MATH 151 Honors, FALL SEMESTER 2011
FINAL EXAMINATION - SOLUTIONS
Name (print):
Instructor’s name: Yasskin
Signature:
Section No:
Part 1 – Multiple Choice (15 questions, 4 points each, No Calculators)
Write your name and section number on the ScanTron form.
Mark your responses on the ScanTron form and on the exam itself
1.
Evaluate
a.
b.
c.
x
2
x2
x
5
2
Correct Choice
6
x
The limit
a.
b.
c.
d.
e.
f
f
f
f
f
lim
2
42
h 0
32 where f x
2 where f x
4 where f x
2 where f x
2 where f x
SOLUTION: f x
3.
6x
4
0
3
SOLUTION: lim x
2.
2
3
d.
e.
3
lim x
h
x2
x
2
6x
4
h 3
h
32
4x 3
4x 3
x3
12x 2
x4
42
Find the line tangent to y
a.
b.
c.
d.
e.
xx 2 x 3
x 2 x 2
lim
x
2
lim
x
2
xx 3
x 2
2
2
2
3
2
5
2
can be interpreted as which of the following?
Correct Choice
h
3
So x
sinh x at x
2 and f x
4x 3 and f 2
32.
1. Its y-intercept is
0
e
2
e
1
e
1
2e
Correct Choice
SOLUTION: f x
sinh x
f x
Tan Line:
y f1 f 1 x 1
e1
b y0
sinh 1 cosh 1 1
cosh x
f1
sinh 1
sinh 1 cosh 1 x 1
e 1
e1 e 1
e 1
2
2
f 1
cosh 1
1
4.
x2
x2
The function f x
shape:
a.
5x
4x
6 has a vertical asymptote at x
4
b.
2. Near x
c.
2, the graph has the
d.
Correct Choice
x2
x2
1 ”
SOLUTION: f x
lim
x 2
5.
x
x
3
2
“
5x
4x
6
4
0
x
2 x 3
x 2 2
x 3
lim
x 2
x 2
Find the absolute minimum value of f x
NOTE:
a.
b.
c.
d.
e.
3
2
0. 806
4
0. 785
x
x
“
0
sin x
2
3
2
1 ”
x on the interval 0,
2
6
1. 570
1
2
3
2
4
1
Correct Choice
2
2
3
2
4
1
2
2
1
0
x
6
2
6
Check critical points in the interval and endpoints:
1
f0
sin
6
2
3
f
sin
0. 806 0. 785
2
2
6
4
2
4
1
f
sin 5
. 5 1. 570
1. 07
6
2
2
2
SOLUTION: f x
6.
cos x
x
3
2
0
absolute minimum
If $1000 is invested at 2% annual interest compounded continuously, how long will it take for the
principal to reach $1000e $2718. 28?
a.
b.
c.
d.
e.
5 years
10 years
25 years
50 years
100 years
SOLUTION: P
2
.
Correct Choice
P 0 e rt
1000e
1000e 0.02 t
e
e 0.02 t
1
0. 02 t
t
50
A rectangular field is surrounded by a fence and divided into
7.
W
3 pens by 2 additional fences parallel to one side. If the total
area is 72 m 2 find the minimum total length of fence.
a. 48
Correct Choice
b. 50
c. 52
d. 60
L
e. 66
SOLUTION: A
4
8.
144
L2
L2
LW
72
36
L
F
6
W
4L
2W
4L
72
6
12
F
144
L
4 6
2 12
144
L2
4
0
48
Let f x be a differentiable function, and suppose f 1
5 and f x
7 for all values of x. Use
the Mean Value Theorem to determine how small f 4 can possibly be.
a.
b.
c.
d.
e.
26
26
Correct Choice
20
35
Not enough information.
SOLUTION: By the MVT, there is a c in 1, 4 where
f4 f1
f c
7
So f 4
f1 74 1
4 1
9.
F
When you prove lim 3x
x 4
a.
b.
c.
d.
e.
7, and pick
5
5
73
26
0, the largest possible
is
Correct Choice
3
2
2
3
SOLUTION: lim 3x
x 4
For all
if 0 |x
0 there is a
0 such that
4|
then | 3x 5 7|
.
Scratch work: | 3x
Proof: Given
if 0
5
0, let
|x
7 means
5
4|
7|
|3x
3
3
.
12|
3|x
4|
|x
4|
or | 3x
5
3
Then
then 3|x
4|
or |3x
12|
7|
.
3
10. When you use a left Riemann sum with 3 intervals of equal length to approximate ln 7
7
1
you discover:
a.
ln 7
b.
ln 7
c.
ln 7
d.
ln 7
e.
ln 7
HINT: Draw a picture.
46
15
46
15
11
6
11
6
23
15
Correct Choice
SOLUTION:
f1
1 dx
x
7
x
2
3
f5
1
3
1 f3
y 1.0
1
0.8
1
5
0.6
From the plot
46
A 1 2 1 2 1 2
5
3
15
The curve is below the rectangles.
0.4
0.2
0.0
0
1
2
3
4
5
6
1 1
3 2
1
3
7
x
/4
11. Evaluate
cos 3x dx
/4
2
a.
b.
c.
3 2
0
2
3
2
d.
e.
Correct Choice
SOLUTION:
/4
cos 3x dx
/4
2
12. If I
sin 3
4
3
/4
sin 3x
3
/4
3
4
sin
3
1
2
2
3 2
2
2 x dx, which of the following is FALSE?
0
a.
b.
c.
d.
e.
2
0
2
2
0
5
0
0
2
2
3 2 x dx
I
3I
32
2
2 x dx
2
2 x dx
2
2 x dx
I
5
2
2
2 x dx
2
2 x
(c) is false.
4
I
I
SOLUTION: (a) 3 f dx
(b) 1
Correct Choice
3 f dx - true
16 on 0, 2 .
So
2
0
1 dx
(d)
2
0
5
2
5
0
0
2
2
2 x dx
- true
2
0
(e)
16 dx. Or 2
0
2
2
0
I
- true
32.- true
2
3
x2
d
dx
13. Compute
x3
sin t 2 dt
sin x 4 2x sin x 6 3x 2
cos x 2 2x cos x 3 3x 2
sin x 4
sin x 6
cos x 2
cos x 3
4
cos x 2x cos x 6 3x
a.
b.
c.
d.
e.
Correct Choice
SOLUTION: Let F t be an antiderivative of sin t 2 .
So d
dx
x2
x3
14. Recall:
Then
x3
sin t 2 dt
d F x2
dx
sin t 2 dt
ex
sinh x
ex
e
4
F x3 .
F x 2 2x
ex
cosh x
2
e
x
x
e
3x
SOLUTION: f x
12 2 12
x
sin x 6 3x 2
Correct Choice
e 3x
ex
sinh x sinh 2x
ex
x
e
4
e
3x
2e x
e
x
e 2x
e
2
2e x
2
cosh x
12 12
x2
4
1
Concave up: 2,
Concave down:
, 2
2, 2
Concave up:
, 2
2,
Concave down: 2, 2
Concave up: 2, 2
Concave down:
, 2
2,
Concave up:
,
Concave down: nowhere
Concave up: nowhere
Concave down:
,
f x
sin x 4 2x
.
2
15. Find the intervals of concavity of the function f x
a.
b.
c.
d.
e.
F x 3 3x 2
sinh x sinh 2x
SOLUTION: cosh x cosh 2x
e 3x
F x2
F x3
x
e
cosh x cosh 2x
sinh 3x
sinh x
cosh 3x
cosh x
cosh x
a.
b.
c.
d.
e.
x2
sin t 2 and
So F t
x2
12 12
x2
3
12
2x2x
2
2x
ex
e
x
e 2x
2
e
2x
2
.
Correct Choice
2x
12
x2
2
2
12
8x 2
12
12
x
x2 2
2 3
12
6x 2 24
12 x 2 3
0
2
Check the signs in each interval:
, 2 :
f
3
0
concave up
2, 2 :
f 0
0
concave down
2, :
f 3
0
concave up
5
Part 2 – Work Out Problems (5 questions. Points indicated. No Calculators)
Solve each problem in the space provided. Show all your work neatly and concisely, and
indicate your final answer clearly. You will be graded, not merely on the final answer, but also on
the quality and correctness of the work leading up to it.
1
12 12 x 2 . (See problem #15.)
Be sure to label all maxima, minima, inflection points and asymptotes.
Be careful with intervals of increase, decrease and concavity.
16. (6 points) Graph the function f x
SOLUTION: increasing: x 0
decreasing: x 0
local and absolute maximum: 0, 1
no minima
Concave up: x
2 or x 2
Concave down: 2 x 2
3
inflection points: 2,
and 2, 3
4
4
no vertical asymptotes,
horizontal asymptotes: y 0 as x
y
1.0
0.5
-5
-4
-3
-2
-1
1
2
3
4
5
x
-0.5
-1.0
17. (8 points) Let g x be the inverse function of f x
SOLUTION: If b
Further f x
6
1
ln x
g 1 then 1
2
1 and f e
x
fb
1 for x
ln x
1 . So b e. So g 1
ln b
1 1
1
e . So g 1
ln e 2 e
0. Find g 1 and g 1 .
e.
1
f e
e.
18. (10 points) The pressure and volume of the gas in an engine cylinder are related by PV 3/2
C
for some constant C. Currently, the pressure is P 15 lb/in and the volume is V 4 in but is
increasing at the rate of dV
2 in 3 /min. Find C and the current value of dP . Is the pressure
dt
dt
currently increasing or decreasing?
2
SOLUTION: C PV 3/2 15 4 3/2 15 8 120
dP V 3/2 0
dP V 3/2
P 3 V 1/2 dV
P 3 V 1/2 dV
2
2
dt
dt
dt
dt
P is decreasing.
3
P 3 dV
V 2 dt
dP
dt
15 3 2
4 2
45
4
19. (6 points) Find a parametric equation for the line perpendicular to the parametric curve
rt
t 2 , t 3 at the point 4, 8 .
SOLUTION: t 2
vt
2t, 3t 2
r tan t
4, 8 t 12, 4
4 12t, 8
v2
4, 12
n
12, 4
4t There is no unique answer.
y
(12 points) A rectangle is inscribed in the upper half of
2
y2
the ellipse x
1 with its base on the x-axis.
16
9
Find the maximum area of such a rectangle.
20.
4
2
-4
-2
2
-2
4
x
-4
SOLUTION: A
A
6 1
1
y
x2
16
3 1
x2
16
x2
16
8
16
2xy with y
6x 1
2
0
3
2
2x
16
2
1 x
16
or
3 1
x 2 . So
16
A
6x 1
Multiply by 1 1
6
0
1
x2
8
0
A
2xy
2 2 2
x
8
x2
16
x2
16
2 2
3
2
12
7