MATH 101 Quiz #5 (v.M1)
Last Name:
Friday, March 18
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Evaluate lim
h 3n3 + n2
n→∞
n3 + 7
i
+ 7e1/n . Simplify your answer completely.
Answer: 10
Solution:
lim
n→∞
h 3n3 + n2
n3 + 7
1/n
+ 7e
i
3 + n1
3+0
= lim
+ 7 = 10.
+ 7 lim e1/n =
n→∞ 1 + 73
n→∞
1+0
n
Marking scheme: 1 for a correct answer in the box.
Short answer questions—you must show your work
∞
X
1
is convergent or divergent. If it is convern + 25
n=0
gent, find its value. A “calculator-ready” answer is acceptable.
2. 2 marks Determine whether the series
Solution: By the Integral Test, this series converges if and only if the integral
converges. Since
Z ∞
Z R
1
1
2 x=R
dx = lim
dx = lim log x +
R→∞ 0 x + 2
R→∞
5 x=0
x + 25
0
5
2
2
− log
= lim log R +
R→∞
5
5
R∞
0
1
x+ 25
dx
diverges, the series diverges .
Marking scheme: 2 marks for a correct argument showing that the series diverges. 1 mark
for giving the answer “divergent” or “diverges” without a correct justification; alternatively, 1
mark for mentioning the Integral Test but not giving a full argument.
P
P∞ 1
1
It is possible to get full marks using a comparison test such as ∞
n=1 n , but
n=0 n+ 2 >
5
P∞
P
∞
mistakes such as the incorrect inequality n=0 n+1 2 > n=0 n1 earn only 1 mark.
5
∞
X
6
3. 2 marks Determine whether the series
is convergent or divergent. If it is convergent,
7n
n=2
find its value. A “calculator-ready” answer is acceptable.
Solution: This is a geometric series with first term a =
the series converges and takes the value
6
72
and ratio r = 71 . As |r| < 1,
a
6/72
6/72
1
=
=
= .
1−r
1 − 1/7
6/7
7
Marking scheme:
• 1 mark for recognizing that it is a geometric series.
• 1 mark for the correct value.
Long answer question—you must show your work
4. 5 marks Find the function y = f (x) that satisfies
dy
= −xy 3
dx
and
1
f (0) = − .
4
Solution: This is a separable differential equation that we solve in the usual way.
Z
Z
y −2
x2
dy
dy
3
= −xy =⇒
=
−
x
dx
=⇒
=
−
+ C =⇒ y −2 = x2 − 2C.
dx
y3
−2
2
To have y = − 14 when x = 0, we must choose C to obey
1 −2
−
= −2C =⇒ −2C = 16 =⇒ y −2 = x2 − 2C = x2 + 16.
4
So y = f (x) = − √
1
. We need to take the negative square root to have f (0) = − 14 .
+ 16
x2
Marking scheme:
• 1 mark for applying the separable differential equation method.
• 1 mark for a correct general solution to the differential equation.
• 1 mark for setting up the equation that imposes the initial condition.
• 1 mark for finding the correct value of the constant, for their form of the general solution.
• 1 mark for solving for y, including taking the correct square root.