MATH 101 Quiz #5 (v.M3)
Last Name: First Name:
Friday, March 18
Student-No:
Grade:
Section:
Very short answer question
1. 1 mark Evaluate lim n →∞ h
3 n 4 n 4
+ n
+ 10
2
+ 5 e
− 1 /n i
. Simplify your answer completely.
Answer: 8
Solution: lim n →∞ h
3 n 4 n 4
+ n
+ 10
2
+ 5 e
− 1 /n i
= lim n →∞
3 +
1 +
1 n 2
10 n 4
+ 5 lim n →∞ e
− 1 /n
=
3 + 0
1 + 0
+ 5 = 8 .
Marking scheme: 1 for a correct answer in the box.
Short answer questions—you must show your work
∞
2. 2 marks Determine whether the series
X n =0
1 n + 1
2 is convergent or divergent. If it is convergent, find its value. A “calculator-ready” answer is acceptable.
Solution: By the Integral Test, this series converges if and only if the integral
R
∞
0 converges. Since
1 x +
1
2 dx
Z
∞
0
1 x +
1
2 dx = lim
R →∞
Z
R
0
1 x +
1
2 dx = lim
R →∞
= lim
R →∞ log x +
1
2 log R +
1
2 x = R x =0
− log
1
2 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.
It is possible to get full marks using a comparison test such as mistakes such as the incorrect inequality P
∞ n =0
1 n +
1
2
> P
∞ n =0
1 n
P
∞ n =0
1 n +
1
2
> earn only 1 mark.
P
∞ n =1
1 n
, but

∞
X
2
3. 2 marks Determine whether the series is convergent or divergent. If it is convergent,
3 n 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
3
2
2 and ratio r =
1
3
. As | r | < 1, a
1 − r
2 / 3 2
=
1 − 1 / 3
=
2 / 3 2
2 / 3
=
1
3
.
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 dx
= 3 x
2 y
3 and f (0) = −
1
2
.
Solution: This is a separable differential equation that we solve in the usual way.
dy dx
= 3 x
2 y
3
= ⇒
Z dy y 3
=
Z
3 x
2 dx = ⇒ y
− 2
− 2
= x
3
+ C = ⇒ y
− 2
= − 2 x
3 − 2 C.
To have y = − 1
2 when x = 0, we must choose C to obey
−
1
2
− 2
= − 2 C = ⇒ − 2 C = 4 = ⇒ y
− 2
= − 2 x
3 − 2 C = − 2 x
3
+ 4 .
So y = f ( x ) = −
1
√
4 − 2 x 3
. We need to take the negative square root to have f (0) = − 1
2
.
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.
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