MATH 101 Quiz #5 (v.M2)
Last Name:
Friday, March 18
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Evaluate lim
n→∞
h 6n2 + 5n
n2 + 1
i
+ 3 cos(1/n2 ) . Simplify your answer completely.
Answer: 9
Solution:
lim
n→∞
i
6 + n5
6+0
+ 3 cos(1/n ) = lim
+ 3 = 9.
lim cos(1/n2 ) =
1 + 3 n→∞
2
n→∞
n +1
1+0
1 + n2
h 6n2 + 5n
2
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 + 34
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
3 x=R
dx = lim
dx = lim log x +
R→∞ 0 x + 3
R→∞
4 x=0
x + 34
0
4
3
3
− log
= lim log R +
R→∞
4
4
R∞
0
1
x+ 34
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+ 3 >
4
P∞
P
∞
mistakes such as the incorrect inequality n=0 n+1 3 > n=0 n1 earn only 1 mark.
4
∞
X
4
3. 2 marks Determine whether the series
is convergent or divergent. If it is convergent,
5n
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
4
52
and ratio r = 51 . As |r| < 1,
a
4/52
4/52
1
=
=
= .
1−r
1 − 1/5
4/5
5
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
y3
=− 2
dx
x
and
f (1) = −1.
Solution: This is a separable differential equation that we solve in the usual way.
Z
Z
dy
y3
y −2
−3
−2
= − 2 =⇒
y dy = − x dx =⇒
= x−1 + C =⇒ y −2 = −2x−1 − 2C.
dx
x
−2
To have y = −1 when x = 1, we must choose C to obey
(−1)−2 = −2 − 2C =⇒ −2C = 3 =⇒ y −2 = −2x−1 − 2C = −2x−1 + 3.
1
So y = f (x) = − p
. We need to take the negative square root to have f (1) = −1.
3 − 2/x
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.