Name:
Instructor:
Math 121 - Sample Midterm Exam (2005)
Problem
Points
1 - 10
50
11
25
12
25
13
25
14
25
15
25
16
25
TOTAL
200
SCORE
Instructions: You will be given 2 hours for this exam. Part A has 10 questions, Part B has 6
problems. The problems are worth a total of 200 points, with each problem weighted as indicated.
Your test should have 7 pages. Write your name at the top of each page. Do not detach this sheet
from the test booklet.
Do all your work in this booklet and circle your final answers and write them in the appropriate
boxes. You are allowed to use a calculator for all parts of the test unless you are instructed to find
answers exactly. Calculator answers must be correct to three decimal places. You are not allowed
to borrow or interchange calculators during the test.
In Part A, no partial credit will be assigned. You do not need to justify your work in this part
of the test. Select the most appropriate answer. Keep in mind that if you use your calculator you
may only get an approximate value, while the choices given in some problems may list the exact
value.
In Part B, partial credit might be given, and answers without sufficient justification might
receive no credit. If you use your calculator, make sure to explain precisely how you used it.
Part A: Multiple Choice Answers. After you are done with Part A you must copy the letters
corresponding to your answers in the space below. Your answers in the space below will be used in
grading Part A of the test; so make sure you copy them correctly.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
correct (5 points each)
Part A Total
1
Sample Midterm Exam page 2
Name:
Part A - Multiple Choice Examination
(10×5=50 points) Select only one answer for each problem. Copy your answers to the cover page!
1. The function f (x) =
sin x + ln x
is continuous on
(x − 2)2
(A) (−∞, ∞)
(D) (−∞, 2) ∪ (2, ∞)
(B) (0, ∞)
(E) (0, 2) ∪ (2, ∞)
(C) (−∞, 0)
(F) None of the above is true.
Answer (E). The functions sin x, ln x, and (x − 2) 2 are all continuous on their own domains. So
f (x) is continuous on its domain. The domain of f (x) is x > 0 (so ln x is defined) and x 6= 2 (so
(x − 2)2 6= 0). In interval form, it is just (0, 2) ∪ (2, ∞).
x3 − x
=
x→1 x − 1
2. The limit lim
(A) 1
(B) 2
(C) −∞
(D) ∞
(E) 0
(F) None of the above is true.
x3 − x
x(x2 − 1)
x(x + 1)(x − 1)
= lim
= lim
= lim x(x + 1) = 1(1 + 1) = 2.
x→1 x − 1
x→1
x→1
x→1
x−1
x−1
Answer (B). lim
3. The limit lim
x→π
(A)
1
π
sin(x + π)
=
x+π
(B) ∞
(C) −∞
(D) 0
(E) 1
(F) None of the above is true.
sin(x + π)
is continuous at x = π,
x+π
sin(x + π)
sin(π + π)
sin(2π)
0
lim
=
=
=
= 0.
x→π
x+π
π+π
2π
2π
Answer (D). Since
4. At x = 0, the function
g(x) =
(A) is not defined.
(D) does not have a limit.

x

 e
1

 sin x
(B) is continuous.
(E) is differentiable
if x < 0
if x = 0
if x > 0
(C) has limit equal to 0.
(F) None of the above is true.
Answer (D). lim− g(x) = e0 = 1, lim+ g(x) = sin 0 = 0. So there is no limit.
x→0
x→0
√
5. Consider the parametric equations x = t, y = 2 ln t, for t > 0. Eliminating t we obtain a
Cartesian equation with y as a function of x of the form:
2
(B) y = e2x (C) y = ln x2 (D) y = ex
(E) 2eln x (F) None of the above is true
√
Answer (A). From x = t we have t = x2 . Substituting into y = 2 ln t, we get
(A) y = 4 ln x
y = 2 ln x2 = 2(2 ln x) = 4 ln x.
2
Sample Midterm Exam page 3
Name:
dy
is
dx
(E) Undefined (F) None of the above is true.
6. If y + x + y 2 = 0, then at the point (0, −1) the value of
(A) 1
(B) −1
(C) 3
(D) 0
Answer (A). First of all, using implict differentiation, y 0 + 1 + 2yy 0 = 0. So y 0 = −
point (0, −1), x = 0, y = −1. So the value of y 0 is −
1
= 1.
1 + 2(−1)
1
. At the
1 + 2y
7. Consider the function f (x) = sin x. Then, the derivative of f at x = π/2 is equal to
sin x − π/2
x − π/2
(D) sin(π/2)
(A) lim
(B) lim
h→0
x→π/2
sin(π/2 + h) − 1
h
(E) 1
Answer (B). f 0 (π/2) = lim
h→0
sin(π/2 + h) − 1
π/2
(F) None of the above is true.
(C) lim
h→0
sin(π/2 + h) − 1
sin(π/2 + h) − sin(π/2)
= lim
.
h
h
h→0
8. If the function f (x) is continuous on the interval [−1, 1] and f (−1) = 2, f (1) = −2, then which
of the following must be true?
(A) −2 ≤ f (x) ≤ 2 for all x in the interval [−1, 1].
(B) f (c) = 0 for some c in the interval (−1, 1).
(C) f (x) > −2 for all x in the interval (−1, 1).
(D) f (x) < 2 for all x in the interval (−1, 1).
(E) f (x) has the derivative for every x in the interval (−1, 1).
(F) None of the above is true.
Answer (B). It is based on the Intermediate Value Theorem.
Use the following data about the functions f and g for problems 9 and 10.
t
0
1
3
5
9. The value of
(A) −9
d
g(3t) at t = 1 is
dt
(B) −3
(C) 1
f (t)
0
1
3
1
f 0 (t)
6
2
0
−1
(D) −1
g(t)
0
5
1
3
g 0 (t)
−1
−2
−3
7
(E) 0
(F) None of the above is true
d
d
0
0
= 3g 0 (3(1)) = 3g 0 (3) = 3(−3) = −9.
Answer (A). g(3t) = g (3t)(3) = 3g (3t). Then g(3t)
dt
dt
t=1
d
(f (t) · g(t)) at t = 5 is
dt
(B) 3
(C) 4
(D) −1.
10. The value of
(A) 5
Answer (C).
Then
(E) 7.
(F) none of the above is true.
d
(f (t) · g(t)) = g(t)f 0 (t) + f (t)g 0 (t).
dt
d
(f (t) · g(t)) = g(5)f 0 (5) + f (5)g 0 (5) = 3(−1) + 1(7) = 4.
dt
t=5
3
Sample Midterm Exam page 4
Name:
Part B - Essay Questions - SHOW YOUR WORK!!
11. (10+10+10=30 points) Consider the function
g(x) = 1 −
3
π
+ .
x x2
(A) Compute the derivative of g.
Since g(x) = 1 − 3x−1 + πx−2 ,
d d
d
d
1 − 3 (x−1 ) + π (x−2 )
1 − 3x−1 + πx−2 =
dx
dx
dx
dx
3
2π
−2
−3
−2
−3
= 0 − 3(−x ) + π(−2x ) = 3x − 2πx = 2 − 3 .
x
x
g 0 (x) =
Answer:
3
2π
− 3
2
x
x
(B) Find all the value(s) of x for which the graph of g(x) has a horizontal tangent line. You have
to give the exact values.
2π
3
We set g 0 (x) = 2 − 3 = 0. By multiplying x3 on both sides
x
x
of the above equation, we have 3x − 2π = 0. So the solution
is x = 2π/3.
Answer:
x = 2π/3
(C) Find the point(s) on the curve of g(x) where the tangent line passes through the point (0, 1).
Show your work.
Suppose that the x-value of the point is a. Then the point is
From the derivative formula the slope at this point is
m1 =
π
3
a, 1 − + 2 .
a a
3
2π
3a − 2π
− 3 =
.
a2
a
a3
On the other hand, since the tangent line passes through both the point
the slope is
m2 =
Clearly m1 = m2 . So we have
1−
3
π
a, 1 − + 2
a a
and (0, 1),
+ aπ2 − 1
−3a + π
.
=
a−0
a3
3
a
3a − 2π
−3a + π
=
.
3
a
a3
By solving this equation we have a = π/2. Hence there is only one point, and the point is (π/2, 1−2/π).
Answer: (π/2, 1 − 2/π)
4
Sample Midterm Exam page 5
Name:
12. (10 points) A continuous and differentiable function f satisfies f (3) = 4 and f 0 (3) = 2. Use
the available data about f to estimate f (3.01).
The linearization of f (x) at x = 3 is
L(x) = f (3) + f 0 (3)(x − 3) = 4 + 2(x − 3) = 2x − 2.
So f (3.01) ≈ L(3.01) = 2(3.01) − 2 = 4.02.
Answer:
4.02
13. (20 points) The volume V and temperature T of a parcel of air rising quickly in the atmosphere
are related via the equation
T V 0.4 = C,
where T is measured in Kelvin degrees, V is measured in cubic meters, and C is an appropriate
fixed constant. Both T and V change as functions of time t as the parcel of air rises, but the
above equation remains true for all time. The Kelvin scale is chosen so the the temperature T
is always positive.
(A) Find a formula for the rate of change in temperature with respect to time as the parcel rises.
Since T and V are the function of time t, and C is a constant, using implicit differentiation we have
d
d
dT
dT
dV
d
(T V 0.4 ) = (C) ⇒ V 0.4
+ T (V 0.4 ) = 0 ⇒ V 0.4
+ T 0.4V −0.6
dt
dt
dt
dt
dt
dt
dT
T dV
From this equation, we have
= −0.4
.
dt
V dt
Answer:
= 0.
dT
T dV
= −0.4
.
dt
V dt
(B) According to this model, if the volume of the parcel of air expands as it rises (that is,
is the temperature increasing or decreasing? Use part (A) to justify your answer.
dV
> 0),
dt
The temperature is decreasing. This is because T > 0 and V > 0, and when dV /dt > 0, from the
formula derived in part (A)
T dV
dT
= −0.4
< 0.
dt
V dt
Since the derivative of T is negative, T is decreasing.
5
Answer:
The temperature is decreasing.
Sample Midterm Exam page 6
Name:
14. (30 points) The curve of the derivative function f 0 (x) is shown in the figure. Suppose f (0) = 0.
On the same set of axes, sketch the graph of function f (x). You need to clearly show the
main characters of the curve, such as the increasing and decreasing properties, concavity, horizontal and vertical asymptotes (if there is any), local maxima and minima,
inflection points.
Local y
Max
Inflection
Point
f(x)
Inflection
Point
f(0)=0
H−asym.−2 1.5
INC
C−up
f ’ (x)
0
DEC
C−down
2
x
Local
Min
INC
C−up
6
Sample Midterm Exam page 7
Name:
15. (10+10+5+5=30 points) A particle moves on a vertical line so that its coordinate at time t is
s(t) = t3 − 12t + 3,
t ≥ 0.
(We assume that the upward direction is the positive direction.)
(A) Find the velocity and acceleration functions.
v(t) = s0 (t) = 3t2 − 12
a(t) = v 0 (t) = 6t.
2
Answer: v(t) = 3t − 12;
a(t) = 6t
(B) When is the particle moving upward and when is it moving downward?
By solving v(t) = 3t2 − 12 = 0, we have t = −2, 2. Since t ≥ 0, we have only one solution t = 2
(v(2) = 0). It is easily seen that v(t) < 0 for 0 < t < 2 and v(t) > 0 for t > 2.
Answer:
Upward when t > 2
Downward when 0 < t < 2
(C) Find the distance that the particle travels in the time interval 0 ≤ t ≤ 3.
Based on the analysis in part (B), the particle is moving downward when 0 < t < 2. The distance
traveled is
|s(2) − s(0)| = (23 − 12(2) + 3) − (03 − 12(0) + 3) = 16.
The particle is moving upward when 2 < t < 3. The distance traveled during this period is
|s(3) − s(2)| = (33 − 12(3) + 3) − (23 − 12(2) + 3) = 7.
Therefore the total distance traveled = 16 + 7 = 23.
Answer:
23
(D) When is the particle speeding up? When is it slowing down?
The particle is speeding up when v(t) > 0 and a(t) > 0, or v(t) < 0 and a(t) < 0. The particle is
slowing down when v(t) > 0 and a(t) < 0, or v(t) < 0 and a(t) > 0.
Because a(t) = 6t > 0 for all t > 0. The particle is speeding up when v(t) > 0, which is t > 2.
The particle is slowing down when v(t) < 0, which is 0 < t < 2.
Answer:
Speed-up when t > 2;
7
Slow-down when 0 < t < 2
Sample Midterm Exam page 8
Name:
16. (15+15=30 points)
(A) For what value(s) of the constant c is the function f (x) continuous on (−∞, ∞)?
f (x) =
(
x + 2 if x < 1
cx2
if x ≥ 1
For any c the function f (x) is always continuous for all x 6= 1. So we only need to choose c such that
f (x) is continuous at x = 1. Because the limit lim f (x) exists if and only if lim f (x) = lim f (x),
x→1
and
lim f (x) =
x→1−
lim f (x) =
x→1+
x→1−
x→1+
lim (x + 2) = (1 + 2) = 3
x→1−
lim (cx2 ) = c(12 ) = c.
x→1+
So only when c = 3, we have lim f (x) = 3. With c = 3 we also have f (1) = c(1 2 ) = 3. Then
x→1
lim f (x) = f (1) = 3. By definition, f (x) is continuous at x = 1 when c = 3.
x→1
Answer:
c=3
(B) A hemisphere-shaped dome of radius 60 ft is to be coated with a layer of rust-proofer before
painting. Use differentials to estimate the amount of rust-proofer needed if the coat is to be 0.01
in thick.
Let V be the volume of the hemisphere with radius r. Then
V =
We have
V0 =
2 3
πr .
3
2
dV
= π(3r 2 ) = 2πr 2 .
dr
3
In this problem, r = 60 and dr = 0.01/12 ft. Then
V 0 (60) = 2π602 = 7200π.
By using the differentials,
∆V ≈ dV = V 0 (60)dr = 7200π(0.01/12) = 6πft3 .
3
Answer: 6πft
8