Mathematical Physics JEST-2016...2018/03/01 · Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ... [email protected] 2 Head office fiziks,

fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES

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Mathematical Physics

JEST-2016

Q1. Given the condition 2 0φ∇ = , the solution of the equation 2 .kψ φ φ∇ = ∇ ∇ is given by

(a) 2

2kφψ = (b) 2kψ φ= (c) ln

2kφ φψ = (d) ln

2kφ φψ =

Ans: (a)

Solution: 2 0φ∇ = ( ). 0φ⇒∇ ∇ = ( ) ˆ ˆ ˆ. 0 x y zφ φ α β γ⇒∇ ∇ = ⇒∇ = + + x y zφ α β γ⇒ = + +

( )2 2 2.k kφ φ α β γ∇ ∇ = + +

If ( )2

2

2 2k k x y zφψ α β γ= = + +

( )2 2 2

2 2 2 22 2 2 k

x y zψ ψ ψψ α β γ∂ ∂ ∂

⇒∇ = + + = + +∂ ∂ ∂

2 .kψ φ φ⇒∇ = ∇ ∇

Q2. The mean value of random variable x with probability density

( ) ( )( )2

2

1 .exp2 2

x xp x

μ

σ π σ

⎡ ⎤+⎢ ⎥= −⎢ ⎥⎣ ⎦

is:

(a) 0 (b) 2μ (c)

2μ− (d) σ

Ans. : (a)

Solution: 2

2 21 exp exp 0

2 22x xx x dx x dxμσ σσ π

∞ ∞

−∞ −∞= − − =∫ ∫





Q3. Given a matrix2 11 2

M⎛ ⎞

= ⎜ ⎟⎝ ⎠

, which of the following represents cos6Mπ⎛ ⎞

⎜ ⎟⎝ ⎠

(a) 1 212 12⎛ ⎞⎜ ⎟⎝ ⎠

(b) 1 131 14

−⎛ ⎞⎜ ⎟−⎝ ⎠

(c) 1 131 14⎛ ⎞⎜ ⎟⎝ ⎠

(d) 1 31

2 3 1

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Ans. : (b)

Solution: We have

2 10

1 2λ

λ−

=−

2 4 3 0 1λ λ λ⇒ − + = ⇒ = or 3λ =

For 1λ =

2 1 1 01 2 1 0

xy

−⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

gives

Thus 0x y x y+ = ⇒ = − . Taking 1x = , the eigenvector associated with 1λ = is

1

11

x ⎡ ⎤= ⎢ ⎥−⎣ ⎦

For 3λ =

1 1 0

1 1 0x

x yy

−⎛ ⎞⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠

Taking 1x = , the eigenvectors associated with 3λ = is 2

11

x ⎡ ⎤= ⎢ ⎥⎣ ⎦

Thus

1 1 1 0 1/ 2 1/ 21 1 0 3 1/ 2 1/ 2

M−⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

01 1 1/ 2 1/ 261 1 1/ 2 1/ 26 0

6

ii M

i

ππ

π

⎡ ⎤⎢ ⎥ −⎡ ⎤ ⎡ ⎤

= ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦





66

6

1 1 0 1/ 2 1/ 21 1 1/ 2 1/ 2

0

ii M

i

ee

e

ππ

π

⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

6 62 26 6

6 62 2 2 2

1 11 1 2 22 21 1 1 1

2 2 2 2

i ii ii i

i ii i i i

e e e ee e

e e e e e e

π ππ ππ π

π ππ π π π

⎡ ⎤⎡ ⎤ + +⎢ ⎥−−⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥= = ⎢ ⎥⎢ ⎥− ⎢ ⎥⎣ ⎦ ⎢ ⎥+ +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎢ ⎥⎣ ⎦

3 3 34 4 4 4cos sin

6 6 3 3 34 4 4 4

i iM Mi

i iπ π

⎡ ⎤+ − +⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎢ ⎥⇒ + =⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠

− + +⎢ ⎥⎣ ⎦

3 3 34 4 4 4cos sin

36 6 3 34 44 4

i iM Mi

i iπ π

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎢ ⎥⇒ + = + ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

Thus

1 13cos1 16 4

Mπ −⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠

Q4. The sum of the infinite series 1 1 11 ...3 5 7

− + − + is

(a) 2π (b) π (c) 2π (d)

4π

Ans. : (d)

Solution: The series for 1tan x− for ,1x > is given by

3 5 7

1 1 1 1tan2 3 5 7

x

x x x xπ− = − + − + + ⋅⋅⋅⋅

Putting 1x = , we obtain

1 1 1 1tan 1 12 3 5 7π− ⎛ ⎞= − − + − + ⋅⋅⋅⎜ ⎟

⎝ ⎠1 1 11

4 2 3 5 7π π ⎛ ⎞⇒ = − − + −⎜ ⎟

⎝ ⎠⋅⋅⋅⋅





Q5. A semicircular piece of paper is folded to make a cone with the centre of the semicircle

as the apex. The half-angle of the resulting cone would be:

(a) o90 (b) o60 (c) o45 (d) o30

Ans. : (d)

Solution: When the semicircular piece of paper is folded to make a cone, the circumference of

base is equal to the circumference of the original semicircle. Let r be the radius of the

base of the core and R be the radius of the semicircle.

Hence, 22Rr R rπ π= ⇒ = .

The stay height of the come will also be R .

Hence, / 2 1sin2

RR

α = =

Thus, 030α =

JEST-2015

Q6. Given an analytic function ( ) ( ) ( ). ,f z x y i x y= +φ ψ , where ( ) yyxxyx 24, 22 +−+=φ .

If C is a constant, which of the following relations is true?

(a) ( ) 2, 4x y x y y Cψ = + + (b) ( ), 2 2x y xy x Cψ = − +

(c) ( ). 2 4 2x y xy y x Cψ = + − + (d) ( ) 2, 2x y x y x Cψ = − +

Ans. : (c)

Solution: ( ) 2 2, 4 2u Q x y x x y y= = + − +

From C.R. equation u vx y∂ ∂

=∂ ∂

u vy x∂ ∂

= −∂ ∂

2 4u xx∂

= +∂

R

r

α R





2 4v xy∂

= +∂

( )2 4v xy y f x= + + (i)

2 2u yy∂

= − +∂

2 2v yx∂

= + −∂

( )2 2v xy x f y= + +

( ) ( )2 4 2 2xy y f x xy x f y+ + = − + (ii)

( ) ( )2 , 4f x x f y y= =

2 4 2v xy y x c= + − +

Q7. If two ideal dice are rolled once, what is the probability of getting at least one ‘6’?

(a) 3611 (b)

361 (c)

3610 (d)

365

Ans: (a)

Solution: Number of point in sample space ( ) 11n S =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1,6 , 2,6 , 3,6 , 4,6 , 5,6 , 6,1 , 6,2 , 6,3 , 6,4 , 6,5 , 6,6⎡ ⎤⎣ ⎦

Number of point in population ( ) 26 36n P = =

Probability that at least one six on face of dice ( )( )

1136

n Sn P

= =





Q8. What is the maximum number of extrema of the function ( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 24

24 xx

k exPxf where

( )∞∞−∈ ,x and ( )xPk is an arbitrary polynomial of degree k ?

(a) 2+k (b) 6+k (c) 3+k (d) k

Ans: (c)

Solution: ( ) ( )4 2

4 2x x

xf x P x e⎛ ⎞

− +⎜ ⎟⎝ ⎠=

( ) ( ) ( ) ( )4 2

2 23x x

x xf x P x P x x x e⎛ ⎞

− +⎜ ⎟⎝ ⎠⎡ ⎤= + − +′ ′⎣ ⎦ ( ) 0f x′⇒ =

( ) ( ) ( )3 0xP x x x P x⎡ ⎤= + − =′⎣ ⎦ is polynomial if order 3k +

From the sign scheme maximum number of extrema 3k= +

Q9. The Bernoulli polynominals ( )sBn are defined by, ( )∑=− !1 nxsB

exe n

nx

xs

. Which one of the

following relations is true?

(a) ( )

( ) ( )∑ +=

−

−

!11

1

nxsB

exe n

nx

sx

(b) ( )

( ) ( ) ( )1

11 1 !

x s nn

nx

xe xB se n

−

= −− +∑

(c) ( )

( )( )∑ −−=−

−

!1

1

1

nxsB

exe n

nnx

sx

(d) ( )

( )( )∑ −=−

−

!1

1

1

nxsB

exe n

nnx

sx

Ans: (d)

Solution: ( )1

xS n

nx

xe xB Se n

=− ∑

Put ( )1S S= − , ( )

( )1

11

x S n

nx

xe xB Se n

−

= −− ∑

( ) ( ) ( )1 1 nnB S B S− = −

( )( ) ( )

1

11

x S nn

nx

xe xB Se n

−

= −− ∑





Q10. Consider the differential equation ( ) ( ) ( )xxkGxG δ=+′ ; where k is a constant. Which

following statements is true?

(a) Both ( )xG and ( )xG′ are continuous at 0=x .

(b) ( )xG is continuous at 0=x but ( )xG′ is not.

(c) ( )xG is discontinuous at 0=x .

(d) The continuity properties of ( )xG and ( )xG′ at 0=x depends on the value of k .

Ans: (c)

Q11. The sum ∑ = ++99

1 11

m mmis equal to

(a) 9 (b) 199 − (c) ( )1991−

(d) 11

Ans: (a)

Solution: 99

1

11m m m= + +

∑

( )99 99

1 1

1 11m m

m m m mm m= =

+ −= + −

+ −∑ ∑

2 1 3 2...... 100 99= − + − + −

100 1 10 1 9= − = − =

JEST-2014

Q12. What are the solutions to ( ) ( ) ( ) 02 =+′−′′ xfxfxf ?

(a) xec x /1 (b) xcxc /21 + (c) 21 cxec x + (d) xx xecec 21 +

Ans.: (d)

Solution: Auxilary equation 0122 =+− DD ( )21 0D⇒ − = 1, 1D⇒ = + +

∵ Roots are equal then ( ) ( )1 2xf x c c x e= + ( ) 1 2

x xf x c e c xe⇒ = +

Q13. The value of ∫2.2

2.0dxxex by using the one-segment trapezoidal rule is close to

(a) 11.672 (b) 11.807 (c) 20.099 (d) 24.119

Ans.: (c)

Solution: 22.02.2 =−=h ( ) ( )2.2 0.2 20.0992hI y y⎡ ⎤⇒ = + =⎣ ⎦ xy xe=∵





Q14. Given the fundamental constants (Planck’s constant), G (universal gravitation

constant) and c (speed of light), which of the following has dimension of length?

(a) 3cG (b) 5c

G (c) 3cG (d)

Gcπ8

Ans.: (a)

Solution: [ ][ ] 21

33

23112

⎥⎦

⎤⎢⎣

⎡−

−−−

TLTLMTML [ ] LL == 2

12

2 1ML T −⎡ ⎤= ⎣ ⎦ , [ ]2312

−−== TLMm

grG

Q15. The Laplace transformation of te t 4sin2− is

(a) 254

42 ++ ss

(b) 204

42 ++ ss

(c) 204

42 ++ ss

s (d) 2042

42 ++ ss

s

Ans.: (b)

Solution: sinatL e bt−⎡ ⎤⎣ ⎦∵( )2 2

bs a b

=+ +

2 sin 4tL e t−⎡ ⎤⇒ ⎣ ⎦ ( )2 2

42 4s

=+ + 2

44 20s s

=+ +

Q16. Let us write down the Lagrangian of a system as ( ) xcxkxxmxxxxL ++= 2,, . What is the

dimension of c ?

(a) 3−MLT (b) 2−MT (c) MT (d) 12 −TML

Ans.: (c)

Solution: According to dimension rule same dimension will be added or subtracted then

dimension of =xMx dimension of Cxx

[ ] [ ]2 2 3ML T C L LT− −⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

[ ] [ ][ ] [ ]MT

MLTMLC == −

−

32

22





Q17. The Dirac delta function ( )xδ satisfies the relation ( ) ( ) ( )∫∞

∞−= 0fdxxxf δ for a well

behaved function ( )xf . If x has the dimension of momentum then

(a) ( )xδ has the dimension of momentum

(b) ( )xδ has the dimension of ( )2momentum

(c) ( )xδ is dimensionless

(d) ( )xδ has the dimension of ( ) 1momentum −

Ans.: (d)

Solution: ( ) ( ) ( )∫∞

∞−= 0fdxxxf δ

( ) ( ) ( ) ( )[ ] ( ) ( )[ ] ( ) [ ]100 −=⇒=⋅⇒= PxfPxxffdxxxf δδδ

Since, ( )[ ] ( )[ ]0fxf =

If ( ) βα += xxF is force [ ]2−TLM

( ) β=0F is also [ ]2TLM

Q18. The value of limit

11lim 6

10

++

→ zz

iz

is equal to

(a) 1 (b) 0 (c) -10/3 (d) 5/3

Ans.: (d)

Solution: 11lim 6

10

++

→ zz

iz 35

610

610lim

610lim

4

5

9

=⇒⇒⇒→→

zzz

iziz

Q19. The value of integral

∫ −=

cdz

zzIπ2

sin

with c a circle 2=z , is

(a) 0 (b) iπ2 (c) iπ (d) iπ−





Ans.: (c)

Solution: ∫ −=

C zzIπ2

sin pole 2

02 ππ =⇒=−⇒ zz

Residue at 2π

=z 2=z∵ so it will be lies within the contour

( ) 22

2

iz

emg C

eI R iz

ππ

= = ×⎛ ⎞−⎜ ⎟⎝ ⎠

∑∫

Res / 2

2

22 22

2

izi

z

z ee i

z

π

π

π

π=

⎛ ⎞−⎜ ⎟⎝ ⎠= = =

⎛ ⎞−⎜ ⎟⎝ ⎠

(taking imaginary part) ; Residue = 21

Now iiI ππ =×= 221

JEST-2013

Q20. A box contains 100 coins out of which 99 are fair coins and 1 is a double-headed coin.

Suppose you choose a coin at random and toss it 3 times. It turns out that the results of all

3 tosses are heads. What is the probability that the coin you have drawn is the double-

headed one?

(a) 0.99 (b) 0.925 (c) 0.75 (d) 0.01

Ans.: (c)

Q21. Compute ( ) ( )2

22

0

ImRelimz

zzz

+→

(a) The limit does not exist. (b) 1

(c) –i (d) -1

Ans.: (a)

Solution: ( ) ( )2 2 2 2 2 2

2 2 2 2 20 0 00

Re Im 2 2lim lim lim 12 2z z y

x

z z x y xy x y xyz x y ixy x y ixy→ → =

→

+ − + − += ⇒ =

− + − +





2 2

2 200

2lim 12x

y

x y xyx y ixy=

→

− +=

− + and

2 2

2 20

2lim2y x

x

x y xy ix y ixy=

→

− += −

− +

Q22. The vector field jyixz ˆˆ + in cylindrical polar coordinates is

(a) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆsincos 22 −++

(b) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆsincos 22 +++

(c) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆcossin 22 +++

(d) ( ) ( ) φρ φφρφφρ ezez ˆ1cossinˆcossin 22 −++

Ans.: (a)

Solution: jyixzA ˆˆ += , , 0x y zA xz A y A⇒ = = =

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆx y zA A e A x e A y e A z eρ ρ ρ ρ ρ= ⋅ = ⋅ + ⋅ + ⋅

( ) ( )cos cos sin sin 0A zρ ρ φ φ ρ φ φ⇒ = + + ( )2 2 ˆcos sinA z eρ ρρ φ ρ φ⇒ = +

( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆˆx y zA A e A x e A y e A z eφ φ φ φ φ= ⋅ = ⋅ + ⋅ + ⋅

( )cos sin sin cosA zφ ρ φ φ ρ φ φ⇒ = − + ⋅ ( ) ˆcos sin 1A z eφ φρ φ φ⇒ = ⋅ −

( ) ( )2 2ˆ ˆ ˆ ˆ ˆcos sin cos sin 1z zA A e A e A e z e z eρ ρ φ φ ρ φρ φ φ ρ φ φ= + + = + + −

Q23. There are on average 20 buses per hour at a point, but at random times. The probability

that there are no buses in five minutes is closest to

(a) 0.07 (b) 0.60 (c) 0.36 (d) 0.19

Ans.: (d)

Q24. Two drunks start out together at the origin, each having equal probability of making a

step simultaneously to the left or right along the x axis. The probability that they meet

after n steps is

(a) 2!!2

41

nn

n (b) 2!!2

21

nn

n (c) !221 nn (d) !

41 nn

Ans.: (a)





Solution: Into probability of taking ' 'r steps out of N steps

1 12 2r

r N r

CN−

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

total steps 2N n n n= = + =

for taking probability of n steps out of N

( ) ( )

2 2

21 1 ! 1 1 2 ! 1 2 !2 2 ! ! 2 2 ! ! 2 ! 4n

n N n n n n n

C n

N n nP NN n n n n n

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Q25. What is the value of the following series?

22

...!5

1!3

11....!4

1!2

11 ⎟⎠⎞

⎜⎝⎛ −+−+⎟

⎠⎞

⎜⎝⎛ −+−

(a) 0 (b) e (c) 2e (d) 1

Ans.: (d)

Solution: 2 4 3 5

cos 1 ..... , sin .....2! 4! 3! 5!θ θ θ θθ θ θ⇒ = − + = − +

11sin1cos!5

1!3

11...!4

1!2

11 2222

=+⇒⎟⎠⎞

⎜⎝⎛ +−+⎟

⎠⎞

⎜⎝⎛ +−⇒ 1cossin 22 =+ θθ∵

Q26. If the distribution function of x is ( ) λ/xxexf −= over the interval 0 < x < ∞, the mean

value of x is

(a) λ (b) 2λ (c) 2λ (d) 0

Ans.: (b)

Solution: ∵ it is distribution function so ( )( )

0

0

.x

x

x xe dxxf x dxx

f x dx xe dx

λ

λ

−∞∞

−∞∞

−∞

−∞

= = ∫∫∫ ∫

2

0

0

2

x

x

x e dx

xe dx

λ

λ

λ

−∞

−∞

⇒ =∫

∫





JEST-2012

Q27. The value of the integral ( )∫

∞

+0 22 1ln dx

xx is

(a) 0 (b) 4π− (c)

2π− (d)

2π

Ans. : (b)

Solution: ( ) ( )2 22 2

0 0

ln ln

1 1

x zdx dzx z

∞ ∞

=+ +

∫ ∫

Let us consider new function ( )2

2

ln1

zf zz

⎛ ⎞= ⎜ ⎟+⎝ ⎠, then

2

20

ln1

zI dzz

∞ ⎛ ⎞= ⎜ ⎟+⎝ ⎠∫

Pole at z i= ± is simple pole of second order.

Residue at z i= is

( ) ( )( ) ( )

22

2 2

ln zd z idz z i z i

= −− +

( )( )

2

2

ln zddz z i

=+

( ) ( ) ( ) ( )

( )

2 2

4

12 ln . ln .2z i z z z iz

z i

+ − +=

+

( ) ( ) ( )

( )

2

3

12ln ln .2z i z zz

z i

+ −=

+

( ) ( )

( )

2

3

12 2 ln ln 2

2

i i ii

i

× − ⋅=

2

4 22 2

8

i i

i

π π⎛ ⎞− ×⎜ ⎟⎝ ⎠=−

2

22

8

i

i

ππ +=

−

2

Res4 16z i

iπ π=

−⇒ = +

Similarly at z i= − ; 2

Res4 16z i

iπ π=−

−= −

2 2 22

20

ln 21 4 16 4 16

zI dz i i i iz

π π π ππ π∞ ⎛ ⎞−⎛ ⎞= = + − − = −⎜ ⎟⎜ ⎟+⎝ ⎠ ⎝ ⎠∫

( ) ( )2

R A B r A B

i f z dz f z dzπ⎛ ⎞ ⎛ ⎞

− = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫ ∫ ∫ ∫ ∫ ; vanish

A B∫ ∫

R

rA

B

•

•





Along path A; z x iε= − + and along path B; z x iε= − −

Thus ( ) ( )( )

( )( )

02

2 20

ln ln1 1A B

x i x ii f z dz dx dx

x i x iε ε

πε ε

∞

∞

⎡ ⎤ ⎡ ⎤⎛ ⎞ − + − −− = = − −⎢ ⎥ ⎢ ⎥⎜ ⎟

− + + − − +⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎣ ⎦∫ ∫ ∫ ∫

( )( )

( )( )

2 2

22 2

0 0

ln ln1 1

x i x ii dx dx

x i x iε ε

πε ε

∞ ∞⎡ ⎤ ⎡ ⎤− + − −⇒ − = −⎢ ⎥ ⎢ ⎥

− + + − − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫

( ) ( )2 2

22 2

0 0

ln ln; 0

1 1x i x i

i dx dxx x

π ππ ε

∞ ∞+ −⎡ ⎤ ⎡ ⎤⇒ − = − →⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

∫ ∫

( )( ) ( )( )( ) ( )

2 2

22 22 2

0 0

ln ln ln41 1

x i x i xi dx ix x

π ππ π

∞ ∞+ − −⇒ − = =

+ +∫ ∫

( )2

220

ln4 41

x iix

π ππ

∞ − −⇒ = =

+∫

Q28. If [x] denotes the greatest integer not exceeding x, then [ ]∫∞ −

0dxex x

(a) 1

1−e

(b) 1 (c) e

e 1− (d) 12 −e

e

Ans.: (a)

Solution: [ ]x

[ ] 010 ==<≤ xx , [ ] 121 ==<≤ xx , [ ] 232 ==<≤ xx

now [ ] [ ] [ ] [ ] [ ] dxexdxexdxexdxexdxex xxxxx −−−−−∞

∫∫∫∫∫ +++=4

3

3

2

2

1

1

00

dxedxedxe xxx ∫∫∫ −−− +++⇒4

3

3

2

2

1

.3.2.10

[ ] ( ) ( ) ....32 43

32

21 +−+−+−⇒ −−− xxx eee

+−+−+−+−⇒ −−−−−−−− 54433221 443322 eeeeeeee

∞++−++⇒ −−−− .....4321 eeee





11

1 1

1

−=

−⇒ −

−

eee

22 1 1

1

er e ee

−− + −

−

⎛ ⎞= = =⎜ ⎟

⎝ ⎠∵

Q29. As x → 1, the infinite series .......71

51

31 753 +−+− xxxx

(a) diverges (b) converges to unity

(c) converges to π / 4 (d) none of the above

Ans.: (c)

Solution: .......753

tan753

1 +−+−=− xxxxx4

1tan 1 π=⇒ −

Q30. What is the value of the following series? 22

.....!5

1!3

11....!4

1!2

11 ⎟⎠⎞

⎜⎝⎛ +++−⎟

⎠⎞

⎜⎝⎛ +++

(a) 0 (b) e (c) e2 (d) 1

Ans.: (d)

Solution: −++++=!3

1!2

11132

1e ,

.....!3

1!2

11132

1 −+−=−e

....!4

1!2

112

1cosh11

+++=+

=−ee

( ) ...!5

1!3

112

1sinh11

+++=−

=−ee

i.e 11sin1cos 22 =− hh

Q31. An unbiased die is cast twice. The probability that the positive difference (bigger -

smaller) between the two numbers is 2 is

(a) 1 / 9 (b) 2 / 9 (c) 1 / 6 (d) 1 / 3

Ans.: (a)

Solution: ( ) ( )( )SnEnp =2





The number of ways to come positive difference ( ) ( ) ( ) ( )[ ]4,6,3,5,2,4,1,3

( )91

3642 ==p

Q32. For an N x N matrix consisting of all ones,

(a) all eigenvalues = 1 (b) all eigenvalues = 0

(c) the eigenvalues are 1, 2, …., N (d) one eigenvalue = N, the others = 0

Ans.: (d)

Solution: 1 1

0, 21 1⎡ ⎤

=⎢ ⎥⎣ ⎦

3,0,0111111111

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

so far NN × matrix one eigen value is N and another’s eigen value is zero

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