AY1112 Sem2 - NUS Mathematics Society

MA1102R
Calculus
AY 2011/2012 Sem 2
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Stefanus Lie
MA1102R Calculus
AY 2011/2012 Sem 2
Question 1
Note that f 0 (x) = −x2 e−x . Then, f is decreasing everywhere and no local maximum or minimum
exists.
Note that f 00 (x) = x(x − 2)e−x . Then, f is concave up in (−∞, 0) and (2, ∞) adn f is concave
down in (0, 2). The inflection points are (0, 2) and (2, 10/e2 )
Question 2
√
√
√
(a) Note that for x > 0, − x3 + x2 + x ≤ x3 + x2 + x · sin πx ≤ x3 + x2 + x. The limit of both
LHS and RHS when x → 0+ is 0. By Squeeze Theorem, the limit must be 0.
(b) Let y = 1/x. Then
x+π
x+e
lim
x→∞
x
x + π = exp lim x · ln x→∞
x+e


ln 1+yπ
1+ye 
= exp  lim
y
y→0+


1+ye π(1+ye)−e(1+yπ)
·
2
1+yπ
(1+ye)

= exp  lim
+
1
y→0
π−e
= exp lim
y→0+ (1 + yπ)(1 + ye)
= ee−π
Question 3
Note that A(−1, 1) and B(2, 4). Let a vertical line through P cuts AB at Q(a, a + 2). The area
2
2
of triangle AP Q is equal to (a+2−a2)(a −1) , by considering P Q as base. In the same way, area of
2
2)
triangle BP Q is (a+2−a2)(4−a
A(a) = 23 (a − a2 + 2).
by considering P Q as base. The area of triangle ABP is therefore
Note that A(a) is a quadratic function, attaining its maximum at a = 1/2, for which P ( 12 , 14 )
Question 4
(a) Using tabular integration, the answer is −x3 cos x + 3x2 sin x + 6x cos x − 6 sin x + C
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MA1102R
Calculus
(b) Let x = 3 tan u, then
dx
du
AY 2011/2012 Sem 2
= 3 sec2 x and
Z
Z
1
1
dx =
cos u du
9
(x2 + 9)3/2
1
= sin u + C
9
x
= √
+C
9 x2 + 9
(c) Let
6x − 2
Ax + B
C
D
= 2
+
+
x4 − 1
x +1
x+1 x−1
Then, multiplying both sides by x4 − 1 and comparing coefficients of power of x, we get the
equations : A + C + D = 0, B − C + D = 0, −A + C + D = 6, and −B − C + D = −2. Solving
this, we get A = −3, B = 1, C = 2, D = 1. Hence,
Z
Z
Z
Z
Z
6x − 2
3
2x
1
2
1
=−
dx +
dx +
dx +
dx
4
2
2
x −1
2
x +1
x +1
x+1
x−1
3
= − ln |x2 + 1| + tan−1 x + 2 ln |x + 1| + ln |x − 1| + C
2
Question 5
(a) The volume is the same if we only consider the upper region revolved about the x-axis. Therefore,
the volume is (using disk method)
Z
π
0
1
x2 2x3 x4
x(1 − x) dx = π
−
+
2
3
4
2
1
= π/12
0
(b) The volume is twice as the volume of the upper region revolved about the y-axis. Therefore, the
volume is (using shell method)
Z
4π
1
Z
xy dx = 4π
0
"
1
3/2
x
0
2x2/5 2x7/2
(1 − x)dx = 4π
−
5
7
#1
= 4π/35
0
Question 6
(a) Consider the Riemann Sum of function f (x) = xp along the interval [0, 1], by dividing into n equal
subintervals and considering the right endpoints. Then
n
X1
1p + 2 p + · · · + np
lim
= lim
f
p+1
n→∞
n→∞
n
n
i=1
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Z 1
i
1
=
f (x)dx =
n
p+1
0
NUS Mathematics Society
MA1102R
Calculus
(b) Let G(t) =
t4
16
√
R
1+u4
du.
u
Then
0
F (x) = x
−1/2
So,
F (x) = x
Thus,
√
1
G(2 x) = √
x
√
00
F 00 (1)
AY 2011/2012 Sem 2
−1/2
· 32x ·
Z
16x2
√
16
1 + 65536x8 x−3/2
−
16x2
2
√
√
= 2 65537 + 0 = 2 65537
1 + u4
du
u
Z
16x2
√
16
1 + u4
du
u
Question 7
(a) Note that
(e−1/x−2 ln |x| )
1
2
dy
+ y( 2 − ) = 1
dx
x
x
Hence,
d(y(e−1/x−2 ln |x| ))
= e−1/x−2 ln |x|
dx
So,
−1/x−2 ln |x|
y(e
Z
)=
e−1/x
dx = e−1/x + C
x2
When x = 1, y = 2. From this, C = 1. So,
y=
e−1/x + 1
e−1/x−2 ln |x|
= x2 (1 + e1/x )
(b) 5 percent of salt enters the tank at the rate of 1 liter per minute means 0.05 liter of salt enters the
tank per minute. Then, 1 liter out of 100 liter of well-stirred mixture leaves the tank per minute
means 0.01 of amount of salt leaves the tank. Thus,
dS
= 0.05 − 0.01S
dt
Note that
dS
= dt
0.05 − 0.01S
Then, −100 ln |0.05 − 0.01S| = t + C or 0.05 − 0.01S = e−0.01(t+C) = D.e−0.01t . Since S = 0 when
t = 0, then D = 0.05. So, S = 5(1 − e−0.01t ).
Question 8
(a) Divide (a, b) into 2012 equal subintervals (x0 , x1 ), (x1 , x2 ), . . . , (x2011 , x2012 ). Here, a = x0 and
b = x2012 . Let also that the length of each subinterval be l. For each i = 0, 1, . . . , 2011, using Mean
)−f (xi )
Value Theorem, there exists ci+1 ∈ (xi , xi+1 ) such that f 0 (ci+1 ) = f (xxi+1
.
i+1 −xi
Now, adding all f 0 (ci+1 ) yields
f 0 (c1 ) + f 0 (c2 ) + · · · f 0 (c2012 ) =
2011
X
i=0
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2011
1X
f (b) − f (a)
f (xi+1 ) − f (xi )
=
f (xi+1 ) − f (xi ) =
=0
xi+1 − xi
l
l
i=0
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MA1102R
Calculus
AY 2011/2012 Sem 2
+···+x2012
(b) Let x = x1 +x22012
. Recall the fact that, since f is concave up, then the tangent line of
f from (x, f (x)) is always lower than the curve of f . Then, for every i = 1, 2, . . . , 2012, f (xi ) ≥
f (x)+f 0 (x)(x−xi ). Summing for all i, then f (x1 )+f (x2 )+· · ·+f (x2012 ) ≥ 2012f (x)+f 0 (x)(2012x−
(x1 + x2 + · · · + x2012 )) = 2012f (x). Q.E.D.
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