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1.1.1.1.1. (a)

2.2.2.2.2. (d)

3.3.3.3.3. (c)

4.4.4.4.4. (a)

5.5.5.5.5. (c)

6.6.6.6.6. (a)

7.7.7.7.7. (b)

15.15.15.15.15. (b)

16.16.16.16.16. (c)

17.17.17.17.17. (c)

18.18.18.18.18. (b)

19.19.19.19.19. (a)

20.20.20.20.20. (c)

21.21.21.21.21. (c)

22.22.22.22.22. (c)

23.23.23.23.23. (a)

24.24.24.24.24. (a)

25.25.25.25.25. (d)

26.26.26.26.26. (c)

27.27.27.27.27. (a)

28.28.28.28.28. (d)

8.8.8.8.8. (d)

9.9.9.9.9. (c)

10.10.10.10.10. (d)

11.11.11.11.11. (c)

12.12.12.12.12. (a)

13.13.13.13.13. (c)

14.14.14.14.14. (d)

GATE-2012

(TEST SERIES-3)

EC: Electronics Engineering

(Maths and G.A)

SlSlSlSlSl..... NNNNNooooo .: 251211.: 251211.: 251211.: 251211.: 251211

A N S W E R SA N S W E R SA N S W E R SA N S W E R SA N S W E R S

29.29.29.29.29. (d)

30.30.30.30.30. (a)

31.31.31.31.31. (b)

32.32.32.32.32. (b)

33.33.33.33.33. (c)

34.34.34.34.34. (a)

35.35.35.35.35. (a)

Note: If you find any discrepancies in the Answer key/Solution, you

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[14] GATE - 2012 (TEST SERIES - 3) ELECTRONICS ENGINEERING

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EXPLANATIONS

Note: If you find any discrepancies in the Answer key/Solution, you are requested to write us at [email protected] within

24 hours, with the correct or suggested solutions from your end. Please specify your Name, Branch of Engineering and phoneno. for future reference with the E mail. Also you are requested to write us about the Feedback of the question paper at

[email protected].

1.1.1.1.1. (a)(a)(a)(a)(a)

Sum of eigen values is equal to the sum of

principle diagonal elements.

2.2.2.2.2. (d)(d)(d)(d)(d)

x 0

1 1lim

sinx x

= x 0

x sinxlim xsinx

=x 0

1 cosxlim

x cos x sin x

+

=x 0

sinxlim

x( sinx) cosx cosx

+ +

=0

00 1 1

=+ +

3.3.3.3.3. (c )(c )(c )(c )(c )x4 is purely algebraic function so it can not

be further expanded.

4.4.4.4.4. (a)(a)(a)(a)(a)

Greens theorem is used to convert line

integral into surface integral.

5.5.5.5.5. (c )(c )(c )(c )(c )

Given equation can be expressed as

2

2d ya dx=

3 / 22

dy1dx

+

Squaring both sides,

22

2

d ya

dx

=

32dy

1dx

+

Hence

order = 2

degree = 2.

6.6.6.6.6. (a)(a)(a)(a)(a)

P.I = x2

1e

D 5D 6+ +Put D = 1

P.I =x

x

2

1 ee

121 5.1 6=

+ +

7.7.7.7.7. (b)(b)(b)(b)(b)

Probability of getting a head = 1

2

Then A can win in 1st, 3rd, 5th, throws. The chances of As winning

=2 4 6

1 1 1 1 1 1 1

2 2 2 2 2 2 2

+ + + +

=2

1/ 2 1 4 2

2 3 311

2

= =

8.8.8.8.8. (d)(d)(d)(d)(d)

Since each desired number is divisible by

5, so we must have 5 at the unit place. So,

there is 1 way of doing it.

Tens place can be filled by any of the

remaining 5 numbers.

So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by

any of the remaining 4 digits. So, there are

4 ways of filling it.

Required number of numbers.=(4 5 1) = 20

9.9.9.9.9. (c )(c )(c )(c )(c )

n(A) = 325

n(B) = 175

n(A B) = 450 50 = 400Required number

= n(A B)= n(A) + n(B) n(A B)= 325 + 175 400 = 100

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10.10.10.10.10. (d)(d)(d)(d)(d)LCM of 252, 308 and 198

= 2772

So, A, B and C will again meet at the starting

point in 2772 second i.e., 46 minimum 12

sec.

11.11.11.11.11. (c )(c )(c )(c )(c )

Let the distance travelled be x km

Then,

x x

10 15 = 2 3x 2x = 60 x = 60 kmTime taken to travel 60 km at 10 km/hr

=60

hrs. 6hrs.10

=

So, Robert started 6 hours before 2 pm i.e.,

at 8 am

Required speed =60

kmph5

= 12 kmph

12.12.12.12.12. (a)(a)(a)(a)(a)

Recondite means difficult for one of ordinary

understanding to comprehend; manifestmanifestmanifestmanifestmanifest

means easily understood or recognized

13.13.13.13.13. (c )(c )(c )(c )(c )

Gracious means to be pleasantp leasantp leasantp leasantp leasant or

considerate in social interactions.

14.14.14.14.14. (d)(d)(d)(d)(d)

In the stem part we have two clues:

1. Unless structural

2. Smoking semantic

Unless here demands a not in the filler

part. Thus alternatives (a) and (c) are wrong.

Alternatives (b) and (d) are structurally right,

but according to smoking the expression

not suffer is wrong, but not recover is

right. Thus only (d) matches.

15.15.15.15.15. (b)(b)(b)(b)(b)The answer is (b), panel because the word

denoting a group of experts is panel.

16.16.16.16.16. (c )(c )(c )(c )(c )

Matrix A is said to be unitary matrix if

A A = A A = IWhere

A is transpose of conjugate of matrix A.

So

A

=

1 1 j1

1 j 13

+

A A =1 1 j 1 1 j1 1

1 j 1 1 j 13 3

+ +

=1 (1 1) (1 j) (1 j)1

3 (1 j) (1 j) 1 1 1

+ + + + + +

=3 01

3 0 3

A A =1 0

I0 1

=

Hence matrix A is a unitary matrix.

17.17.17.17.17. (c )(c )(c )(c )(c )

The given equations can be written as

1 1 1 x

1 2 3 y

1 2 z

=

6

10

Hence

A =

1 1 1

1 2 31 2

and B =

6

10

Hence augumented can be written as

[A : B] =

1 1 1 6

1 2 3 10

1 2

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=2 2 1

3 3 1

1 1 1 6R R R

0 1 2 4R R R

0 1 1 6

There are infinite solutions if

(A) = (A : B) = 2So

3 = 0 or = 3 and 10 = 0 or = 10

18.18.18.18.18. (b)(b)(b)(b)(b)

We know that

/2m n

0

sin xcos xdx

=K(m 1)(m 3) (n 1)(n 3)

(m n)(m n 2) (m n 4)

+ + +

Where

K = 1 ; if either m or n or both are odd.

K = ;2

if both m and n are even.

/215 3

0 sin xcos xdx

=

(15 1)(15 3)(15 5)(15 7)

(15 9)(15 11)(15 13) (3 1)1

(18)(18 2)(18 4)(18 6)(18 8)

(18 10)(18 12)(18 14)(18 16)

=14 12 10 8 6 4 2 2

18 16 14 12 10 8 6 4 2

=2 1

18 16 144=

19.19.19.19.19. (a)(a)(a)(a)(a)

Here

F

= (x y)i (2x z)j (y z)k+ + + +

Curl F

=

i j k

x y z

x y 2x z y z

+ +

= 2i k+

Also equation of the plane through A, B, C(vertices of triangle in figure shown below) is

x y z

2 3 6+ + = 1

3x + 2y + z = 6

0

x

y

z

C (0, 0, 6)

B (0, 3 0)

A (2, 0, 0)

Vector N

normal to this plane is

(3x + 2y + z 6) = 3i 2j k+ +

N = 3i 2 j k

9 4 1+ ++ +

=1 (3i 2j k)14

+ +

Hence

[ ]c

(x y)dx (2x z)dy (y z)dz+ + + + =c

F dr

=c

curl F.Nds

where s is the area of triangle ABC

[Applying Stokes theorem]

=c

3i 2j k (2i k) ds14

+ ++

=s

1(6 1) ds

14+

=7

(Area of ABC)14

=7

3 1414

= 21

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20.20.20.20.20. (c )(c )(c )(c )(c )f(x) = 3x4 2x3 6x2 + 6x + 1

f(x) = 12x3 6x2 12x + 6= 6(x2 1) (2x 1)

f(x) = 0 when x = 11,2

So in the interval (0, 2), f(x) can have

maximum or minimum at

x =1

or 1.2

Now

f(x) = 36x2 12x 12So that

1f

2

= 9

f(x) has a maximum value at x = 12

Thus maximum value

=1

f2

=4 3 2

1 1 1 13 2 6 6 1

2 2 2 2

+ +

=39

16

21.21.21.21.21. (c )(c )(c )(c )(c )

dy(x 1) y

dx+ = e3x(x + 1)2

Dividing throughout by (x + 1),

dy ydx x 1

+

= e3x(x + 1)

Here

P =1

x 1

+

and

Pdx = dxx 1 + = ln(x + 1)

= ln(x + 1)1

I.F = Pdxe =1n( x 1)e

+l=

1

(x 1)+Thus the solution of given differential

equation is

y(I.F) = 3xe (x 1) (I.F)dx C + + or

y

x 1+= 3x 3x

1e dx C e C

3+ = +

or y =3x1e C (x 1)

3

+ +

22.22.22.22.22. (c )(c )(c )(c )(c )

Let

f(z) =2

z 3

z 2z 5

+ +

The poles of f(z) are given by z2 + 2z + 5 =

0

i.e. by

z =2 4 20

1 2i2

=

Equation of circle is

z (1 + i) = 2Hence only one pole z = 1 + 2i lies inside

the circle C. Therefore, f(z) is analytic within

C except at this pole.

So residue of f(z) at z = 1 + 2i is

Res f(1 + 2i) = { }z ( 1 2i)

lim z ( 1 2i) f(z) +

+

=z ( 1 2i)

(z 3)lim

(z 1 2i) +

+ +

=4 2i 1

i4i 2

+ = +

Hence by residue theorem

c

f(z)dz = [ ]2 i Resf( 1 2i) +

=1

2 i i2

+

= (i 2)

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23.23.23.23.23. (a)(a)(a)(a)(a)Poles of f(z) are given as

z2 2z = 0

z(z 2) = 0 z = 0 and 2.Now residue at z = 0 is

Res { }f(0) =z 0lim z f(z)

=2

2z 0

z 1lim z

z 2z

+

=2

z 0

z 1 0 1 1lim

z 2 2 2

+ += =

Also residue at z = 2 is

{ }Res f(2) =z 2lim (z 2) f(z)

=2

2z 2

z 1lim (z 2)

z 2z

+

=

2

z 2z 1 5lim

z 2 + =

24.24.24.24.24. (a)(a)(a)(a)(a)

The probability that A can solve the problem

is1

.2

The probability that A cannot solve the

problem is1 1

1 .2 2

=

Similarly the probabilities that B and C

cannot solve the problem are

1 11 and 1 .

3 4

The probability that A, B and C cannot solve

the problem is1 1 1

1 1 1 .2 3 4

Hence the probability that the problem will be

solved, i.e. at least one student will solve it

=1 1 1

1 1 1 12 3 4

=1 2 3 1 3

1 12 3 4 4 4

= =

25.25.25.25.25. (d)(d)(d)(d)(d)

f(x) = 3x cos x 1

f(x) = 3 + sin xNewtons iteration formula gives

xn + 1 = nnn

f(x )xf (x )

xn + 1 =n n

nn

3x cosx 1x

3 sin x

+

xn + 1 =n n n

n

x sinx cosx 1

3 sinx

+ ++

Putting n = 0, the first approximation x1 is

given by

x1 =o o o

o

x sinx cosx 1

3 sinx

+ ++

=(0.6)sin(0.6) cos(0.6) 1

3 sin(0.6)+ +

+= 0.6071

26.26.26.26.26. (c )(c )(c )(c )(c )

Given

I =2t

0

te sintdt

or it can also be written as

I = st0t e sintdt

where s = 2

So I = L(t sin t), by definition

I =2

d 1( 1)

ds s 1

+

I =2 2 2 2

2s 2 2

(s 1) (2 1)

=+ +

=4

25[Putting s = 2]

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27.27.27.27.27. (a)(a)(a)(a)(a)

(A + B)s 1 days work =1

5

(B + C)s 1 days work =1

7

(A + C)s days work =1

4

Adding, we get

2(A + B + C)s 1 days work = 1 1 15 7 4

+ +

=83

140

(A + B + C)s 1 days work =83

280

As 1 days work =83 1 43

280 7 280

=

Bs 1 days work =83 1 13

280 4 280

=

Cs 1 days work =83 1 27

280 5 280

=

Thus time taken by A, B, C is280

days,43

28days,

13280

days27

respectively.

Clearly, the time taken by A is least.

28.28.28.28.28. (d)(d)(d)(d)(d)II.II.II.II.II. Let the present ages of Arun and his

son be 11 x and 6 x years respectively.

I.I.I.I.I. 5 years ago, Aruns age

= 2 his sons age

III.III.III.III.III. 5 years hence,Arun's age 12

Son's age 7=

Clearly, any two of the above will give Aruns

present age.

29.29.29.29.29. (d)(d)(d)(d)(d)Let number of students appeared from

school A = 100

Then, number of students qualified from

school A = 70

Number of students appeared from school

B = 120Number of students qualified from school

B =150

70 105100

=

Required percentage=

105100 %

120

= 87.5%

30.30.30.30.30. (a)(a)(a)(a)(a)

Finesse(n.) is the subtle, skilful handling

of a situation, diplomacy, tact; refined or

delicate performance or execution.

31.31.31.31.31. (b)(b)(b)(b)(b)

The correct choice is answer (b) Something

characterized by preponderance is not

scarce. Something characterized by agility

is not stiff.

32.32.32.32.32. (b)(b)(b)(b)(b)

Average percent profit earned for the given

years.

= [ ]1

40 55 45 65 70 606

+ + + + +

=335 5

556 6

=

33.33.33.33.33. (c )(c )(c )(c )(c )

Let the expenditure in 1997 be x.

Then, expenditure in 2000.

= x + (25% of x)

=5

x4

Also, let the incomes in 1997 and 2000 be

I1

and I2

respectively.

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Then, for the year 1997, we have

45 = 1I x

100x

45

100= 1

I1

x

I1 =145x

1.45x100

=

Also, for the year 2000, we have

60 =2

5I x

4100

5x

4

60100

= 24I

15x

I2 =160 5x

2 x100 4

=

Difference between the two incomes= (2x 1.45 x) = 0.55 x

Percent by which I1

is less than I2

=0.55x

100 %2x

= 27.5%

34.34.34.34.34. (a)(a)(a)(a)(a)

Laplace equation:

2 = 0

i.e.x y z

+ + = 0

[2ax + 2by + 2cz] = 0 2a + 2b + 2c = 0

or a + b + c = 0

35.35.35.35.35. (a)(a)(a)(a)(a)

a + b + c = 1 3 + 2 = 0

!!!!

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