Iit-jee 2012 Fst2 p2 Solns

25

♣ Brilliant Tutorials Pvt. Ltd. IIT/B.MAT2/CPM/P(II)/Solns

CHEMISTRY

TEST-II

PART-I

SECTION–I

Single Correct Choice Type

1. (B) 2 4

4 2MnSO MnO+ +

→

The equivalent weight of 4

mol wtMnSO

2=

i.e., 0.2 M MnSO4 ≡ 0.4 N MnSO4

Normality of KMnO4 solution = 0.05 N

Therefore 20 × 0.4 = 0.05 V

20 0.4

V 160 mL0.05

×= =

2. (D) 2.303 56 2.303

K log log 460 14 60

= =

12.3032 0.3010 min

60

−= × ×

0

dNKN

dt− =

rate = decay constant × initial number of nucleides on substitution

0

2 2.303 0.301056 N

60

× ×= ×

0

56 60 1680N

2 2.303 0.3010 0.693

×= =

× ×

= 2400 (approx)

3. (C) Molecular weight of phenol in solution f

f

1000 K w

W T=

⋅ ∆

1000 5.12 20

1 1000 0.69

× ×=

× ×

= 148.4

Actual mol. Wt. of phenol ( )6 5C H OH 94=

Hence association of phenol in benzene takes place.

( )6 5 6 5 22 C H OH C H OH

1 − α 2

α

26


Number of particles before association = 1

Number of particles after association 12

α= − α +

12

α = −

1Normal mol.wt 942

obs.mol.wt 1 148.4

α−

= =

148.4 − 74.2 α = 94

148.4 94 54.4

0.73374.2 74.2

−α = = =

% association = 73.3%

4. (A)

3CH

( )3

3 2

CrO

CH CO O→

( )3 2CH OCOCH

3H O+

∆→

CHO

5. (A) β-keto acids on heating easily undergo decarboxylation via six-membered cyclic

transition state

OC

2CH

C

O

H

O

2CO−→ 6 5C H C−

OH

2CH

6 5 3C H C CH→ − −

O6 5C H

6. (B) 2CH C

2CH O

O

3CH OH→

2CH C

2CH O

O−

3O CH−

H

+

H⊕−→ 2CH 3C O CH− −

2CH O

O

→2 3CH C OCH− −

2CH O−−

O

H2 2 3HO CH CH COOCH

⊕

→ − − −

7. (A) Trans ( ) 22Co en Cl

+ does not exhibit optical isomerism others exhibit optical

isomerism.

8. (D) As the size of the cation decreases its acidic strength increases. As the positive

charge decreases, the acidic strength increases. Hence the strength of acids follow

the order 3 < 1 < 2 or ( ) ( ) ( )3 2 3

2 2 26 6 6Al H O Fe H O Fe H O

+ + + < <

27


SECTION–II

Multiple Correct Choice Type

9. (A, D)

22 5 2 5 2 2CO

C H CH COOH C H CH NO∆

−− − → − −

2NO

(optically active)

(optically inactive)

O O does not exhibit tautomersim as the α − H atom is attached to sp2 carbon.

Cyclopenta dienyl cation (4n rule) is anti aromatic. In E1CB mechanism, the first step is the

ionization of the substrate giving H+ and carbanion.

10. (B, C)

2 2(g)2 2H O 2e H 2 OH− × + → +

(cathode)

2 2(g)4 OH 2H O O 4e− → + +

(anode)

Add

2 2(g) 2(g)2H O 2H O→ +

is cell Redox reaction

11. (A, B, D)

2 6 0Fe 3d 4s+ 4 unpaired e−

3 6 0Co 3d 4s+ 4 unpaired e−

3 7 0Ni 3d 4s+ 3 unpaired e−

3 4 0Mn 3d 4s+ 4 unpaired e−

Fe+2

, Co+3

and Mn+3

have the same no. of unpaired electrons and hence have the same

magnetic moment n(n 2)+ i.e., 4 6 4.85 BM× =

12. (A, B, D)

HClO is the strongest oxidising agent and weakest oxy acid. HClO4 is the strongest oxy acid

and weakest oxidising agent. Solubility in water decreases from BeSO4 to BaSO4 due to

decrease in hydration energy while lattice energy of these salts is almost constant.

28


SECTION−−−−III

Integer Type

13. Cr(24) 1s2 2s

2 2p

6 3s

2 3p

6 4s

1 3d

5

Electron in the d-orbital has = 2 value.

Ans. 5

14. Number of chiral carbon atoms in the molecule 3 3CH CHCl CHBr CHCl CH− − − − is 3.

This molecule has plane of symmetry. Hence meso isomer is possible. Two meso forms are

possible.

3 3CH C CH C CH− − − −

H H H

Cl Br Cl

and

3 3CH C C C CH− − − −

H

H

H

Cl

Br

Cl

are meso isomers.

Ans. 2

15. PM

dRT

= or dRT

PM

=

x

x

dx RTP

M= and y

y

dy RTP

M=

yx

y x

MP dx3 2 6

P dy M

= = × =

Ans. 6

16. In diamond carbon atoms are present in corners, face centres and 50% of tetrahedral sites of

a unit cell. Hence effective number of carbon atoms per unit cell is 1 1

8 6 4 88 2

× + × + =

Ans. 8

29


17.

P

O

OO

P

OP

O

P

O

O O

O

O

There are four shorter P − O bonds in P4O10 which have double bond character.

Ans. 4

18.

C C=Br

Cl

F

IC C=

Br

Cl F

I

C C=Br

ClF

I

C C=Br Cl

F I

C C=Br

Cl

F

IC C=

Br Cl

FI

Ans. 6

SECTION−−−−IV

Matrix-Type

19. (A) – (r) ; (B) – (p), (q), (s) ; (C) – (p), (q), (s) ; (D) – (s), (t)

20. (A) – (s) ; (B) – (r), (t) ; (C) – (p) ; (D) – (q)

30


PHYSICS

PART-II

SECTION–I


21. Let us suppose that the rate of flow (volume flowing per second) V depends on (i) coefficient

of viscosity η (ii) radius of cross-section r and (iii) pressure gradient P

, where is the

length and P is the pressure difference.

zx y P

V K r

= η

, K-dimensionless constant.

3 1 x x x y z 2z 2zL T M L T L M L T

− − − − − =

x z x y 2z x 2zM L T+ − + − − −= ⋅ ⋅

∴ x + z = 0, − x + y − 2z = 3 and − x − 2z = − 1

i.e., x + 2z = 1 or x + z + z = 1

∴ z = 1, x = − 1, y = x + 2z + 3 = 4

4

1 4 P KrV K r P−

∴ = η = ⋅ η

Viscous resistance 1

4

PR K

V r

η= =

∴∴∴∴ (C)

22. In (a), Q is given by

Q = Q0 (1 − e− αt

)

n Q = n Q0 + n (1 − e− αt

)

⇒ No log graph (neither b nor c)

In (d),

Q = Q0eαt

, n Q = n Q0 + αt ⇒ (C)

∴∴∴∴ correct choice is (D)

23. Let at t = 0, train T be just crossing O and

car be at C.

1c

90 5v 90 kmph 25 ms

18

−×= = =

1 1t

5v 126 kmph 126 ms 35 ms

18

− −= = × =

Velocity of train w.r.t car : tc t cv v v= −

60°

tv

cvxC

O

y

T

31


1ctv (35 cos 60 , 35 sin 60 ) , v ( 25, 0) ms−= ° − ° = −

1tcx

85v 35 cos 60 25 42.5 ms

2

−∴ = ° + = =

1tcy

3v 35 sin 60 35 30.3 ms

2

−= − ° = − × = −

2 2 1tcv 42.5 30.3 52.2 ms−∴ = + =

At t = 0, position of the car C [It takes 3 min to cross O]

= (4.5 km, 0)

Position of train T = (0, 0)

(5 min. after crossing O)

At t = 8 min, position of train T w.r.t. car is

x = 42.5 × 8 × 60

= 20.4 km

y = − 30.3 × 8 × 60 = − 14.5 km

2 2d x y 25 km∴ = +

∴∴∴∴ (D)

24. Max. acceleration

2 22v dva 2

r dt

= + =

22 2dv v4

dt r

= −

4

2

1 v4 1

4 r

= − ⋅

4dv v

2 1dt 4r2

∴ = −

Now, 4 4

2 6

v 256 10

4r 4 10

×=

× = 0.64

Hence maximum rate of decrease 22 1 0.64 1.2 ms−= − =

∴∴∴∴ (A)

25. FBD of B : 2TF

Ba0.5 kg

B

F − 2T = mBaB … (1)

FBD of A:

T

Aa

2 kg

Agm

Am

T − mAg = mAaA … (2)

32


By constraints, aA = 2aB

∴ T − mAg = 2mAaA or 2T − 2mAg = 4mAaB … (3)

Adding (1) and (3): F − 2mAg = (4mA + mB) aB

A A B BF 2m g (4m m ) a∴ = + + … (4)

Given 2

2B

v 16a 16 ms

2s 2 0.5

−= = =×

∴ F = 40 + (8.5) × 16 = 40 + 136.0 = 176 N

∴∴∴∴ (B)

26. Resistance of the wire AB : 2

Rr

ρ=

π

, ρ -resistivity of the material

dR d 2dr 0.05 0.5

2R r 0.5 5

∴ = − = + ×

= 0.3 … (1)

For unstrained wire,

Voltage at D : o oD

v R vv

R R 2

×= =

+

Voltage at B : oB

vv

2=

∴ P.D. between B and D = 0

For strained wire,

o o oD B

v v (R dR) v (R dR)v , v

2 R R dR (2R dR)

+ += = =

+ + +

A

B

C

D

RR

RR

ov

oB D

vv v v [2R 2dR 2R dR]

2(2R dR)∆ = − = + − −

+

o o

dR

v dR v R

dR dR22R 2 2

R R

= = ⋅

+ +

∴ % change o

v 0.3 100 15100 6.5%

v 2(2 0.3) 2.3

∆ ×= × = = =

+

∴∴∴∴ (D)

27. 15 = 10I + 10 (I − I1)

= 20I − 10I1

∴ 3 = 4I − 2I1

or 13 2II

4

+= … (1)

10 Ω10 Ω

10 Ω

LR15 V

1(I I )−

I 1I

33


(10 − RL) I1 − 10(I − I1) = 0 … (2)

11 L

(3 2I )I (10 R 10) 10I 10

4

++ + = =

I1 (40 + 2RL) = 15 + 10Il

O1

L L O L

E15 7.5I

30 2R 15 R R R= = =

+ + +

∴ EO = 7.5 V

RO = 15 Ω

∴∴∴∴ (C) 7.5 V

15 Ω

LR

28. Let rm be the minimum distance of the

satellite (at A) in the orbit from the centre

of the earth and Vmax be the maximum

velocity. At the point P of launching, let

velocity Vo = 104 ms

− 1 make an angle φo

with the radius vector joining the satellite

to the centre of earth at a distance ro .

or oφ

maxv

E

mrP

ov

A

By the conservation of energy and angular momentum at point A and P,

2 2max 0

min o

1 GMm 1 GMmmV mV

2 r 2 r− = − … (1)

min max o o omr V mr v sin= φ

omax o o

min

rV V sin

r= × φ

ro = (6400 + 400) km = 6.8 × 106 m

rmin = (6400 + 200) km = 6.6 × 106 m

4 4max o o

6.8V 10 sin 1.03 10 sin

6.6∴ = × φ = × φ

2 8 2 8 2

o

min o

1 1 1 11.03 10 sin 10 gR

2 2 r r

∴ × × × φ = × + −

Here, GM = gR2 = 10 × 6.4

2 × 10

12

2 12

2o 2 8 6

1 20 6.4 10 0.2sin

1.03 10 6.8 6.6 10

× ×∴ φ = + ×

× ×

2

10.037 0.94 0.037

1.03= + = +

0.98

sin φo = ± 0.9899 ⇒ φo = 82° or 180° = 90° ± 18°

∴∴∴∴ (D)

34


SECTION–II


29. T + qE − mg = ma1

3 6 61

5010 10 20 10 500 10 10 ma

0.1

− − −× + × × − × × =

3 3 3 6110 10 10 10 5 10 500 10 a− − − −× + × − × = ×

qE

1a

mg

T

+

3

21 4

15 10a 30 ms 3g

5 10

−−

−

×∴ = = = ↑

×

OR : mg + qE − T = ma2 ↓

3 3 3 4 22 25 10 10 10 10 10 5 10 a , a 10 ms g− − − − −× + × − × = × = = ↓

∴∴∴∴ (B, C)

30. Vde = 60 V

1I in 10 6A∴ Ω =

2I in 20 3AΩ =

3I in 30 2AΩ =

∴ current incoming at a = current outgoing at b

∴ I = I1 + I2 + I3 = 11 A

To find Vcd,

4(I4) = 8 (11 − I4) = Vcd

∴ 12I4 = 88

4

88 22I A

12 3= =

4 Ω

10 Ω20 Ω30 Ω

8 Ω

6 Ω

e

d

c

1 2 3I I I I= + +a

b

1I2I3I

5I 4I

I

e

c

cd

22 88V 4 29 V

3 3∴ = × =

ab ac cd de ebV V V V V= + + +

88 88 378

0 60 663 3

+= + + + =

466

3= = 155 V

∴∴∴∴ (B, C, D)

35


31. Process a → b : V ∝ T : isobaric heating

(P = constant)

a

bc

TO

V

ab

c

TO

P

a b

c

VO

P

Process b → c : isochoric cooling (V = constant) P ∝ T

Process c → a : isothermal (T = constant) compression

PV = constant : P is increasing with V decreases

∴ correct graphs are (C) and (D)

∴∴∴∴ (C, D)

32. When S is in position 1 : 11

di10 0.5 50 i

dt= +

i.e., 11 1

di20 100 i 100 (i 0.2)

dt= − = − −

100 t11

1

di100 dt i (t) 0.2 Ke

i 0.2

−∴ = − ⇒ = +−

At t = 0, i1 = 0.

Hence, K = − 0.2 or i1(t) = 0.2 (1 − e− 100 t

)

50 Ω

0.5 H2

10 V10 V

1

S

At t = 10 ms 1

s100

= , i1 = 0.2 (1 − e− 1

) = 0.126 A = 126 mA

∴ P.D. across resistor just before t = 10 ms = 50 × 0.126 = 5 × 1.26 = 6.30 V

When S is in position 2 (at t′ = (t − 10) ms = 0)

22

di50 i 0.5 10

dt+ = −

′

100 t2i (t ) 0.2 K e

′−′ ′∴ = − +

At t′ = 0, i2 = 0.126 A

∴ K′ = 0.326 A

i2(t′) = 100 t0.326 e 0.2

′− −

36


Now i2(t′) = 0

when 100 t 0.326e 1.63

0.2

′= =

100t′ = n 1.63 = 0.49

0.49

t s 4.9 ms (t 10) ms100

′∴ = = = −

∴ t = 14.9 ms

Further, as t′ → ∞, i2 = − 0.2 A = − 200 mA

After closing S in position 2, P.D. across resistor = 6.3 V

∴ P.D. across inductor = − 10 − 6.3

= − 16.3 V

Alternate

100 t

2i 0.326 e 0.2′− = −

2

t 0

L di100 0.326 0.5 16.3 V

dt ′ =

= − × × = −

∴∴∴∴ (A, B, C, D)

SECTION −−−− III

Integer Type

33. Potential in central field V(r) = kr3

If the particle describes circular motion of radius a, then attractive force d dV

[mV(r)] mdr dr

− = −

should be equal to centrifugal force in magnitude. If v is the speed of the particle,

2

2mv3 mka

a∴ = or 3v 3 ka=

Angular velocity 3v 3ka

3 kaa a

ω = = =

∴ angular momentum 2 2L mr ma 3ka= ω =

Let the particle be displaced slightly from its circular orbit, so that r = a + x, x << a

Total energy 2

2

2

1 LE mr mV(r)

2 2mr= + +

Let 2 4

3 3

2

m r 3kr 5U (r) mkr mkr

22mr

×′ = + =

Hence 2 31 5E mr mkr

2 2= +

37


Differentiating w.r.t. time,

215mr r mk.r r 0

2⋅ + =

or 215r kr 0

2+ = r a x, r x= + =

2 2 2 2r (a x) a 2ax x= + = + +

= a2 + 2ax, x << a

215x 15 kax ka 0

2∴ + + =

Let 1

X x ka2

= + X 15kaX 0⇒ + =

∴ angular frequency 15 ka 3 nkaω = =

∴ n = 5

Ans.: 5

34. Charges on C1 and C2 are q1 = C1V, q2 = C2V

After the connection, Q = (C1 − C2) V.

Let 1

q′ and 2

q′ be the final charges on C1 and C2 respectively, such that

1 21 1 22

1 2 1 2 1 2 1 2

q q q q (C C )QV

C C C C C C (C C )

′ ′ ′ ′+ −= = = =

+ + +

1 1 21

1 2

C (C C )q V

(C C )

−′∴ =

+ and 2 1 2

21 2

C (C C )q V

(C C )

−′ =

+

Now, electrostatic energy before connection

21 1 2

1E (C C ) V

2= +

After connection 22

21 22

1 2 1 2

(C C )1 Q 1E V

2 (C C ) 2 C C

−= =

+ +

Loss in energy ( ) ( )2

2 2

1 2 1 2 1 2

1 2

1 VE E C C C C

2 (C C ) − = + − − +

2

21 2 1 2

1 2 1 2

2C C C C VnV

(C C ) 2 (C C )= ⋅ = ⋅

+ +

∴ n = 4

Ans.: 4

38


35. Let h be the height of the cliff. Parallel incident

ray AB and reflected ray (from sea) CDB from

distant star interfere with each other in the field

of view of telescope.

The geometrical path difference between the

rays

δ = DB − EB = DB − DB sin 16°

= DB (1 − sin 16°)

But h

sin 37 0.6DB

= ° =

D

cliff

B

E

A

C

37°

h

37°

16°

sea

h 7.2

(1 0.28) h 1.2 h0.6 6

∴δ = − = ⋅ =

Condition for destructive interference is (Taking reflection from denser medium)

δ = nλ, n = 0, 1, 2, …..

Now, n = 0 corresponds to h = 0. Hence for n = 1,

1.2 h = 240

240

h 200 m1.2

= =

= 0.2 km x

km10

=

∴ x = 2

Ans.: 2

36. Maximum kinetic energy of photoelectrons

Kmax = hν − φ

Photon energy 34 14

19

6.62 10 6 10h eV

1.6 10

−

−

× × ×ν =

×

2.5 eV

∴ Kmax = 2.5 − 2 = 0.5 eV

Positive potential to be developed at the surface = 5 n

0.5 V V V10 10

= =

∴ n = 5

[If V > 0.5 V, no electrons leave the surface]

Ans.: 5

39


37. N0 = 82 − 10 = 72 cps

N1 = 19 − 10 = 9 cps

3

0

N 1 1

N 8 2∴ = =

Let T be the half life. Then, t = 3T = 180

T = 60 s = 1 min

Ans.: 1

38.

2

2 1Z R L

C

= + ω −

ω

i

22

VI

1R L

C

=

+ ω −

ω

O2ViV , ω

R LC

O1V

Output

2 2 2i

O12 2 20

V R L CV lim

RC ( LC 1)ω→

+ ω ⋅ ω=

ω + ω −

= ViRC ⋅ ω

Output iO2 i

2 2 20

V CV lim V

C RC ( LC 1)ω→

⋅ ω= =

ω ω + ω −

O1

O2

VRC

V∴ = ⋅ ω

∝ ω = ωn ⇒ n = 1

Ans.: 1

SECTION–IV

Matrix Type

39. (A) – (s), (B) – (p, q), (C) – (t), (D) – (r)

(p) F = αx, F ⋅ dx = dK 21K x

2∴ = α

K versus x is a parabola, because

dK

F (B)dx

= O

K

x

(q) 2 UU 500 60x F 120 x

x

− ∂= + ⇒ = = −

∂

120

a x 24 x5

= − = −

For SHM, 2 1a x 24 rad. s−= − ω ⇒ ω = (B).

40


(r) For conservative electric field, C

E d 0⋅ =∫

and for non-conservative field, C

E d 0⋅ ≠∫ (D)

(s) 1 2

1 1 1 1( 1) , f

f R R ( 1)

= µ − − ∝

µ −

V R

R V

f ( 1)1

f ( 1)

µ −= <

µ −, because R Vµ < µ (A)

(t) F qV B F V= × ⇒ ⊥

No work done, ∆K = 0

But P mV=

changes (in direction) (C)

due to F

on charged particle.

40. (A) – (q, r, s); (B)–(r); (C)–(p, r, s, t); (D)–(p, q)

1S

2S

10 Ω 5 Ω

5 mH60 VI

A

B

200 Fµ

(p) S1 closed, S2 open: AB

60t 0 , I 6A, V 0

10

+= = = =

6 3RC 10 200 10 2 10 s− −τ = = × × = × = 2 ms

(q) S1 closed, S2 open: ABt , I 0, V 60 V→ ∞ = =

τ = 2 ms

(r) S1 open, S2 closed: 60

t , I 4 A15

→ ∞ = = , VAB = 4 × 5 = 20 A

3L 5 10

sR 5

−×τ = =

= 1 ms

(s) S1 and S2 closed: at steady state : AB

60I 4A, V 20 V

15= = =

(t) At t = 0+ s (S1 and S2 closed): AB

60I 6A, V 0

10= = =

41


MATHEMATICS

PART-III

SECTION – I


41. (B) α, β are the roots of x2 + x + 1 = 0 ⇒ α, β are the complex cube roots of unity so

that α + β = − 1, αβ = 1.

∴ sum of the roots of the required equation = 2(α4 + β4

) + β + α

= 2(α + β) + (β + α) 3 3( 1)α = β =∵

= − 3

Product of the roots = (2 ) ( 2 )α + β α + β

2 22( ) 5= α + β + αβ

( ) 2

2 2 5 2 1 2 5= α + β − αβ + αβ = − + = 3

Hence the required equation is x2 − ( − 3) x + 3 = 0

⇒ x2 + 3x + 3 = 0

42. (D) Let x[x]

f (x)sin | x |

=

Then left hand limit of f(x) as x → − 1

h 0

( 1 h) [ 1 h] 1 ( 2) 2Lt (h 0)

sin | 1 h | sin 1 sin 1→

− − − − − × −= > = =

− −

Right hand limit h 0

( 1 h) [ 1 h] 1 1 1Lt (h 0)

sin | 1 h | sin 1 sin 1→

− + − + − × −= > = =

− +

≠ left hand limit

∴ limit does not exist.

43. (B) Clearly f(x) does not exist at x = 0 and x = − 3

∴ the domain of 2

2

x 2xf (x)

x 3x

−=

+ is R − − 3, 0

and x 2 5

f (x) 1x 3 x 3

−= = −

+ +

Now 1f (f (x)) x− =

1

51 x

f (x) 3−⇒ − =

+

1 5f (x) 3

1 x

− + =−

( )1

2

d 5f (x)

dx (1 x)

− =−

42


44. (A)

( )

12 9 3 6

3 3 35 3

2 5

2 5dx

2x 5x dtx xdx

t1 1x x 1 1x x

+ + = = −

+ + + +

∫ ∫ ∫ where 2 5

1 1t 1

x x= + +

( )

10

2 25 3

1 1 xc

22t x x 1

= = ++ +

( )

( )

12 91

35 30

2x 5x 1 1 1dx 0

2 9 18x x 1

+ ∴ = − =

+ +∫

45. (D) Let 1 2 3a a i a j a k= + +

Then 1a i a⋅ = , 2 3a j a , a k a⋅ = ⋅ =

( ) ( ) ( ) ( ) ( )1 2 3a i a i a a i a a j a a k∴ ⋅ × = × + × + ×∑

( )1 2 3a a i a j a k a a 0= × + + = × =

46. (C) 2 2 2 2a b c 8R+ + =

2 2 2sin A sin B sin C 2⇒ + + = ( )a 2R sin A etc.,=∵

1 cos 2A 1 cos 2B 1 cos 2C

22 2 2

− − −⇒ + + =

⇒ cos 2A + cos 2B + cos 2C = − 1

⇒ 2 cos (A + B) cos (A − B) + 2 cos2 C − 1 = − 1

⇒ − 2 cos C cos (A − B) + cos (A + B) = 0 ( )cos C cos (A B)= − +∵

⇒ cos A cos B cos C = 0 ⇒ one of A, B, C is a right angle.

47. (C) 21z i z iz 1 0

z− = ⇒ − − =

2z iz 1 0⇒ − + + =

2(iz) (iz) 1 0+ + =

∴ iz = ω or ω2 where ω is a complex cube root of unity.

∴ iz = ω ⇒ z = − iω

( )

2005 2005

2005 2005

1 1z ( i )

z i∴ − = − ω −

− ω

21i i i

i= − ω + = − ω − ω

ω

( )2i i 1 i= − ω + ω = − × − =

43


48. (B) y divides z ⇒ z = ky where k is a positive integer.

2xz 2 1008 ky

yx z 1008 ky

× ×∴ = =

+ +

1008

1008 ky 2016 k k2016 y

⇒ + = ⇒ =−

⇒ k is any divisor of 1008 except 1

Now 4 21008 2 3 7= ⋅ ⋅

⇒ the number of divisors of

1008 = (4 + 1) (2 + 1) (1 + 1) = 30

But this includes 1 also.

∴ the required number of sequences = 30 − 1 = 29

SECTION – II


49. (B, D)

[sin θ + 1] + [cos θ ] = 0 ⇒ [sin θ] + [cos θ] = − 1

⇒ if both [sin θ] and [cos θ] are negative: then RHS = − 2 which is impossible.

∴ one of them is negative and the other is zero.

3

, or , 22 2

π π ⇒ θ∈ π π

50. (A, C)

Consider the numbers a a b b b c c c c

, , , , , , , ,2 2 3 3 3 4 4 4 4

Their

12 3 4 9a b c a b c

A.M G.M9 4 27 256

+ +≥ ⇒ ≥

× ×

2 3 4 9a b c 2 4 27 256⇒ ≤ × × ×

and the greatest value is obtained when

a b c a b c

22 3 4 2 3 4

+ += = = =

+ +

i.e., when a = 4, b = 6, c = 8

51. (B, D)

n n n n 2 n n0 1 2 n(1 x) C C x C x ..... C x+ = + + + +

Also n n n n n 1 n0 1 n(x 1) C x C x ..... C−+ = + + +

∴ coefficient of xn − 1

in n n n n n n n n0 1 1 2 n 1 n(1 x) (x 1) C C C C ..... C C−+ + = ⋅ + ⋅ + + ⋅

i.e., coefficient of xn − 1

in (1 + x)2n

= n n n n n n

0 1 1 2 n 1 nC C C C ..... C C−⋅ + ⋅ + + ⋅

i.e., 2n

n 1RHS C −= 2n

n 1 n 1=

− +

44


n 1

n

2n 2 n 1 n 1s

s n n 2 2n

+ + − +∴ = ⋅

+

(2n 1) (2n 2) 15

n (n 2) 4

+ += =

+

⇒ n2 − 6n + 8 = 0

⇒ (n − 2) (n − 4) = 0

⇒ n = 2 or 4

52. (A, B, C)

P(E1) = probability that the first digit is 4 = 2 1

4 2=

similarly 2 3

2 1P(E ) P(E )

4 2= = =

P(E1 ∩ E2) = probability that the first and second digits are 4 1

4=

similarly 1 3 2 3

1P(E E ) P(E E )

4∩ = ∩ =

1 2 1 2

1P(E E ) P(E ) P(E )

4∴ ∩ = = ∩ etc.

Hence (A), (B), (C) are true.

1 2 3

1P(E E E )

4∩ ∩ = ≠ P(E1) ⋅ P(E2) ⋅ P(E3)

SECTION – III

Integer Type

53. 2

2

cot 2x tan 2x 1 tan 2x2

cot 4x (1 tan 2x)tan 2x

2 tan 2x

− −= =

−⋅

Ans. 2

54. 2

xlog (x 5)x 16

− = ⇒ x > 0, x ≠ 1 and (x − 5)2 = 16

⇒ x − 5 = ± 4 ⇒ x = 1 or 9.

But x ≠ 1 ⇒ x = 9.

Ans. 1

55. Let a and r be the first term and the common ratio of the G.P.

Then p 1 q 1 r 1x aR , y aR , z aR− − −= = =

q r r p p q (q r) (r p) (p q) (p 1) (q r) (q 1)(r p) (r 1)(p q)x y z a R− − − − + − + − − − + − − + − −∴ ⋅ ⋅ =

= a0R

0 = 1

Ans. 1

45


56. The equation of the circle with centre at ( )0, 5 is

( )2

2 2x y 5 r+ − =

i.e., 2 2 2x y 2 5 y r 0+ − − =

For x, y to be rational y = 0 and r2 should be rational.

When y = 0, x = ± r

∴ the points on the circle with rational coordinates are (r, 0) and ( − r, 0)

Ans. 2

57. Let y = mx + c be a tangent to 2 2x y

16 2

+ =

∴ c2 = 6m

2 + 2 … (1)

Given 3 m

tan 451 3m

−= °

+ = 1

3 m

11 3m

−⇒ =

+ or − 1 ⇒

1m

2= or − 2

For each of these values of m there are two values of c from (1)

∴ the number of tangents with the given condition = 4

Ans. 4

58.

n

r 1

n n2 2

rr 1 r 1

n2 2

r 1

1 n n

2 r n n 1 n n 72

(2r 1) n n n 1

=

= =

=

∆ = + + + =

− + +

∑

∑ ∑

∑

2 2

2 2 2

n n n

n(n 1) n n 1 n n 72

n n n n 1

⇒ + + + + =

+ +

2 2 2

2 2 2

1 1 1

n n n n n 1 n n 72

n n n n 1

⇒ + + + + =

+ +

2

2

1 0 0

n n n 1 0 72

n 0 n 1

⇒ + =

+

2 2 1

3 3 1

C C C

C C C

→ − → −

⇒ n(n + 1) − 72 = 0 ⇒ (n + 9) (n − 8) = 0

⇒ n = 8

Ans. 8

46


SECTION – IV

Matrix Type

59. (A) −−−− s; (B) −−−− s; (C) −−−− p; (D) −−−− r

Sol.: (A) 1 14 2 4 2cos cos sin sin 2

3 3 3 3

− −π π π π + = π − + π −

= π

(B) 1 1 1 1 2 3

tan 1 tan 2 tan 3 tan4 1 2 3

− − − − π ++ + = + π +

− ×

1 1 1 a btan a tan b tan if ab 1

1 ab

− − − ++ = π + >

− ∵

1tan ( 1)4

−π= + π + −

4 4

π π= + π − = π

(C) 1 11 1 2cos 2 sin 2

2 2 3 6 3

− − π π π+ = + × =

(D) Let 1 1

2 2

1 1tan tan .... to n terms

2 1 2 2

− −+ + =⋅ ⋅

Then n n

1 1

2 2r 1 r 1

1 2tan tan

2r 4r

− −

= =

= =

∑ ∑

n

1

r 1

(2r 1) (2r 1)tan

1 (2r 1) (2r 1)

−

=

+ − −=

+ + − ∑

n

1 1

r 1

tan (2r 1) tan (2r 1)− −

=

= + − −∑

1 1tan (2n 1) tan 1− −= + −

1

2r 1

1tan

2 4 42n

∞−

=

π π π ∴ = − =

∑

60. (A) −−−− s; (B) −−−− p; (C) −−−− r; (D) −−−− q

Sol.: (A) 2

2

(ydx xdy)xdy ydx xy dx xdx

y

− −− = ⇒ =

2x x x

d xdx cy y 2

⇒ − = ⇒ − + =

2x y 2 (cy x)⇒ = −

47


(B) 2

3 2 2

2

dy y xyx dy x ydx xy dx

dx x

−+ = ⇒ =

put y = vx

2dvv x v v

dx∴ + = −

2

dv dx 1 1 dxdv 2

x v 2 v xv 2v

⇒ = ⇒ − =

−−

2 2v 2log log cx y 2x cx y

v

− ⇒ = ⇒ − =

(C) ( )3 dy2x 10y y 0

dx− + =

2

3

dy y dx 2x10y

dx dy y10y 2x⇒ = ⇒ + =

−

which is a linear equation in y.

∴ the solution is

2 2dy dy

2y yxe e 10y dy∫ ∫

= ⋅∫

i.e., 2 4 5xy 10 y dy 2y c= = +∫

(D) The equation of the tangent at (x, y) on y = f(x) is

dy

Y y (X x)dx

− = −

The Y-intercept of this tangent is obtained by putting X = 0

2dy dyY y x i.e., 2xy y x

dx dx∴ = − = −

i.e., 2

ydx xdy2xdx

y

−=

2x

d x cy

⇒ = +

∫

2x y x cy⇒ = −

Iit-jee 2012 Fst2 p2 Solns

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