SSEA Math Module
1
Diagnostic Test
August 2012
(9 points)
(a) Compute cot(2π/3) + eln(5) .
Solution.
1
+5
tan(2π/3)
1
= √ +5
− 3
1
= − √ + 5.
3
cot(2π/3) + eln(5) =
(b) Find all values of x that satisfy 2 ≤ x2 − 2 < 7.
Solution.
2 ≤ x2 − 2 < 7 =⇒ 4 ≤ x2 < 9
=⇒ 2 ≤ x < 3 or − 3 < x ≤ −2.
(c) What is f(g(h(x))), where f (x) = x2 + 1, g(x) = 1 − 2x, and h(x) = sin(x)?
Solution.
f (g(h(x))) = f (g(sin(x)))
= f (1 − 2 sin(x))
= (1 − 2 sin(x))2 + 1
= 4 sin2 (x) − 4 sin(x) + 2.
SSEA Math Module
Diagnostic Test
Compute the derivatives of the following functions.
2 (9 points)
(a)
f (x) = 4x3 + 7 +
1
x3
Solution.
f 0 (x) = 12x2 −
3
.
x4
(b)
q
f (x) = tan2 (x) + 3
Solution. We use the chain rule, twice.
0
1
f 0 (x) = p 2
tan2 (x) + 3
2 tan (x) + 3
0
1
2 tan(x) tan(x)
= p 2
2 tan (x) + 3
1
= p 2
2 tan(x) sec2 (x)
2 tan (x) + 3
1
=p 2
tan(x) sec2 (x).
tan (x) + 3
(c)
f (x) = ex ln(x) +
ex
x2 + 1
Solution. We use the product rule and the quotient rule.
f 0 (x) = ex
1
ex (x2 + 1) − ex (2x)
+ ex ln(x) +
.
x
(x2 + 1)2
August 2012
SSEA Math Module
3 (12 points)
Diagnostic Test
August 2012
(a) Compute the limit or explain why it does not exist.
x2 − 9
lim
x→−3 x2 + 2x − 3
Solution.
x2 − 9
(x + 3)(x − 3)
=
lim
x→−3 x2 + 2x − 3
x→−3 (x + 3)(x − 1)
(x − 3)
= lim
x→−3 (x − 1)
−6
=
−4
3
= .
2
lim
(b) Compute the limit or explain why it does not exist.
2 + x + 10x3 − 3x4
x→∞
x2 + 2x4
lim
Solution.
2 + x + 10x3 − 3x4
2/x4 + 1/x3 + 10/x − 3
=
lim
x→∞
x→∞
x2 + 2x4
1/x2 + 2
−3
=
.
2
lim
SSEA Math Module
Diagnostic Test
(c) Compute the limit or explain why it does not exist.
August 2012
lim cos(1/x2 )
x→0
Solution. This limit does not exist. As x approaches 0 from either side, the
function cos(1/x2 ) oscillates faster and faster between -1 and 1. A similar function
is analyzed and graphed on page 98 of Stewart’s book.
(d) Compute the limit or explain why it does not exist.
lim x2 cos(1/x2 )
x→0
Solution. Note that
−x2 ≤ x2 cos(1/x2 ) ≤ x2 .
Since
lim −x2 = 0 = lim x2 ,
x→0
x→0
by the Squeeze Theorem on page 110 of Stewart we have
lim x2 cos(1/x2 ) = 0.
x→0
SSEA Math Module
4 (12 points)
Diagnostic Test
August 2012
(a) If 1200 cm2 of material is available to make a box with a square base and an open
top, find the largest possible volume of the box.
Solution. Let b be the side length of the square base, and let h be the height
of the box. Hence the surface area of the box is b2 + 4bh. Since the surface area
of the box is also 1200 cm2 , we have
1200 = b2 + 4bh =⇒ 1200 − b2 = 4bh
1200 − b2
=⇒
= h.
4b
Hence the volume of the box is
V = b2 h = b2
1200 − b2
b3
= 300b − .
4b
4
To maximize the volume, we take the derivative and set it equal to zero. We
get
0 = V 0 = 300 −
3b2
4
which implies
3b2
= 300 =⇒ b2 = 400 =⇒ b = 20.
4
One can check that this is indeed a maximum by using the second derivative test,
3
for instance. We plub b = 20 into the formula V = 300b − b4 to see that the
largest possible volume of the box is
300(20) −
8000
(20)3
= 6000 −
= 4000 cm3 .
4
4
SSEA Math Module
Diagnostic Test
August 2012
dy
2
(b) Suppose y is a function of x and that dx = x y. If y(0) = −5, find y(2).
Solution. This is a separable differential equation. When y 6= 0 we can write
dy
= x2 dx
y
and integrate to get
ln |y| = x3 /3 + C.
We take the exponential of each side to get
|y| = eC ex
3 /3
=⇒ y = ±eC ex
3 /3
.
We can write this more succinctly as
y = Aex
3 /3
,
where A is an arbitrary constant (note when A = 0 we get the constant solution
y = 0). We plug in the initial condition y(0) = −5 to see
3 /5
−5 = Ae(0)
=⇒ A = −5.
Hence
y = −5ex
3 /3
and
3 /5
y(2) = −5e(2)
= −5e8/3 .
SSEA Math Module
Diagnostic Test
Consider the function
5 (10 points)
August 2012
f (x) = x3 + 3x2 + 1.
(a) (6 points) Sketch the graph of f using calculus techniques. Hint: Where are the
local maxima and minima of f ? Where is f concave up and concave down?
Solution. The first derivative of f is
f 0 (x) = 3x2 + 6x = 3x(x + 2).
So the derivative is zero when x = −2 or 0, the derivative is positive when x < −2
or x > 0, and the derivative is negative when −2 < x < 0. So we have a local
maximum at x = −2 and a local minimum at x = 0.
The second derivative of f is
f 00 (x) = 6x + 6 = 6(x + 1).
Hence function f is concave down when x < −1, the function is concave up when
x > −1, and we have an inflection point at x = −1.
We compute f (−2) = 5, f (0) = 1, and f (−1) = 3. We plot these three points
and then sketch the graph of f using the derivative and concavity information
that we have determined above.
SSEA Math Module
Diagnostic Test
August 2012
(b) (4 points) Estimate f (0.99) by using a linear approximation to f (x) at x = 1.
This is also called a tangent line approximation or a linearization.
Solution. The linear approximation formula is
f (x) ≈ f (a) + f 0 (a)(x − a),
for x near a. See page 241 of Stewart. We apply this formula with x = 0.99 and
a = 1. Note that
f (1) = (1)3 + 3(1)2 + 1 = 5
and
f 0 (1) = 3(1)2 + 6(1) = 9.
We get
f (0.99) ≈ f (1) + f 0 (1)(0.99 − 1)
= 5 + 9(−.01)
= 4.91.
SSEA Math Module
6 (12 points)
Diagnostic Test
August 2012
(a) Evaluate the integral or explain why it diverges.
Z
x3 + 2 dx
Solution.
Z
1
x3 + 2 dx = x4 + 2x + C.
4
(b) Evaluate the integral or explain why it diverges.
Z
x sin(x) dx
Solution. We use the integration by parts formula
Z
Z
u dv = uv − v du.
with u = x and dv = sin(x) dx. Note du = dx and v = − cos(x). We get
Z
Z
x sin(x) dx = x − cos(x) −
− cos(x) dx
Z
= −x cos(x) + cos(x) dx
= −x cos(x) + sin(x) + C.
SSEA Math Module
Diagnostic Test
(c) Evaluate the integral or explain why it diverges.
Z 2
2
xex +3 dx
August 2012
0
Solution. We use the technique of substitution. Let u = x2 + 3. Note that
du = 2x dx. We must also be careful with the limits of integration. As x varies
from 0 to 2, u = x2 + 3 varies from 3 to 7. Hence
Z 2
Z
1 2
2
x2 +3
xe
dx =
2xex +3 dx
2 0
0
Z
1 7 u
e du
=
2 3
1 7
= e u 3
2
1
= (e7 − e3 ).
2
(d) Evaluate the integral or explain why it diverges.
Z 3
1
dx
0 x−1
1
Solution. Since the function x−1
is not continuous at x = 1, this is an improper
integral. We need to split it into two parts.
Z 1
Z 3
Z 3
1
1
1
dx =
dx +
dx.
0 x−1
1 x−1
0 x−1
Let’s analyze the two parts separately. For the first part, we have
Z t
Z 1
1
1
dx = lim−
dx
t→1
0 x−1
0 x−1
t = lim− ln |x − 1|0
t→1
= lim− ln |t − 1| − ln | − 1|
t→1
= lim− ln |t − 1| − e
t→1
= −∞.
Since this first part diverges, we know already that the entire integral
Z 3
1
dx
0 x−1
diverges, and we do not need to analyze the second part at all.
SSEA Math Module
7 (6 points)
Diagnostic Test
August 2012
(a) Determine whether the series is convergent or divergent. If it is convergent, find
the sum.
9 27 81
4−3+ −
+
− ...
4 16 64
Solution. This is a geometric series with ratio −3/4 and first term 4. Since
| − 3/4| < 1, the series is convergent and its sum is
4
4
16
=
= .
1 − (−3/4)
7/4
7
See page 567 of Stewart.
(b) Determine whether the series is convergent or divergent. If it is convergent, find
its sum.
∞
X
n2 − 1
n2
n=1
Solution. Note that
n2 − 1
= lim = lim 1 − 1/n2 = 1 6= 0.
2
n→∞
n→∞
n→∞
n
lim
Since the terms of this series do not tend to zero, this series is divergent. See page
570 of Stewart.