UNIVERSITY OF SOUTHAMPTON MATH1055W1
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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.
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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.
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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.
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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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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.
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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.
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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
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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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(b) [7 marks] Find C�1 when ↵ = 4.
out of 7
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(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
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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
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(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
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(c) [5 marks] By differentiating the result in (b) verify that
d
dx
�sinh
�1 x�=
1px2 + 1
out of 5
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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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out of 11
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(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
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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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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
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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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