AY1011 Sem1 - NUS Mathematics Society

MA1102R
Calculus
AY 2010/2011 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Theo Fanuela Prabowo
MA1102R Calculus
AY 2010/2011 Sem 1
Question 1
ε
(a) Given ε > 0. Take δ = min{1, 19
}. Then if 0 < |x + 2| < δ, we have
|x3 + 8| = |x + 2| x2 − 2x + 4
≤ |x + 2| |x|2 + 2|x| + 4
(by triangle inequality)
(∵ |x + 2| < δ ≤ 1)
ε
(∵ |x + 2| < δ ≤ )
19
< 19|x + 2|
< ε.
Hence, by definition, lim x3 = −8.
x→−2
(sin x − a)(cos x − b)
= 5.
x→0
ex − 1
(b) Since f is continuous at x = 0, we have lim f (x) = f (0) ⇔ lim
x→0
Note that
(sin x − a)(cos x − b)
× lim (ex − 1) = 5 × 0 = 0.
x→0
x→0
ex − 1
lim (sin x − a)(cos x − b) = lim
x→0
Since the function (sin x−a)(cos x−b) is continuous on R, we have (sin 0−a)(cos 0−b) = a(1−b) = 0.
Since lim (sin x − a)(cos x − b) = 0 = lim (ex − 1), we may apply L’Hopital’s Rule to obtain
x→0
x→0
(sin x − a)(cos x − b)
cos x(cos x − b) − sin x(sin x − a)
= lim
= 1 − b = 5.
x
x→0
x→0
e −1
ex
lim
Thus, b = −4.
Recall that a(1 − b) = 0. Hence, a = 0.
Question 2
(a)
lim x
1
ln(x3 +1)
x→∞
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ln(x)
= lim exp
x→∞
ln(x3 + 1)
ln(x)
= exp lim
x→∞ ln(x3 + 1)
1/x
= exp lim
(by L’Hopital’s Rule)
x→∞ 3x2 /(x3 + 1)
x3 + 1
= exp lim
x→∞ 3x3
!
1 + x13
1
= exp lim
= exp
.
x→∞
3
3
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NUS Mathematics Society
MA1102R
Calculus
AY 2010/2011 Sem 1
(b) Since
lim
x→0+
2 + e1/x sin x
+
|x|
1 + e4/x
!
= lim
x→0+
2/e4/x + 1/e3/x
1/e4/x + 1
!
+ lim
x→0+
sin x
=0+1=1
x
and
lim
x→0−
2 + e1/x sin x
+
|x|
1 + e4/x
it follows that lim
x→0
!
= lim
x→0−
2 + e1/x sin x
+
|x|
1 + e4/x
2 + e1/x
1 + e4/x
!
+ lim
x→0−
sin x
−
x
=
2+0
− 1 = 1,
1+0
!
= 1.
Question 3
(a) First, we note that f is not defined when x < 0.
88 5/3 11 8/3
x − x .
3
3
440 2/3 88 5/3 88 2/3
x − x = x (5 − x).
f 00 (x) =
9
9
9
A necessary condition for inflection point is f 00 (x) = 0, so that x = 0 (omitted since f is not defined
when x < 0) or x = 5. Observe that f 00(x) > 0 when x ∈ (0, 5) and f 00 (x) < 0 when x ∈ (5, ∞).
Thus, the point (5, f (5)) = 5, 150 · 52/3 is an inflection point.
f 0 (x) =
Remark: According to a more general definition of exponentiation, i.e. z a := exp (a ln |z| + ia arg(z)),
f (x) is actually defined to be a non-real number for negative value of x. Please refer to the definition
used by your lecturer.
(b) It suffices to show that g 0 (x) = 0 for all x ∈ R.
Given any x ∈ R. Note that
g(y) − g(x) g(y) − g(x)
≤ |y − x|.
y − x ≤ |y − x| ⇒ −|y − x| ≤
y−x
Since lim −|y −x| = 0 = lim |y −x|, by the Squeeze Theorem, lim
y→x
y→x
y→x
g(y) − g(x)
= g 0 (x) = 0. Hence,
y−x
g is a constant function.
(c) Let u = x − t. Then
Z 3x
Z −2x
d
d
sin (x − t)2 dt =
− sin(u2 )du
dx 0
dx x
Z x
d
=
sin(u2 )du
dx −2x
Z x
Z −2x
d
d
=
sin(u2 )du −
sin(u2 )du
dx 0
dx 0
Z −2x
d(−2x)
d
2
= sin(x ) −
·
sin(u2 )du
dx
d(−2x) 0
= sin(x2 ) + 2 sin (−2x)2
= sin(x2 ) + 2 sin(4x2 ).
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NUS Mathematics Society
MA1102R
Calculus
AY 2010/2011 Sem 1
Question 4
2
Since the volume of the outside cylinder is 16π ft3 , we have 16π = π r + 21 (h + 1), so that
h = 161 2 − 1.
(r+ 2 )
!
!
16r2
16
2
2
, r > 0.
V (r) = πr
2 − 1 = π
2 − r
r + 12
r + 12
Thus,
1 2
2
− 16r2 · 2 r +
V 0 (r) = π
4
r + 21
3 2πr r + 12 8 − r + 21
=
.
4
r + 12
16 · 2r r +
1
2
!
− 2r
Critical points are attained when V 0 (r) = 0 or V 0 (r) is undefined, i.e. when r = − 21 or r = 0, or
r = 23 . Since r > 0, the only critical point is r = 32 . Note that V 0 (r) > 0 when r ∈ (0, 23 ) and
V 0 (r) < 0 when r ∈ ( 32 , ∞). Thus, V (r) is maximized when r = 23 . Recall that h = 161 2 − 1.
(r+ 2 )
Thus, h = 3. Hence, the radius and the height of the inside cylinder are 1.5 ft and 3 ft respectively.
Question 5
2
x)
(a) Let u = tan(ln x) ⇒ du = sec (ln
dx. Thus,
x
Z p
Z p
sec2 (ln x)
tan(ln x) sec4 (ln x)
dx =
tan(ln x) 1 + tan2 (ln x) ·
dx
x
x
Z
√
=
u(1 + u2 )du
Z =
u1/2 + u5/2 du
2
2
= u3/2 + u7/2 + C
3
7
2
2
= (tan(ln x))3/2 + (tan(ln x))7/2 + C.
3
7
Z
(b) First, we will solve the indefinite integral
3x
dx.
1 + 32x
Let u = 3x ⇒ du = ln 3 · 3x dx. Thus,
Z
Z
3x
1
du
1
1
dx =
=
tan−1 (u) + C =
tan−1 (3x ) + C.
2x
2
1+3
ln 3
1+u
ln 3
ln 3
Therefore
Z
∞
0
3x
dx = lim
t→∞
1 + 32x
Z
0
t
3x
dx
1 + 32x
1
tan−1 (3t ) − tan−1 (1)
t→∞ ln 3
1 π π −
=
ln 3 2
4
π
=
.
4 ln 3
= lim
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Calculus
AY 2010/2011 Sem 1
Question 6
(a)
s
ln 2
Z
2πy
Area =
1+
0
dy
dx
2
ln 2
Z
cosh x
dx = 2π
p
1 + sinh2 xdx
0
ln 2
Z
cosh2 xdx
= 2π
0
ln 2 x
e
Z
= 2π
0
+ e−x
2
2
dx
ln 2 1 2x 1 1 −2x
= 2π
dx
e + + e
4
2 4
0
1 2x 1
1 −2x ln 2
= 2π e + x − e
8
2
8
0
15
=
+ ln 2 π.
16
Z
(b) Note that
Z
1
2
2
2 2
(k(4x − 3x )) − (kx )
V1 = π
1
Z
16x2 − 24x3 + 8x4 dx
0
8 5 1
2 16 3
4
= πk
x − 6x + x
3
5
0
14 2
= πk
15
dx = πk
0
2
and
Z
1
2
x k(4x − 3x ) − kx
V2 = 2π
2
0
Z
dx = 2πk
0
Since V1 is half of V2 , then
14
2
15 πk
1
1
1
4(x − x )dx = 8πk x3 − x4
3
4
2
3
= 31 πk. Since k > 0, it follows that k =
1
0
2
= πk.
3
5
14 .
Question 7
(a) Integrating factor = e
R
cos xdx
= esin x .
dy sin x
·e
+ y cos x · esin x = 2x
dx
d sin x ⇒
e
· y = 2x
dx
Z
Z
⇒ d esin x · y = 2xdx
⇒esin x · y = x2 + C.
Substituting x = π and y = 0 to the last equation, we get C = −π 2 . Hence, esin x · y = x2 − π 2 , or
equivalently
x2 − π 2
y = sin x .
e
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NUS Mathematics Society
MA1102R
Calculus
AY 2010/2011 Sem 1
(b) (i)
dQ
= k(50 − Q)(100 − Q)
Zdt
Z
dQ
⇒
= k dt
(Q − 50)(Q − 100)
Z
Z
dQ
dQ
1
= kt + C
−
⇒
50
Q − 100
Q − 50
1
⇒ (ln |100 − Q| − ln |50 − Q|) = kt + C
50 100 − Q = 50kt + 50C
⇒ ln 50 − Q 100 − Q
⇒
= ±e50C · e50kt
50 − Q
100 − Q
⇒
= A · e50kt
50 − Q
, A = ±e50C .
Plugging in Q = t = 0 to the last equation, we get A = 2. Hence,
100 e50kt − 1
100 − Q
50kt
= 2e
⇒Q=
.
50 − Q
2e50kt − 1
Remark: Here we omit the trivial solutions Q(t) = 50 and Q(t) = 100 since they do not satisfy the
initial condition.
(ii) Let f (Q) = k(50 − Q)(100 − Q) = kQ2 − 150kQ + 5000k. We want to find the value of Q that
minimizes f (Q). Note that f 0 (Q) = 2kQ − 150k. Critical point: f 0 (Q) = 0 ⇒ Q = 75. Observe
that f 0 (Q) < 0 when Q < 75 and f 0 (Q) > 0 when Q > 75. Hence, f (Q) is minimized when Q = 75.
Question 8
Lemma: For the function f defined in the question, we have
Z 1
Z 1 2
x −x
f (x)dx =
f 00 (x)dx.
2
0
0
Proof of Lemma:
Z
0
1
f (x)]10
Z
1
f (x)dx = [x ·
x · f 0 (x)dx
−
0
Z 1
=−
x · f 0 (x)dx
0
Z 1
Z
1 1 0
0
=−
x · f (x)dx +
f (x)dx
2 0
0
Z 1
1
=−
x−
f 0 (x)dx
2
0
2
1 Z 1 2
x −x
x −x
=−
f 0 (x) +
f 00 (x)dt
2
2
0
0
Z 1 2
x −x
=
f 00 (x)dx.
2
0
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MA1102R
Calculus
AY 2010/2011 Sem 1
Proof of Question 8:
Since f 00 is continuous on [0, 1], by the Extreme Value Theorem, it attains minimum and maximum
values. Let m = f 00 (a) and M = f 00 (b) where a, b ∈ [0, 1] be the minimum and maximum values
respectively. Then,
m ≤ f 00 (x) ≤ M
2
2
2
x −x
x −x
x −x
00
≤
f (x) ≤ m
⇒M
2
2
2
Z 1 2
Z 1 2
Z 1 2
x −x
x −x
x −x
⇒M
f 00 (x)dx ≤ m
dx ≤
dx
2
2
2
0
0
0
Z 1 2
M
x −x
m
⇒−
f 00 (x)dx ≤ −
≤
12
2
12
0
Z 1 2
x −x
f 00 (x)dx ≤ M
⇒m ≤ −12
2
0
Z 1
⇒m ≤ −12
f (x)dx ≤ M
0
Z 1
f (x)dx ≤ f 00 (b).
⇒f 00 (a) ≤ −12
∀x ∈ [0, 1]
∀x ∈ [0, 1]
0
Since f 00 is continuous on [0, 1] and a, b ∈ [0, 1] , by the Intermediate Value Theorem, there exists
Z 1
Z 1
1
00
c ∈ [0, 1] such that f (c) = −12
f (x)dx, or equivalently
f (x)dx = − f 00 (c).
12
0
0
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