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JAN ISSUE - MATHEMATICS - IIT JEE PAPER

1.(a) 2 2

1 1f K

n m

= −

k = constant

[ ]1 1,f K n m∴ = = = ∞Q

[ ]2

31; 2

4f K n m= = =

seris limit of balmer series

2

1 1

2 4

kf K

= − = ∞ 1 2f f= −

2.(c) B.E. of 17 7.75 17 131.75O MeV= × =

B.E. of 16 7.97 16 127.52O MeV= × =

∴ Ans is 131.75 127.52 4.23MeV− =

3.(a) Survival probability is = ( ) 0

0 0

tN t N e

N N

λ−

=

te

λ−= 1e

−= 1

at tλ

=

= 1e

4.(d) minkαλ λ λ∆ = −

when min

λ becomes two times but kαλ remains

the same.

( )1

min2 2

k kα αλ λ λ∆ = ∆ − = ∆ − ∆

∴ ( )1 2λ λ∆ < ∆

5.(a)21

22

mx mv v x= ⇒ =

6.(b) Phase change of π in 50 oscillationsPhase change of 2π in 100 oscillations

∴ Frequency difference is 1 in 100

7.(a) No change in temperature.

8.(a) ( )1/ 3

1/ 31/ 3 14 564

Ar A A

= ⇒ = ⇒ =

∴ 56 30 26Z = − =

The frequency of kα x-ray from the atom will be

- ( )2 18

2 2

1 11 1.55 10

1 2f RC z HZ

= − − = ×

P H Y S I C S

9.(a)

a

o s1

Q

s2

2a

θθ

Path difference 1 3 sinS Q a θ∆ =� .........

for Maxima 3 sin 0, , 2a θ λ λ= ............

1 2 150, , ...........

15 15 15Sinθ = ( )sin 1θ ≤Q

∴ In each quadrant we have 15 maximas

∴Total 60 maximas.

10.(d)

11.(a)

12.(c)

13.(b)9 4 12 1

4 2 6 0Be He C N+ → +

14.(a) No. of neutrons per sec 610=No. of α particles per sec

6 94000 10 4 10= × = ×

∴ Activity of Rn 94 10 / sec.A = ×∴ Intial activity of

Rn9

0 2 4 2 10 / sec.A A= = ×

2 20 00

22

d dA AA → →

∴ 0

04 2 10Nλ = ×

9

0

ln 24 2 10

4 24 3600N⇒ = ×

× ×

11

0

4 2 96 36 10

ln 2N

× × ×⇒ =




15.(c) The activity of radon after 4 days = 1

2

(activity after 2 days)

∴ No. of neutrons produced per sec. after 4days

1

2= No. of neutrons produced per sec. after

2 days

16.(a) Optical path difference between beems arrivingat P;

( )2 1sinx d θ∆ = − +l l

For maxima x nλ∆ =

[ ]2 1

1sin ( )x

dθ⇒ = ∆ − −l l

1

1 1sin

2n l

dθ λ

⇒ = − −

9 6

6

1[ 500 10 20 10 ] 2 1

10 10 40

nn

− −

−

= × × − × = − ×

⇒ 1sin 2 1

40

nθ −

= −

17.(a) | sin | 1θ ≤

1 2 1 140

n − ≤ − ≤

20 40 20n⇒ − ≤ − ≤ 20 60n⇒ ≤ ≤

∴ no. of maxima = 41

18.(d) At C, phase difference

φ ( ) 6

1 1 9

2 220 10 80

500 10l l

π πφ π

λ−

−

= − = × × =

×

∴ There is maxima at C.For minima at C

( 1)2

tλ

µ − =9

500 100500

2( 1) 2 0.5t nm

λ

µ

−×⇒ = = =

− ×

19.(a) ω1 = 100 π ; f

1 =

π

ω

2 = 50 Hz, f

2 =

π

π

2

92 = 46 Hz

Beat frequency = f1 = f

2 = 4

20.(a) v = ω/K = 100 π / (0.5 π) = 200

21.(c) At x = 0y = y

1 + y

2 = 2A cos 96 πt cos 4πt

For y = 0, cos 96 πt = 0 or 4πt = 0⇒ 96 πt = (2n + 1) (π/2) and 4πt = (2m + 1) π/2For 0 < t < 1

– 2

1 < n < 95.5 and –

2

1 < m < 3.5

Here n and m are integers, therefore netamplitude becomes zero 100 times.

22. (A) →p, (B) →p; q, (C) →p; r, (D) →s

(i) 2 2 2 2

1 2

1 1 9 9

2 4H H

R Rn n

− = −

Putting 1 2

6; 12n n= = makes both sides equal.

(ii) Energy of infrared radiation is less than theenergy of UV radiation

∴ (b) is correct

4 3E E→ 1 12 2

1 1 7

3 4 9 16hv E E

= − = ×

12 6E E→ 1 12 2

1 1 108

6 12 36 144hv E E

= − = ×

∴ 12 6 4 3E E→ →<

∴ (a) is also correct(iii) a,c

(iv) (d) 2 n

n

n

rT

v

π=

nn

n

TV

γ⇒ ∝ 3

nT n⇒ ∝

23. (A) →p, (B) →q, (C) → r, (D) →s

(i) (a) k

mω =

(ii) (b) Mean position is mg

k below the unstretched

position of the spring ∴ mg

Ak

=

2 20k

fm

ω π π= = = ∴ 2

1

400

m

k π=

( )max 2

120 /

400 2

gV A m sω π

π π

= = =

(iii) (c) sin( )x A tω φ= +

at 0t = / 2x A=

∴ 1

sin2

φ= / 6, 5 / 6φ π π⇒ =

since velocity is negative ∴ 5 / 6φ π=

(iv) (d)2 A

v AT

πω= = ;

2 4 24

/ 2 2

A A Vv V

T T π π< > = = = =




C H E M I S T R Y

24.(c)

D|

CHCHCHCH 23 =−−

→ 2D

D|

CHCHCHCH||DD

23 −−−

+

DD||

DCHCHCHCH 23 −−−

25.(b) Greater the stability of carbonium ion higher the

rate of precipitation :-

II.

+

Aromatic → 6 eπ most stable

III.

+

Aromatic → 2 eπ less stable

than II.

IV. 3 2CH CH CH= −

+Resonance stabilised

less stable than III.

I.

+

Unstable

26.(d)

III.

2 2CH CH−

OH

3CH CH−

OH

+ +1,2 hydride

shift (p)

1° 2° more stable

II.

2 3CH CH−

OH

2 3CH CH−

OH

+

+1,2 hydride

shift (r)

2° 3° more stable

27.(c)

)(OH|

CHCHCHCH 323

+

−−− + CH3 – CH

2 – CH

2 – CH

2 – OH

+ )(

COOHCHPh|CH3

±−−

+(H )

CH – CH – CH – CH – O – C – C – Ph3 2 2 2

|| |O CH3

(±) enantiomers (racemic mixture)

and

Ph – CH – C = O|

CH3

|O|

CH – CH – CH – CH3 2 3

(++) and (+–)(two diastereomers)

Total no. of fractions = 3.

28.(c) In compound ‘C’ the two COOH groups aretrans and there is no possibi l i tyof

intramolecular H bonding.

29.(a) In case of chloroderivative of benzene stabilityof carbonium ion is decided mainly on the basisof inductive effect.

∴ Cl 2CH+

Benzyl ic carbonium ion has only onechlorophenyl hence it is most stable.




30.(b)

+

2

1 mol is 11200 ml of gas at STP..

31.(d)

32.(c)

3 2CH C MgBr Br CH− − + − −

3CH

3CH

3CH

3CH

3C CH− −

33.(b) both reactions takes place through carboniumion hence.

∴ reason is not correct.

34.(a)2CH CH CH

•

= − → conjugation between odd

electron and = bond

3 2CH CH CH•

− − → no conjugation no reso-

nance (unstable)

35.(b) Both 1 & 2 are true but 2 is not the correctexplanation of 1.

36.(d)

C O O C O O C= + = = + =3CH

H

D

H

Product of reductive ozonolysis

D

HC C C= =

3CH

H

37.(b) Y → Gives tollen’s test and forms acetic acid

∴ Y is acetal dehyde

38.(a)

Zn

O3 → +

HCCH||O

3 −− →reagents'Tollen

→ CH3COOH 'Z'

39.(a)

Brno reaction

NaCN

CNΘ

attack is not possible at arylic place

Hence,

Br MgBr,Mg ether

COOH2CO

H +

40.(b) Grignard reagent formation is not favoured withthe molecule having functional group e.g., ester,nitro, etc.

41.(d)

2CH Br−

2CH Br−

2CH

2CH

,Mg ether

vicinal halide Alkene

2CH CN

2CH CN

2CH COOH

2CH COOH

2 ,H O HCl

Heat

2NaCN

Sol. 42, 43, 44Explanation :NBS- gives selective bromination,following reaction is elimination.

Hence, 8 10

C H (unsaturation - 4) is

2 3CH CH




2 3CH CH

3CHBr CH−

NBS→

2CH CH=

t BuO−−→

' 'A ' 'B ' 'C

3C CH− 2 3CH CH−

PCC←

CH O=

3

3

CH MgBr

H O+←

' 'F ' 'E ' 'D

O OH

HCH O+ =

reductionozonolysis reductive

42.(c) 2 2,Br H O is good oxidising agent. Aldhyde is

oxidised to acid but ketone is not oxidised evenalcohol is oxidised to carbonyl group.

43.(b)

44.(b)

2 2CH CH=

3

2 2

1.

2. ,

BH THF

H O OH−

⋅→

2 2CH CH OH− −

45. (A) →s,(B) →p, (C) →s, (D) →s

3 2

3 3

A CH CH COOH

CH COOH

→ −

Both will give positive litmus test

s

2

3 2

B CH CH COOH

CH CH COOH

→ = −

− React with 2Br water p

2CH CH COOH Br OH+

= − + − →

2HO CH CH COOH− − −

Br

3 2 2CH CH COOH Br+ 3

|CH CHCOOH

Br

→ −

α − bromo acid

Hell-volhard-zelinsky ( )HVZ reaction

C Ph OH

Ph COOH

→ −

−

Both acidic, Both will give positive litmus test

s}

3 2D CH NH→ −

3 2CH C NH− −

O

Both basic, Both will give positive litmus test

s}

46. (A) → r,, (B) →s, (C) →p, (D) →q

3A CH MgBr→ − + 2 2Ph C CH CH Cl− − − −

O

2 2Ph C CH CH Cl− − − −

3CH

OMgBrH OH

2 2Ph C CH CH Cl− − − −

3CH

OH

(r)

(B) → s

2 2 2 2 3CuLi PhCCH CH Cl PhCCH CH CH+ →

( )3 2CH

O O

(C) → p

( )4

2 2 2 34LiAlH

CPhCCH CH Cl H PhCHCH CH+ →

O OH

(D) → q

2 2PhCCH CH Cl/

2 2 2Clemensen'sReduction

Zn Hg HCl PhCH CH CH Cl− ∆→




M A T H E M A T I C S

47.(a) Put ( ))xn(nn lll = t

∫e

1t

dt =

e

1tnl = 1

48.(b)

2 21

0( )

(2 )

x tf x dt

t

+=

−∫

2 21 1

0 0(2 ) (2 )

x tdt dt

t t= +

− −∫ ∫

21 1

2

0 0

1

(2 ) 2

tx dt dt

t t= +

− −∫ ∫

( )1 1 112

0 0 0 0

4log 2 2

2x t t dt dt dt

t= − − − − + −∫ ∫ ∫

[ ] [ ] [ ]1

21 12

0 0

0

log1 log2 2 4 log(2 )2

tx t t

= − − − − − −

2

2

1( ) log 2 2 4( log 2)

2f x x= − − − −

2 5( ) log 2 4log 2

2e ef x x= + −

( ) 2 5( ) log 2 ( 4)

2ef x x= + −

2( 4)y K x K′⇒ = + +

it is a parabola.

49.(d) 2.5

10

cos

12

xI dx

x

π

π

=

+

∫

0

22

cos

12

x dxI

xπ

π

−=

+

∫

2 2.5 0

1 20 2 2

cos cos cos

1 1 12 2 2

x dx x dx x dxI I

x x x

π π

π π

π π π

−+ = + +

+ + +

∫ ∫ ∫

Now

2 2.5

1 22 2

cos cos

1 12 2

x dx x dxI I

x x

π π

π π

π π

−+ = +

+ +

∫ ∫

Q x is not integral multiple of π

so cos

12

x

x

π

+

is an odd function

2

2

cos0

1

2

xdx

x

π

π

π

−∴ =

+

∫ 2.5

1 22

cos

12

xdxI I

x

π

π

π

⇒ + =

+

∫

2.5

2

cos

12

2

xdx

π

π

=

+∫

2.5

2

2 2cos

5 5x dx

π

π

= =∫

50.(b)3

y x x= −

2(1 )y x x= −

-1 10

( )a

3

1

1x x d x− =∫

a2 4

1

12 4

x x ⇒ − =

⇒

2 4 11

2 4 4

a a− − =

⇒2 4

1142 4

a a− = ± +

⇒2 2

11 142 2

a a − = ± +

puting 2a z=

35142 2 4

z zor

⇒ − = −

⇒ 22 5 3z z or− = −

22 5 0z z⇒ − + = ( has no solution )

2 2 3 0z z⇒ − − =

( 3)( 1) 0z z⇒ − + =

2 3a⇒ =

3a⇒ = ±

51.(d) The equation of the tangent at the point R(x,f(x))

is - ( ) ( )( - ). Y f x f x X x′=

The coordinates of the point P are

(0, ( ) - ( ))f x xf x′




The slope of the perpendicular line through P is

( ) ( )( )

' 1

'1

f x xf x

f x

−= −

−

( ) ( ) ( )( )2

' ' 1f x f x x f x⇒ − =

1dx

dyx

dx

ydy2

=

−⇒ is D.E. of the curve.

52.(d)

(0, 2)

(1, 1)

(2, 0)

P

B

A

2 3 1 0x y+ + =

The required point will lie on the perpendicularbisector of AB & simultanously on the line

2 3 1 0x y+ + = Equation of AB will be 2x y+ = .

Therefore the perpendicular bisector of AB has

equation y x= ⇒ Point of intersection of

0&2 3 1 0x y x y− = + + = will be 1 1

,5 5

− −

, which

is the required point makes

min . . 0PA PB i e− =

53.(b) Equation of line passing through ( )3,0 is

y – 0 = m(x – 3)

If above line will touch parabola 2y x= then

mx – 3m = x2

x2 – mx + 3m = 0now D = 0m2 – 12m = 0m = 0 or m = 12y = 12x – 36

54.(c) 2 2

1 4 7 0S x y x= + + + =

2 2

2

3 5 90

2 2 2

x yS x y= + + + + =

2 2

3 0S x y y= + + =

radical axis of S1 & S

2 is

S

1 - S

2= 0

i.e. 1 0x y− + =

similarly radical axis of 2 3&S S is S2-S

3=0

i.e. 3 0x y+ + =

∴ required point is intersection of 1 0x y− + = &

3 0x y+ + =

i.e.(-2,-1)

55.(b) a1e

1 = a

2e

2

b2 = a12(1 – e

12) ⇒ b2 = a

22(e

22 – 1)

Equating b2 we get 2e

1

e

12

2

2

1

=+

56.(d) Assertion is wrong because 0,4

xπ

∀ ∈

tan tannx x< but reason is correct. because for

1b a> >n

x x> .

57.(a) Solution of order 3 should have 3 arbirtraryconstants and Equation of all circles in a plane

is given by 2 2 2 2 0x y gx fy c+ + + + =where g,f,c are an arbitrary . Hence reason istrue and also since if the 3 points are noncollinear points then the system formed in g , f,c obtained satisfying by these points have uniquesolution therefore there is only one circle through

3 non collinear points .

58.(a) Equation of director circle of

2 2

2 21

x y

a b+ = is

2 2 2 2x y a b+ = + . Also given ellipse can be

written as

2 2

116 9

x y+ = ⇒ Its director circle is

2 2 25x y+ = . Since ( )3,4 lies on it. Hence

angle between the tangents drawn form it to

the ellipse is 90°

59.(a) ( )2 24 0 4 2 3P A P B= = − + − = ; circum

centre of triangle PAB is mid point of PO as

090 ;PAO PBO∠ = ∠ =

so, ( )4 0 0 0

, 2, 02 2

− + + = −




D

θθ 2θ

O

32

60.(d) The required ratio is 1 : 1

61.(a) P’ be a point on the circle. 62

3π

=θ⇒π

=θ

Area of the rhombus ( ) 363332 =×= .

Dθθ 2θ

O

32

62.(a) 2 2 2( 1) ( ) 2f x f x x+ = +

( 1) ( ) 2f y f y y⇒ + = +

( 1) ( ) 2f y f y′ ′⇒ + = −

( 1) ( )f y f y′′ ′′⇒ + =

( )f y⇒ is a polynomial not more than

degree 2.

2( )

(0) 0 0

(1) 0 (1) 0

(2) 4 2

2 4 2

2 1

Let f y ay by c

Given f c

also f a b a b f

f a b c

a b

a b

∴ = + +

= ⇒ =

= + ⇒ + = =

= + +

= +

⇒ + =

Q

1

1

a

b

=

∴ = −

2( )f y y y∴ = −

3 2( )g x x x∴ = −

which is not differentiable at 1x = only

63.(c)

1 1

3 2

0 0

1( ) ( )

12g x dx x x dx= − = −∫ ∫

0A A∴ + =

64.(b) Min { }( ) 0

( ), ( )( ) 0

g x xf x g x

f x x

≤=

≥

∴ Required area

2

2

1

5( )

6x x dx= − =∫

( )g x( )f x

2x =1

0

PASSAGE :

(x - y) f(x + y) - (x + y) f(x - y) = 4xy (x2 - y2)

⇒2 2f (x y) f (x y)

(x y) (x y)x y x y

+ −− = + − −

+ −

⇒2 2f (u) f (v)

u vu v

− = −

⇒ 2 2f (u) f (v)

u v cu v

− = − =

⇒ f(u) = cu + u3 ⇒ f(x) = cx + x3

As f(1) = 2, c = 1 ⇒ f(x) = x + x3

65.(d) f(x) = x3 + x

11 5 3

1 1

16( )

5 3 15

x xx f x dx

− −

= + =

∫

66.(d) Required area =

0 1

1 0

3( ) ( )

2f x dx f x dx

−

+ =∫ ∫ sq. units

67.(a) (1) 4f ′ =Equation of tangent at (1, 2) is4x - y -2 = 0Hence area = 1/2 sq. units.




68. (A) →s, (B) →p,(C) →q, (D) → r

(A) OA = 1 + 4cot θOB = 4 + tan θOA + OB = 5 + 4 cot θ + tanθ

5 2 4cot .tanθ θ≥ + = 5 + (2 x 2) = 9

B

D

O C A

(1.4)

θ

θ

(B) Reflection of Q(4, -1) in y = x is P (-1, 4)

( ) ( )2 2

4 1 1 4 50 5 2PQ = + + − − = =

(C) 2 2AB =

2OE =

Max. value of d OF=

2 2 2 3 2= + =

B(0,2)

E

C

F

X45

o

A (2, 0)

d

O

y

D

(D)1 1

4 2 9.2 2

yx y x

+ = + ⇒ = −

Length of intercept made by x-axis is 9

2

69. (A) → r, (B) →s,(C) →p,(D) → r

(A)A

B

C7,1

O (0,0)

2

2045

2

2

7 4x y+ =equation of AB=

D

required area = 2 area ∆(AOC)- sector area AOBD

=22 2 1

2 22 2 2

π× − × ×

4 π= −

(B)

1

2O A

BC

required Area = Area of rectangle OABC -

2

π=

42

2 2

π π−− =

(C) The conic will be either ellipse or hyperbola

because both are symmetric about &x y axis

and order of its differential equation will be 2.

so 2

2

K

k

ππ

− =− =

(D)

b

a




Area of ellipse

2

204 1

a xb dx

a= −∫

2 2

0

4 aba x dx

a= −∫

22 2 1

0

4sin

2 2

a

b x a xa x

a a

− = − +

214

sin (1) 02

b a

a

− = +

=

24

2 2

b a

a

π× ×

abπ=

⇒ 4ab ab Kabπ− = ⇒ 4 kπ− =

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