Solutions for Math 113/114 H1 Final 2014 Fall
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Multiple Choices:
V1. DACBBBCCDA
V2. ACDBBBCDAC
1. (15 points) Evaluate the following integrals:
Z
(a) (5 points)
3
|x2 − 3x + 2|dx
0
Solution. Since
(
x2 − 3x + 2 ≥ 0 for x in [0, 1] ∪ [2, 3]
x2 − 3x + 2 ≤ 0 for x in [1, 2],
Z
3
Z0 1
|x2 − 3x + 2|dx
Z
2
Z
3
(x2 − 3x + 2)
(x − 3x + 2) +
(x − 3x + 2)dx −
2
1
0
3
1 3
2
x
x
3 2
3 2
=
−
− x + 2x
− x + 2x
3
2
3
2
0
1
3
3
x
5
3
1
5
11
+
= − (− ) + =
− x2 + 2x
3
2
6
6
6
6
2
Z 1
ex
dx
(b) (5 points)
x
0 e +2
Solution. Subsitute t = ex + 2:
Z 1
Z 2+e
ex
dt
t=ex +2
dx ====x==
x
dt=e dx
t
0 e +2
3
i2+e
= ln |t|
= ln(2 + e) − ln 3
=
2
2
3
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http://www.math.ualberta.ca/˜xichen/math11314f/113H1 FINAL-2014 FALLSOLUTIONS.pdf
1
√
cos( x)
√
dx
(c) (5 points)
x
√
Solution. Substitute t = x:
√
Z
Z
√
cos( x)
t= x
√
dx ====== 2 cos tdt = 2 sin t + C
dx
x
2dt= √
x
√
= 2 sin( x) + C
Z
2. (20 points) Let f (x) =
00
f (x) =
2 + x − x2
x−5
. Given that f 0 (x) =
and
2
(x − 1)
(x − 1)3
14 − 2x
, find each of the following:
(x − 1)4
(a) (2 points) The domain of f and intercepts with x and y axes.
(b) (4 points) Vertical and horizontal asymptotes.
(c) (4 points) The intervals of increase and decrease.
(d) (4 points) The local maximum and minimum values.
(e) (4 points) The intervals of concavity and inflection points.
(f) (2 points) Sketch the graph of f .
Show all your work and justify each answer.
Solution. (a) The domain of f (x) is {x 6= 1} = (−∞, 1) ∪ (1, ∞).
Solving 2 + x − x2 = 0, we obtain x = −1 and x = 2. So y = f (x)
has two x-intercepts: (−1, 0) and (2, 0). And it has one y-intercept
(0, f (0)) = (0, 2).
(b) Since
2 + x − x2
=∞
x→1 (x − 1)2
lim
y = f (x) has a vertical asymptote x = 1. Since
2
+ x1 − 1
2 + x − x2
x2
=
lim
= −1
x→∞ (x − 1)2
x→∞ (1 − 1 )2
x
lim
2
+ x1 − 1
2 + x − x2
x2
=
lim
= −1
x→−∞ (x − 1)2
x→−∞ (1 − 1 )2
x
lim
2
y = f (x) has a horizontal asymptote y = −1.
(c) Since f 0 (x) > 0 for x > 5 or x < 1 and f 0 (x) < 0 for 1 < x < 5,
f (x) is increasing on (−∞, 1) and [5, ∞) and decreasing on (1, 5].
(d) Since f 0 (x) changes from negative to positive at 5, f (x) has a local
minimum f (5) = −9/8 at x = 5.
(e) Since f 00 (x) > 0 for x < 7 and f 00 (x) < 0 for x > 7, f (x) is concave
upward on (−∞, 1) and (1, 7) and concave downward on (7, ∞). It has
an inflection point at (7, −10/9).
(f)
y
2
−1
1
−1
3
2
5
7
(5, − 9
)
8
(7, − 10
)
9
x
3. (10 points) A kite 100 ft above the ground moves horizontally at a
speed of 8 ft/s. At what rate is the angle between the string and the
ground changing when 200 ft of string has been let out?
K
8 ft/s
100
G
x
y
Solution. As in the figure above, K is the position of the kite and G
is the ground end of the string. Let x be the angle between the string
and the ground (in rad), y be the horizontal distance between the end
of the string and the kite (in ft) and t be the time variable (in s). Then
d
d y dx
1 dy
y
⇒
cot x =
⇒ − csc2 x
=
cot x =
100
dt
dt 100
dt
100 dt
2
dx
sin x dy
⇒
=−
dt
100 dt
When the length of the string is |GK| = 200, sin x = 100/200 = 1/2
and
dx
(1/2)2
1
=−
8=− .
dt
100
50
So the angle is decreasing at a rate of 1/50 rad/s.
4. (10 points) You need to fence in a rectangular play zone for children
to fit into a right-triangular plot with sides measuring 4 m and 12 m.
Two sides of the rectangle should be on the sides and one vertex on the
hypotenuse of the triangle. What is the maximum area for this play
zone?
4
A
4−y
C0
x
B0
4
y
C
B
12
Solution. Let x and y be the length and width of the rectangle as in
the figure (in m). Then the area of the rectangle is A = xy and
x
4−y
x
=
⇒y =4−
12
4
3
since ∆ABC and ∆AB 0 C 0 are similar. Therefore,
x
x2
A = x(4 − ) = 4x −
= f (x)
3
3
and it suffices to find the maximum of f (x) on [0, 12]:
• Solve f 0 (x) = 0:
f 0 (x) = 0 ⇒ 4 −
2x
= 0 ⇒ x = 6.
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• f (6) = 12.
• f (0) = f (12) = 0.
• Comparing f (6), f (0) and f (12), we conclude that f (x) takes the
maximum 12 when x = 6.
So the maximum area of the rectangular play zone is 12 m2 when the
play zone has dimension 6m × 2m.
5. (15 points) Do the following:
5
3
x2
(a) (7 points) Evaluate the integral
+ 1 dx using the defini9
0
tion of definite integrals. Use right end points. No marks will be
given if the definition is not used.
n
X
n(n + 1)(2n + 1)
.
Hint:
i2 =
6
i=1
Z
Solution. Using right endpoint approximation, let ∆x = 3/n and
xi = x0 + i∆x = 3i/n. Then
n
X
x2
+ 1 dx = lim
f (xi )∆x
n→∞
9
0
i=1
!
!
2
n
n
n
X
3 1 3i
3 X 2 3X
= lim
+ 1 = lim
i +
1
3
n→∞
n→∞
n
9
n
n
n
i=1
i=1
i=1
3 n(n + 1)(2n + 1)
+3
= lim
n→∞
n3
6
1
1
= lim 1 +
1+
+ 3 = 4.
n→∞
n
2n
Z
3
Z
(b) (2 points) Evaluate the integral
0
3
x2
+ 1 dx using the Fun9
damental Theorem of Calculus.
Solution. Using FTC, we have
Z
0
3
3
3
x2
x
+ 1 dx =
+x
=4
9
27
0
(c) (6 points) Show that the equation x5 + 2x + 1 = 0 has exactly one
solution.
Proof. Let f (x) = x5 + 2x + 1. Clearly, f (x) is differentiable on
(−∞, ∞) and f 0 (x) = 5x4 + 2.
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Since f (−1) = −2 < 0 and f (0) = 1 > 0, f (x0 ) = 0 for some x0
in (−1, 0) by Intermediate Value Theorem. So f (x) = 0 has at
least one solution in (−1, 0).
If f (x) = 0 has two solutions f (x1 ) = f (x2 ) = 0 for some x1 < x2 ,
then
f (x2 ) − f (x1 )
=0
f 0 (x3 ) =
x2 − x 1
for some x3 in (x1 , x2 ) by Mean Value Theorem. But
f 0 (x) = 5x4 + 2 > 0 for all x.
Contradiction. So f (x) = 0 has no more than one solution.
In conclusion, f (x) = 0 has exactly one solution.
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