Mathematical Physics JEST-2016...2018/03/01 · Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR...
Transcript of Mathematical Physics JEST-2016...2018/03/01 · Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR...
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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
∞ ∞
−∞ −∞= − − =∫ ∫
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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
⎡ ⎤⎢ ⎥ −⎡ ⎤ ⎡ ⎤
= ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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 ⎛ ⎞⇒ = − − + −⎜ ⎟
⎝ ⎠⋅⋅⋅⋅
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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
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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
= =
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
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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
−
= −− ∑
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
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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=∵
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
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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
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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−
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
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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→ → =
→
+ − + − += ⇒ =
− + − +
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
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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)
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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
−∞
−∞
⇒ =∫
∫
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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
•
•
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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
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Head office fiziks, H.No. 40 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498
Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16
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