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UNIVERSITY OF SOUTHAMPTON MATH1055W1
SEMESTER 2 EXAMINATION 2016/17
MATH1055 Mathematics for Electronic and Electrical Engineering
Duration: 2hrs
Answer all questions. The total number of marks available is 100.
Please WRITE your student number.
Your Student Number:
B1 B2 B3—————————————————
—————————————————
Part A consists of 25 multiple-choice questions. The questions 1-20 are each worth 2marks and questions 21-25 are each worth 3 marks. For each question, exactly one of the5 answers (a)-(e) is correct. (Sometimes that could be “(e) none of the above”.) Find thecorrect answer and mark it on the designated answer sheets AS1/MATH1054-1055/2017and AS2/MATH1054-1055/2017. No answer is worth 0 marks, and wrong answers �0.5marks. You can use a standard University answer booklet for your workings.
Part B consists of 3 questions. Write your answers in the boxes provided on the questionpaper. The blue answer books are for ROUGH WORKING ONLY AND WILL NOT BEMARKED. Write your name in a blue answer book and attach it to this document. You canuse the blank pages at the end of this booklet in case you run out of space in a dedicatedanswer box. The number of the problem and nature of any work on these blank pagesshould be clearly stated.
Formula sheet FS/1054-55/17 will be provided.
Only University approved calculators may be used.
A foreign language word to word R� translation dictionary (paper version) is permittedprovided it contains no notes, additions or annotations.
Page 1 of 18
Copyright 2017 c� University of Southampton Page 1 of 18
2 MATH1055W1
PART AAnswers to section A should be filled in on Answer Sheets
AS1/MATH1054-1055/2017 and AS2/MATH1054-1055/2017. If you do not haveany answer sheets, ask an invigilator for them NOW.
1. [2 marks] The definite integralZ
⇡
0x sin(x) dx equals
(a)⇡
2
, (b)⇡
3
, (c)⇡
4
, (d)⇡
6
,
(e) none of the above.
2. [2 marks] If z = 2 + j then
(a) z̄ � 3z = �4� 4j and
����1
z
���� =1p5
,
(b) z̄ � 3z = �4� 2j and
����1
z
���� =1p5
,
(c) z̄ � 3z = �4� 4j and
����1
z
���� =1
5
,
(d) z̄ � 3z = �4� 2j and
����1
z
���� =1
5
,
(e) none of the above.
3. [2 marks] Euler’s formula states that
(a) ej↵ = cos(↵) + j sin(↵), (b) cos(↵ + j�) = cos(↵) + j sin(�),
(c) sin(↵ + j�) = cos(↵)� j sin(�), (d) ej↵ = sin(↵) + j cos(↵),
(e) none of the above.
4. [2 marks] An integrating factor for the differential equation t3dx
dt+ 2xt2 = sin(t) is
(a) t2, (b) exp�23t
3�, (c) t, (d) exp(2t2), (e) none of the above.
Copyright 2017 c� University of Southampton Page 2 of 18
3 MATH1055W1
5. [2 marks] Which of the following is a particular integral to the equation?
d2x
dt2� 5
dx
dt+ 6x = e3t
(a) e3t, (b) te3t, (c) e2t, (d)1
2
e3t, (e) none of the above?
6. [2 marks] The function f(x) = ex2+a
(a) does not have stationary points,
(b) has a maximum at x = ea,
(c) has a minimum at x = 0,
(d) has a maximum or a minimum at x = 0 depending on the sign of a,
(e) none of the above.
7. [2 marks] The derivative of f(x) =1
cos(x2)is equal to
(a)2x sin(x2)
cos
2(x2)
, (b)2x sin(x2)
cos(x2), (c)
sin(2x)
cos
2(x2)
, (d)sin(2x)
cos(x2),
(e) none of the above.
8. [2 marks] The solution to the differential equationdy
dx= xy satisfying y(0) = 1 is
(a) y = ln(x), (b) y = ln(|x|) + 1, (c) y = ex
2
2 , (d) y = 2ex2+ 1,
(e) none of the above.
9. [2 marks] The determinant of the matrix C =
0
@1 3 0
2 6 4
�1 0 2
1
A is
(a) �36, (b) 36, (c) �12, (d)12, (e) none of the above.
Copyright 2017 c� University of Southampton
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Page 3 of 18
4 MATH1055W1
Answers to section A should be filled in on Answer SheetsAS1/MATH1054-1055/2017 and AS2/MATH1054-1055/2017. If you do not have
any answer sheets, ask an invigilator for them NOW.
10. [2 marks] The eigenvalues of the matrix C =
✓5 3
3 5
◆are
(a) �1 = 5 + 3j and �3 = 5� 3j, (b) �1 = 2 and �2 = 8 , (c) �1 = �2 = 8,
(d)�1 = 1 and �2 = 3,
(e) none of the above.
11. [2 marks] The definite integralZ 1
0
2
3 + 4x2dx is equal to
(a)1p3
arctan
✓2p3
◆, (b)
1p3
arctan
✓1p3
◆,
(c)1
2
arctan
✓1p3
◆, (d)
1
2
arctan
✓2xp3
◆,
(e) none of the above.
12. [2 marks] The indefinite integralZ
(x3 � 1)
2 dx equals
(a)1
3
(x3 � 1)
3+ C , (b)
2x
3
(x3 � 1)
3+ C
(c)x7
7
� x4
2
+ x+ C , (d)1
3
✓x4
4
� x
◆3
+ C ,
(e) none of the above.
13. [2 marks] The value of the double integralZ 1
y=0
Zy
2
x=1(x2y + 1) dx dy
equals to
(a) 5/9, (b) 5/18, (c) 5/6, (d) 7/6,
(e) none of the above.
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5 MATH1055W1
14. [2 marks] The improper integralZ 1
0x�
23 dx is
(a) not defined, (b) equal to 1, (c) equal to 3, (d) equal to1
3
,
(e) none of the above.
15. [2 marks] The derivative of 2x with respect to x is equal to
(a) 2x ln x, (b) 2x ln x+ 2
x, (c) 2x ln 2, (d) 2x�1ln x,
(e) none of the above.
16. [2 marks] The derivative of ln(cosh x) with respect to x is equal to
(a) cosh x, (b) ln(cosh x) + sinh x, (c) tanh x, (d) cosh x+ sinh x,
(e) none of the above.
17. [2 marks] The inverse of the function f(x) = 1� e�x, x 2 R, is
(a) g(x) = � ln(1� x) for x < 1, (b) g(x) = ex + 1 for x 2 R,
(c) g(x) = ln(1 + e�x
) for x 2 R, (d) g(x) = � ln(x) + 1, for x > 0
(e) none of the above.
18. [2 marks] The inverse Laplace transform ofe�2s
s2 + 4
is
(a)1
2
H(t� 2) sin(2(t� 2)), (b)1
2
H(t+ 2) sin(2(t+ 2)),
(c)1
2
H(t+ 2) cos(2(t+ 2)), (d) 2H(t+ 2) cos(2(t+ 2)),
(e) none of the above.
Copyright 2017 c� University of Southampton
TURN OVER
Page 5 of 18
6 MATH1055W1
Answers to section A should be filled in on Answer SheetsAS1/MATH1054-1055/2017 and AS2/MATH1054-1055/2017. If you do not have
any answer sheets, ask an invigilator for them NOW.
19. [2 marks] The partial derivatives@f
@xand
@f
@yof the function
f(x) = sin
2(x+ y) are
(a)@f
@x= 2 cos(x) and
@f
@y= 2 cos(y),
(b)@f
@x= 2x sin(x+ y) cos(x+ y) and
@f
@y= 2y sin(x+ y) cos(x+ y),
(c)@f
@x= 2 sin(x+ y) cos(x+ y) and
@f
@y= 2 sin(x+ y) cos(x+ y),
(d)@f
@x= 2 sin(x+ y) cos(x) and
@f
@y= 2 sin(x+ y) cos(y),
(e) none of the above.
20. [2 marks] If1
2
a0 +1X
n=1
(an
cos(nt) + bn
sin(nt))
is the Fourier series of the periodic function f(t) = t if �⇡ < t < ⇡ andf(t+ 2⇡) = f(t), then, for all n > 0,
(a) an
= 0 and bn
6= 0, (b) an
6= 0 and bn
= 0,
(c) an
= 0 and bn
= 0, (d) an
6= 0 and bn
6= 0,
(e) none of the above.
21. [3 marks] The following improper integralZ ⇡
2
0
cos(x)
(sin(x))13
dx ,
(a) is not defined, (b) is equal to 1, (c) is equal to2
3
, (d) is equal to3
2
,
(e) none of the above.
Copyright 2017 c� University of Southampton Page 6 of 18
7 MATH1055W1
22. [3 marks] For the complex number z = �1�p3j, we have that
(a) z6 = e�14⇡j/3 and z�2= e4⇡j/3, (b) z6 = 64e�14⇡j/3 and z�2
=
14e
2⇡j/3,(c) z6 = 1� 6
p3j and z�2
= 1 + 2
p3j, (d) z6 = 64 and z�2
=
14e
�2⇡j/3,
(e) none of the above.
23. [3 marks] The component of the vector i+ j+ k in the direction of 2i+ 3j� 6k isequal to
(a) 1, (b) �1, (c) �1/7, (d) �3/4,
(e) none of the above.
24. [3 marks] The differential equation
dy
dx=
x cos x
y
that satisfies y = 2 when x = 0 has the solution
(a) y =
p2(x sin(x) + cos(x) + 1), (b) y =
px cos(x) + 2,
(c) y =
px sin(x) + cos(x) + 1, (d) y = 2 cos(x),
(e) none of the above.
25. [3 marks] Consider the plane that is parallel to the vectors i+ k and i+ 2j� 2kand passes through the point (�1,�1, 1). The perpendicular (shortest) distancefrom the point (1, 1, 1) to this plane is equal to
(a)3
5
, (b)3p7
, (c)2p17
, (d)3p11
,
(e) none of the above.
END OF PART A
Copyright 2017 c� University of Southampton
TURN OVER
Page 7 of 18
8 MATH1055W1
PART B(Write your answers in the boxes provided)
1. [Total 15 marks]
(a) [4 marks] Evaluate the determinant of the matrix
C =
0
@1 2 3
0 1 ↵5 6 0
1
A
and state the value of ↵ for which the inverse of C does not exist.
out of 4
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9 MATH1055W1
(b) [7 marks] Find C�1 when ↵ = 4.
out of 7
Copyright 2017 c� University of Southampton
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Page 9 of 18
10 MATH1055W1
(c) [4 marks] Hence, or otherwise, solve the set of linear equations
x+ 2y + 3z = 8
y + 4z = 6
5x+ 6y = 17
out of 4
Copyright 2017 c� University of Southampton Page 10 of 18
11 MATH1055W1
2. [Total 15 marks]
(a) [4 marks] If y = sinh
�1 x, use the exponential definition of sinh(y) to show thaty satisfies the equation
e2y � 2xey � 1 = 0
out of 4
Copyright 2017 c� University of Southampton
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Page 11 of 18
12 MATH1055W1
(b) [6 marks] Using u = ey, rewrite the above equation as a quadratic and hencededuce that
sinh
�1 x = ln
�x+
px2 + 1
�
out of 6
Copyright 2017 c� University of Southampton Page 12 of 18
13 MATH1055W1
(c) [5 marks] By differentiating the result in (b) verify that
d
dx
�sinh
�1 x�=
1px2 + 1
out of 5
Copyright 2017 c� University of Southampton
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14 MATH1055W1
3. [Total 15 marks]
(a) [11 marks] Calculate the integralZ
x2 � 2x� 5
(x+ 3)(x2 + 2x+ 2)
dx
Hint: You can first apply partial fractions
out of
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15 MATH1055W1
out of 11
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16 MATH1055W1
(b) [4 marks] The area bounded by the curve y = x(x� 1), the x-axis, and thelines x = 0 and x = 1, is rotated by the x-axis through one complete revolution.Find the volume of the solid of revolution.
out of 4
Copyright 2017 c� University of Southampton Page 16 of 18
17 MATH1055W1
Extra space if needed (Use blue books for rough working. If you have to usethis space for answers, specify clearly the problem number):
out of
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18 MATH1055W1
Extra space if needed (Use blue books for rough working. If you have to usethis space for answers, specify clearly the problem number):
out of
END OF PAPER
Copyright 2017 c� University of Southampton Page 18 of 18
19 MATH1054W1
B3 (a) Zx
2 � 2x� 5
(x+ 3)(x
2+ 2x+ 2)
dx.
Partial fractions gives
x
2 � 2x� 5
(x+ 3)(x
2+ 2x+ 2)
=
A
x+ 3
+
Bx+ C
x
2+ 2x+ 2
=
A(x
2+ 2x+ 2) + (x+ 3)(Bx+ C)
(x+ 3)(x
2+ 2x+ 2)
[up until here 3 points]
so equating numerators we obtainx
2 � 2x� 5 = A(x
2+ 2x+ 2) + (x+ 3)(Bx+ C). Letting x = �3 we have
10 = A(5) + 0(Bx+ C) ) A = 2. Letting x = 0 we obtain�5 = A(2) + 3C ) C = �3. Any other value of x will do. Choose x = 1, then�6 = A(5) + 4(B +C) ) �6 = 10 + 4B � 12 ) B = �1. The integral is nowZ
2
x+ 3
� x+ 3
x
2+ 2x+ 2
dx
=
Z2
x+ 3
dx�Z
x+ 1
x
2+ 2x+ 2
dx�Z
2
x
2+ 2x+ 2
dx
[until here another 3 points]
(i)
Z1
x+ 3
dx, (ii)
Zx+ 1
x
2+ 2x+ 2
dx, (iii)
Z1
x
2+ 2x+ 2
dx.
[integrals (i) and (ii) 1 point each; integral (iii) 3 points]
(i) ln |x+ 3|+ C . (ii) let u = x
2+ 2x+ 2, then the integral becomes
1
2
Z1
u
du =
1
2
ln |u|+ C =
1
2
ln |x2 + 2x+ 2|+ C
. (iii) partial fractions no good, as denominator doesn’t factorize over R. Letu = x+ 1 then we haveZ
1
(x+ 1)
2+ 1
dx =
Z1
u
2+ 1
du = tan
�1u+ C = tan
�1(x+ 1) + C .
using these results from we have the final solution
= 2 ln |x+ 3|� 1
2
ln |x2 + 2x+ 2|� 2 tan
�1(x+ 1) + C .
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