x − 8)5 − ((x − 8)5 )0 (ex + 1) x2 + 2
=
dx
((x − 8)5 )2
√
√
ex x2 + 2 + (ex + 1) 2√2x
(x − 8)5 − 5(x − 8)4 (ex + 1) x2 + 2
x2 +2
=
(x − 8)10
√
√
ex x2 + 2 + (ex + 1) √xx2 +2 (x − 8) − 5(ex + 1) x2 + 2
=
(x − 8)6
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AY 2009/2010 Sem 1
Question 3
(a) f 0 (x) = 3x2 − 18x + 24.
let f 0 (x) > 0, we obtain that x > 4 or x < 2.
let f 0 (x) < 0, we have 2 < x < 4.
Therefore, f (x) is increasing on (−∞, 2) ∪ (4, ∞), and decreasing on (2, 4).
(b) x ∈ R for f 0 (x), let f 0 (x) = 0, we can find out all the critical points at x = 2, and x = 4.
Since f 0 (x) > 0 on (−∞, 2) ∪ (4, ∞), and f 0 (x) < 0 on (2, 4).
By First Derivative Test, we obtain that f (x) have a local maximum at x = 2, and a local minimum
at x = 4.
(c) f 00 (x) = 6x − 18.
let f 00 (x) > 0, then x > 3.
let f 00 (x) < 0, then x < 3.
Thus, f (x) is concave up on (3, ∞), and concave down on (−∞, 3).
(d) x ∈ R for f 00 (x), let f 00 (x) = 0, we obtain its unique inflection point at x = 3.
f (3) = 33 − 9 × 32 + 24 × 3 − 7 = 11.
Hence, the coordinates of its inflection point is (3, 11).
Question 4
Let x denotes the length of the side facing the main road in meters, f (x) denotes the total cost.
Hence, x > 0,
1200
f (x) = 6x + 3 x +
×2 .
x
7200
f (x) = 9x +
.
x
7200
f 0 (x) = 9 − 2
x
√
2.
let f 0 (x) = 0, we have x = 20 √
√
In addition, f 0 (x) > 0 on (0, 20 2), and f 0 (x) < 0 on√(20 2, ∞).
Hence, f (x) attains its absolute minimum at x = 20 2.
Hence,
√in order to minimize the cost of the fence, the length of the side facing the main road should
be 20 2 meters.
Question 5
(a)
Z
Z
cos x ln(sin x) dx = sin x ln(sin x) −
sin x
1
sin x
cos x dx
Z
= sin x ln(sin x) −
cos x dx
= sin x ln(sin x) − sin x + C where C is a constant
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MA1102R
(b) Let a =
Calculus
√
AY 2009/2010 Sem 1
2 − x, x = 2 − a2 , then a ∈ (0, 1), and dx = −2a da.
Z 2
Z 1
√
x 2 − x dx =
(2 − a2 )a(−2a) da
1
0
Z 1
2a4 − 4a2 da
=
0
2 5 4 3 1
=
a − a
5
3
0
2 4
= −
5 3
14
=−
15
Question 6
(a)
3
Z
s
s=
1+
1
3
Z
s
=
1+
1
3
dy
dx
2
dx
x2
1
− 2
2
2x
2
dx
r
1
1
x4
+ 4 − dx
1+
4
4x
2
1
s
Z 3 2
x
1 2
=
+ 2
dx
2
2x
1
Z 3 2
x
1
=
+ 2 dx
2
2x
1
3
3
x
1
+
=
6
2x 1
Z
=
=4
(b)
Z
2
(2x2 )2 dx
a
4 5 2
=π x
5
a
128π 4π 5
=
−
a
5
5
r 2
Z 2a2
y
2
V2 = π
a −
dy
2
0
2 2a2
2 2a2
y
=π a y 0 −π
4 0
V1 = π
= 2a4 π − a4 π
= a4 π
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Calculus
AY 2009/2010 Sem 1
thus,
V = V1 + V2
4π
128π
= − a5 + a4 π +
5
5
V 0 = −4a4 π + 4a3 π, a ∈ (0, 2), then V 0 = 0 at a = 1.
In addition, V 0 > 0 in (0, 1), V 0 < 0 in (1, 2). By First Derivative Test, for a ∈ (0, 2), V attains its
maximum value at a = 1.
Question 7
(a)
dy
1
+ 2y =
dx
x + x3
dy
2
1
+ y= 2
dx x
x + x4
x
Therefore, p(x) = x2 , the integrating factor is
e
R
p(x) dx
=e
R
2
x
dx
= e2lnx
= x2 .
x2
2
x2
dy
+ yx2 = 2
dx x
x + x4
d 2
1
x y=
dx
1 + x2
R 1
2 dx + C
y = 1+x 2
,
x
arctan x + C
y=
,
x2
C∈R
C∈R
(b)
dP
= 0.0008P (100 − P )
dt
1
dP = dt
0.0008P (100 − P )
Z
Z
1250
dP
=
dt
100P − P 2
Z
1
dP = t + C1
1250
502 − (P − 50)2
1250
P
1250
P
ln
=
ln
= t + C1
100
|P − 100|
100
100 − P
P
e0.08t+C2 =
100 − P
100
P (t) =
1 + C3 e−0.08t
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Calculus
AY 2009/2010 Sem 1
where C1 , C2 , C3 are some real numbers. Since P (0) = 20, we obtain C3 = 4. Hence,
P (t) =
100
1 + 4e−0.08t
Question 8
(a) Assume g(c) = 0 for some c ∈ (a, b)
By Mean Value Theorem, there exist two values m ∈ (a, c), and n ∈ (c, b) and
g(c) − g(a)
=0
c−a
g(b) − g(c)
g 0 (n) =
=0
b−c
g 0 (m) =
By Mean Value Theorem, there exists a value x ∈ (m, n) such that g 00 (x) =
contradicts with g 00 (x) 6= 0. Thus, g(x) 6= 0 for all x ∈ (a, b).
g 0 (n)−g 0 (m)
n−m
= 0 which
(b) Let h(x) = f (x)g 0 (x) − f 0 (x)g(x)
Thus, we obtain that h(a) = 0 and h(b) = 0.
By Mean Value Theorem,
there exists a value c ∈ (a, b) such that h0 (c) = 0.
Hence,
h0 (c) = f (c)g 00 (c) − f 00 (c)g(c) = 0
Since c ∈ (a, b), we have g(c) 6= 0 and g 00 (c) 6= 0,
(c)
we obtain that there exists c ∈ (a, b) such that fg(c)
=
f 00 (c)
g 00 (c) .
Question 9
Consider xg(x) for x ∈ R.
Z
1
Z
x
xf (xt) dt =
f (u) du where u = xt
0
0
Rx
f (u) du
⇒ g(x) = 0
for x ∈ R \ {0}
x
xg(x) =
Hence g(x) is differentiable for all x ∈ R \ {0} and
Rx
xf (x) − 0 f (u) du
f (x) − g(x)
0
g (x) =
=
2
x
x
Now, let limx→0
f (x)
x
for x ∈ R \ {0}.
= M.
f (x)
⇒ lim f (x) = lim
·x=
x→0
x→0 x
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f (x)
lim
x→0 x
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lim x = M · 0 = 0
x→0
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MA1102R
Calculus
Hence, by the continuity of f , f (0) = 0. Furthermore, notice that g(0) =
AY 2009/2010 Sem 1
R1
0
f (0) dt = f (0) = 0.
g(x) − g(0)
x→0
Rx x
f (u) du
= lim 0
x→0
x2
f (x)
= lim
by L’Hˆopital’s rule
x→0 2x
M
=
2
f (x) − g(x)
lim g 0 (x) = lim
x→0
x→0
x
f (x)
g(x)
= lim
− lim
x→0 x
x→0 x
M
M
=M−
=
2
2
⇒ g 0 (0) = lim
Therefore, g 0 is continuous at 0.
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Errata: Calculus
AY 2009/2010 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
ERRATA FOR PAST YEAR PAPER SOLUTIONS
MA1102R Calculus
AY 2009/2010 Sem 1
Compiled November 22, 2014
Written by
Lee Kee Wei
Question 1a
8+8−8
= 12 .
16
Audited by
Henry Morco
The original solution evaluated the limit to be
8+8−8
−16
= − 21 . It should have been
R1
Question 5b The original solution gave the definite intergal to be 0 (2 − a2 )a(−2a)da after the
R0
substitution. The substitued intergal should have been 1 (2 − a2 )a(−2a)da. This would lead the answer
14
14
to be 15
instead of the given − 15
.
Question 6a The original solution evaluated the integral as:
R3 2
3
1 3
have been: 1 x2 + 2x12 dx = [ x6 − 2x
]1 = 14
3 .
R3
1
x2
2
3
1 3
+ 2x12 dx = [ x6 + 2x
]1 = 4. It should
END OF ERRATA
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