university of southampton math1055w1

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.

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

,


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(↵),


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 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,


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),


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,


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

TURN OVER

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,


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

◆,


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 ,


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,



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

,


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,


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,


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


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)),



TURN OVER

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),


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,


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

,



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,


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,


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),


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

,


END OF PART A


TURN OVER

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


9 MATH1055W1

(b) [7 marks] Find C�1 when ↵ = 4.

out of 7


TURN OVER

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


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


TURN OVER

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


13 MATH1055W1

(c) [5 marks] By differentiating the result in (b) verify that

d

dx

�sinh

�1 x�=

1px2 + 1

out of 5


TURN OVER

of 18

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


15 MATH1055W1

out of 11


TURN OVER

of 18

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


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


TURN OVER

of 18

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


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 .


TURN OVER

Page 19 of 20

university of southampton math1055w1

Documents