1
University of Sharjah
College of Sciences
Department of Mathematics
Date: 24 /06 /2018
Time: 8:00 – 9:15
Semester: Summer 2017/2018
Midterm Exam: Mathematics for Engineers (1440262)
Name:
Instructor:
I. D. #:
Section #:
Question
Possible
Mark
#
Mark
Obtained
1
4
2
4
3
4
 Notes and books are NOT ALLOWED.
4
2
 Attempt ALL questions.
5
7
 Attempt each question in the space provided. If
more space is needed, use the back of the page.
6
5
 Dictionaries of any type are NOT ALLOWED.
7
4
Total
30
Instructions to Candidates
 JUSTIFY ALL ANSWERS.
 Programmable calculators and watches are NOT
ALLOWED.
 Time allowed: 75 min.
 If you have questions, ask only the course
instructor(s).
 The exam paper consists of 5 Pages (including the
cover page)
2
𝑥 + 2𝑦 = 0
Q1:[4 Marks]Use Cramer's rule to solve the system { 2𝑥 + 𝑦 − 𝑧 = 1 for 𝒚 only
2𝑥 + 𝑦 − 2𝑧 = −1
Solution:
1 2 0 𝑥
0
𝑦
The matrix form of the system is [2 1 −1] [ ] = [ 1 ]
2 1 −2 𝑧
−1
1 2 0
1+ 2− 0+
1 −1
2
● 𝐷 = |2 1 −1| = | 2
| − 2|
1 −1| = 1 |
1 −2
2
2 1 −2
2
1 −2
1+
● 𝐷2 = | 2
2
0− 0 +
1
1 −1| = 1 |
−1
−1 −2
−1
| = −1 + 4 = 3
−2
−1
| = −3
−2
⟹𝑦=
𝐷2 −3
=
= −1
𝐷
3
Q2: [4 Marks] Use inverses to solve the system
𝑥 − 2𝑦 = 5
{
2𝑥 + 5𝑦 = −8
Solution:
1 −2 𝑥
5
] [𝑦 ] = [ ]
2 5
−8
1
1 5 2
5 2
𝑨−1 =
[
]= [
]
det 𝐴 −2 1
9 −2 1
𝑥
1 5
1
9
1
5
2 5
∴ The solution is [𝑦] = 𝑨−1 [ ] = [
][ ] = [
]=[ ]
9 −2 1 −8
9 −18
−2
−8
The matrix form of the system is [
3
𝑥−𝑦+𝑧 =1
Q3: [4 Marks] For which values of 𝑎 and 𝑏 will the system { 𝑦 + 2𝑧 = 1
𝑥 − 3𝑦 + 𝑎𝑧 = 𝑏
have no solution? exactly one solution? infinitely many solutions?
Solution:
The augmented matrix is
1 −1 1 1 −𝑅1+𝑅3 1 −1
[0 1 2 1] ⇒
[0 1
1 −3 𝑎 𝑏
0 −2
1
1
1
1 2𝑅2 +𝑅3 1 −1
[0 1
2
1 ]
2
1 ]⇒
0 0 𝑎+3 𝑏+1
𝑎−1 𝑏−1
⟹No solution if 𝑎 = −3, 𝑏 ≠ −1
Infinitely many solutions if 𝑎 = −3, 𝑏 = −1
Unique solution if 𝑎 ≠ −3
Q4: [2 Marks] Find the sum of the series ∑∞
𝑛=0
22𝑛
7𝑛+1
Solution:
(22 )𝑛
∑∞
𝑛=0 7𝑛 71
=
1 4 𝑛
∑∞
𝑛=0 7 (7)
=
1
7
4
1−
7
=
1
3
4
Q5: [7 Marks] Determine whether the series is convergent or divergent
1 𝑛
(1) ∑∞
𝑛=1 [1 − ]
𝑛
1𝑛
𝑎𝑛 = [1 − ] ⇒
undetermined form 1∞ , so we consider
as
𝑛→∞
𝑛
1 𝑛
ln[1− ]
𝑛
1
𝑛
ln 𝑎𝑛 = ln [1 − ] = 𝑛 ln [1 − ] =
𝑛
[
⇒′
lim ln 𝑎𝑛 = lim
L Hopital Rule 𝑛→∞
1
𝑛
1
𝑛→∞
⇒
as 𝑛→∞
1
1
1 ][𝑛2 ]
1−
𝑛
−1
𝑛2
undetermined form
= lim
−1
1
𝑛→∞ 1−𝑛
0
0
= −1
⟹ we deduce that lim 𝑎𝑛 = 𝑒 −1
𝑛→∞
Since lim 𝑎𝑛 ≠ 0, then, by the divergence test, the series diverges.
𝑛→∞
−𝑛
(2) ∑∞
𝑛=1 𝑛𝑒
2
2
The function 𝑓(𝑥) = 𝑥𝑒 −𝑥 is continuous and positive on the interval [1, ∞).
2
𝑓 ′ (𝑥) = 𝑒 −𝑥 [1 − 2𝑥 2 ] is negative on the interval [1, ∞) ⟹ 𝑓 is decreasing on the
interval [1, ∞). Now, we evaluate the improper integral
∞
∞
2
∫1 𝑓(𝑥)𝑑𝑥 = ∫1 𝑥𝑒 −𝑥 𝑑𝑥
[use the substitution 𝑢 = −𝑥 2 ]
1 −𝑥 2 ∞
1
1
1
1
= [− 𝑒 ] = [− 𝑒 −∞ ] − [− 𝑒 −1 ] = 0 + 𝑒 −1 = 𝑒 −1
2
2
2
2
2
1
Since the improper integral converges then by the integral test the series converges
5
Q6: [5 Marks] Determine whether the series
(−1)𝑛
∑∞
𝑛=1 3
√𝑛2 +𝑛
is absolutely convergent, conditionally convergent, or divergent.
Solution:
(−1)𝑛
● ∑∞
𝑛=1 | 3
√𝑛2 +𝑛
| = ∑∞
𝑛=1 3
1
√𝑛2 +𝑛
∞
⟹ we take ∑∞
𝑛=1 𝑏𝑛 = ∑𝑛=1 3
⟹
1
√𝑛2
lim
𝑎𝑛
𝑛⟶∞ 𝑏𝑛
1
3 2
√𝑛 +𝑛
1
𝑛⟶∞ 3
√𝑛2
= lim
the leading term in the numerator
=
the leading term in the denominator
1
3
√𝑛2
and find
3
= lim
√𝑛 2
3
𝑛⟶∞ √𝑛2 +𝑛
3
= lim √
𝑛⟶∞
𝑛2
𝑛2 +𝑛
3
= √1 = 1
∞
Since the limit is a positive finite number and ∑∞
𝑛=1 𝑏𝑛 = ∑𝑛=1
1
2
is a divergent
𝑛3
2
1
3
√𝑛2 +2𝑛+3
𝑝-series because 𝑝 = ≤ 1, then, by the Limit Comparison Test, ∑∞
𝑛=1 3
is divergent ⟹ ∑∞
𝑛=1 3
(−1)𝑛
√𝑛2 +2𝑛+3
is not absolutely convergent
● But, for the series itself ∑∞
𝑛=1 3
(−1)𝑛
√𝑛2 +2𝑛+3
i) {𝑏𝑛 } is decreasing
,we find that 𝑏𝑛 =
1
3
√𝑛2 +2𝑛+3
satisfies
ii) lim 𝑏𝑛 = 0
𝑛⟶∞
⟹ by the Alternating Series Test, the series converges.
The series is convergent but not absolutely convergent⟹ conditionally convergent
Q7: [4 Marks] Find the radius of convergence for the power series ∑∞
𝑛=1
(𝑥−1)𝑛
𝑛2 5𝑛
Solution:
|𝑎𝑛+1 |
|𝑥 − 1|𝑛+1
𝑛2 5𝑛
1
𝑛 2
1
|𝑥
|𝑥 − 1|
lim
= lim
×
=
−
1|
lim
=
(
)
𝑛⟶∞ |𝑎𝑛 |
𝑛⟶∞ (𝑛 + 1)2 5𝑛+1
𝑛⟶∞ 𝑛 + 1
|𝑥 − 1|𝑛 5
5
By the Ratio Test, the series is absolutely convergent when
1
5
|𝑥 − 1| < 1 ⟹ |𝑥 − 1| < 5 ⟹ The radius of convergence is 𝑅 = 5