Iit 2012 Pti Pi Solns

8/3/2019 Iit 2012 Pti Pi Solns

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Brilliant Tutorials Pvt. Ltd. IIT/PT 1/CPM/P(I)/Solns

B.MAT PART TEST 1FOR OUR STUDENTS

TOWARDS

IIT----JOINT ENTRANCE EXAMINATION, 2012

SECTION I

1. (D) Work function = Energy required to just dislodge the electron in the ground state

= =

00

hch

=

34 8

96.63 10 3 10

330 10

=26

719.89 10

3.3 10

= 6 1019 J

2. (C) For H-bond formation, hydrogen must be covalently bonded to oxygen, nitrogenor fluorine which is more electronegative. In CH3F, hydrogen is attached to

carbon but not to fluorine. Hence there is least tendency for H-bond formationin CH3F.

3. (A) According to Boyles Law

1PV

at constant temperature

P =constant

V

IIT-JEE 2012PT1/CPM/P(I)/SOLNS

PAPER I SOLUTIONSCHEMISTRY PHYSICS MATHEMATICS

PART A: CHEMISTRY


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log P = log (constant) log V (represents graph C)

or log V = log constant log P (represents graph D)

P 1

V. Hence P vs V is a curve but not a straight line.

4. (A) Conc. of final solution (NaOH) = 10 mg per ml

= 10 g per litre

=1040

equivalent per litre

= 0.25 N

Conc. of initial solution of NaOH = 0.5 N

V1 N1 = V2 N2

or V2 =1 1

2

V NN

=500 0.5

0.25

= 1000 ml

V2 V1 = 1000 500 = 500 ml

Hence 500 ml of water should be added to 500 ml of 0.5 N NaOH to get theNaOH solution with conc of 10 mg/ml.

5. (C) The gas with higher value of van der Waals constant a undergoesliquefaction readily. Thus

NH3 (a = 4.17) undergoes liquefaction readily. It is very difficult to liquefy O 2

[a = 1.3]. Thus the order of readiness with which these gases can be liquefied is

NH3 > CH4 > N2 > O2

a = 4.17 2.253 1.39 1.3

6. (D) Oxidation number of the metal in amalgams and metal carbonyl is zero.

7. (B) Since the formula of the oxide is M2O3 the valency of the element M is 3

Weight of oxide = 0.559 g

Weight of the element M = 0.359 g weight of oxygen = 0.200 g

Eq. wt. of element M =0.359 8

14.360.2

=


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Atomic weight of M = Eq. wt valency

= 14.36 3

= 43.08

43

SECTION II

8. (A), (B)

Kinetic energy absolute temperature

Kinetic energy = 0 at zero kelvin temperature

At constant temperature, change in pressure of the gas does not affect kinetic

energy of the gas.

9. (A), (C)

Critical temperature, TC =8a

27Rb

Boyles temperature, TB =a

Rb

Inversion temperature, Ti =2aRb

Unit of gas constant, R = 0.082 lit. atm K

1

mol

1

= 8.31 joules mol1 K1

= 1.987 cal mol1 K1

= 8.31 kPa dm3 K1 mol1

10. (B), (C)

4SF One lone pair of electrons

4XeF Two lone pair of electrons

2Cl O Two lone pair of electrons

[ ]6IF One lone pair of electrons


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11. (A), (D)

SS

O

OO

ThiosulphateOxidation states of S are zero and +4

O

S

OOO S O

O

O

Pyrosulphate

Both sulphur atoms are in + 6 state

O

S

OOO S O

O

O

O

Perdisulphate

Both sulphur atoms are in + 6 state

OS

O

SO S O

O

OS

Tetrathionate

Oxidation states of sulphur are 0 and + 5

SECTION III

12. (B) orbital no. of nodal planes

(= )

2s 0

2p one

3d two

13. (D) The number of spherical node = n 1

For 3p orbital, n 1 = 3 1 1 = 1

For 3d orbital n 1 = 3 2 1 = 0

14. (B) 22 21 2

1 1 1RZ

n n

=

Z = 2 for He+ ion

For Lyman series limiting line n1 = 1 and n2 =

=

2 2 21 1 1R 2 1

=2

1R 4 1 4R

=


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15. (C) En =2

213.6 Z

n

eV/atom for H-like ions.

Ionization energy of He+ ion 2He e He+ +

, Z for He = 2

2

213.6 2

4 13.6 eV 1

= =

Ionization energy of Be3+ ion 3 4Be e Be+ +

, Z for Be = 4

2

213.6 4

1

=

= 16 13.6 eV

The ratio of ionization energies of He+ ion and Be3+ ion

= 4 13.6 : 16 13.6

= 1 : 4

16. (B) E =

hc=

27 10

86.62 10 3 10

4000 10

erg/photon

=19.86

4 1012 erg/photon.

= 4.965 1012 erg/photon

= 4.965 1019 Joules

No. of photons having 2 Joule of energy

=19

2

4.965 10

=1820 10

4.965

4 1018

SECTION IV

17. (2) An+5

3A O

+

(oxidation)

change in oxidation state = (5 n)

4MnO H

+

Mn2+


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change in oxidation no: 7 2 = 5

No. of equivalents of An+ = 0.268 (5 n)

No. of equivalents of 4MnO = 0.161 5

Both are same. Hence

0.268 [5 n] = 0.161 5

1.340 0.805 = 0.268 n

or n =0.535

20.268

18. (6) 25

Br 5e 5Br2

+

Add5

2 31

Br 5e B O2

+

2 33Br 5Br BO + ; no. of electrons involved is 5.

E = Eq. wt. of Br =6[A] 6[A]

5no. of e=

5E = 6[A]; compared with 5E = x[A]

or x = 6

19. (6) Fe (26) = 3d

6

4s

2

Fe2+ (24 e) = 3d6 4s0

No. of electrons in 3d orbital in Fe2+ ion = 6

20. (4) Density of gas = d =PMRT

For the gas A, dA = A AP M

RTor PA=

A

A

d RTM

For the gas B, dB = B BP M

RTor PB =

B

B

d RTM

Now A BB A

P MdA 3 2P dB M 1.5 1

= =

= 4


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21. (2) SO3 has sp2 hybridised sulphur central atom

S[16] 1s2 2s2 2p6 3s2 3p4 3s 3p

S* hybridisation 3s 3p 3d

3p2sp 3d

The three single electrons present in the three sp2 orbitals of S are involved information of bonds with the three oxygen atoms. Hence SO

3contains one

p p bond and two p d bonds.

22. (3) 2 2 24 3 42MnO 5SO 6H 2Mn 5SO + + + + + + 3H2O

or 2 2 24 3 43MnO 7.5SO 9H 3Mn 7.5SO + + + + + + 4.5H2O

no. of moles of 4MnO required to react with 7.5 moles of 23SO

completelyin the acid medium is 3.

23. (2) () =150 150

4V 37.5

= =

or = 2


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SECTION I

24. (A) v = a + bt +c

d t+

Hence [a] = [LT1

], [b] = [LT2

]

[d] = [T], [c] = [L]

[Q] =abd

c

=1 2LT LT T

L

= [LT2

]

unit of Q = unit of acceleration.

25. (B) F = Krn

, n > 0.

[K] = [M] [L]n + 1

[T]2

.

The coefficient of viscosity = kMx

Ey

Kz, k dimensionless quantity

Dim. = [M] [L1

][T1

] = [M]x

[M]y

[L]+2y

[T]2y

[M]z

[L](n + 1)z

[T]2z

= [Mx + y + z

] [L2y + (n + 1)z

] [T2y 2z

]

Hence x + y + z = 1, 2y + (n + 1)z = 1, 2y + 2z = 1

(i.e.,) z(2 n 1) = 2 or z =2

1 n.

And y =

1 1 4

[1 2z] 12 2 1 n

=

1 1 n 4

2 1 n

=

(n 3)

2(n 1)

+=

Hence

(n 3)

2(n 1)E

+

. But E , absolute temperature.

(n + 3)/2(n 1)

p

, p > 0

Hence p =(n 3)

02(1 n)

+>

n < 1 and further n > 0.

0 < n < 1

PART B: PHYSICS


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26. (B) 0

= 1 m, 0

= 0.01 m, =v

0.6c

,

= v

0.02c

Now, = 2

0 2

v1

c.

Taking log, n = n 0

+1

2n

2

2

v1

c

Differentiating

= +

2

20

20

2

vd 1

cdd 1

2 v1

c

=

02

0

2

d v dv 1c c v

1c

d 0.6( 0.02)0.01

(1 0.36)

=

Max. fractional error in the measurement of is

d 0.6 0.02 1.20.01 0.01

0.64 64

= + + = +

= 0.01 + 0.01875 = 0.02875

Now, 1 1 0.36= = 0.80 m.

d = 0.023000 = 0.023

length of metre scale = [0.80 0.023]m

27. (A) OA 2i, AB 4 2 [cos i sin j]= =

4i 4 j=

= =

3BC r , OC 4i

OC OA AB BC= + +

= = +

3BC r 4i 2i 4i 4 j

2i 4 j= +

3r 4 16 2 5 km,= + = making an

angle = tan1

2

4

west of north

1r

2

r

A

B

3r

CE45

O

N

W

S


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28. (B) Initial velocity u = 144 kmph =3

1144 10 40 ms

60 60

=

Let m be the mass of the aircraft in kg

Resistive force F = (900 + V2)

mN

1000.

= + 2dV m

m (900 V )dt 1000

or2dV dV (900 V )

Vdt dx 1000

+= =

2

VdV dx,

1000900 V =

+

let V2

+ 900 = y2

2VdV = 2ydy

0x30

250 0

ydy 1dx

1000y= ,

500

30

x dy 5n

1000 y 3

= =

, V = 40 ms

1, y = 50

V = 0, y = 30

x0

=5

n3

km = 0.5 km

= =

25 5n 2 n 1.02

9 3

29. (B) With respect to the truck moving with acceleration a, the block is about to slide

down the incline (for minimum a).Let m be the mass of the block

and f the force of friction and =

30.

may

+ f mg sin = 0, f = N.

(along the incline)

+ ma

+ N mg cos = 0,

ay

= a cos .

a

= a sin

(normal to the incline)

N = m(g cos a sin )

ma cos + m (g cos a sin ) mg sin = 0

a(cos + sin ) = g(sin cos )

FBD of the block

mg =

30

ma

N f

ve+ve+

a


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a =g( cos sin )

( sin cos )

+

(i.e.,) a =

3 110 0.64

2 2

1 30.64

2 2

+

= 0.45 ms2

30. (B) Let 1 2 3C C i C j C k= + +

, F 2i j k= +

, F C j k = = +

1 2 3F C 2C C C 4 = + =

(1)

1 2 3

i j k

F C 2 1 1

C C C

=

= ( ) + + +

2 3 3 1 2 1

i C C j(2C C ) k(2C C ) (2)

Comparing it with

,

C2

+ C3

= 0, 2C3

+ C1

= 1, 2C2

C1

= 1

[Relation C2

+ C3

= 0 becomes redundant]

C3

= 1(1 C )

2

+ , C

2=

+ 1(1 C )

2 (3)

Substituting (3) in (1): 1 11(1 C ) (1 C )

2C 42 2

+ ++ + =

14 1

C 13

= =

Hence C1

= 1, C2

= 1, C3

= 1 C i j k= +

SECTION II

31. (A), (C)

Let u be the initial speed, making angle with the horizontal.

u

O

B

10 m

15 ms

5 mB


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Max. height h = 15 m =2 2u sin

2g

u2

sin2 = 2gh = 2 10 15 = 300 (1)

By the time, bird flies horizontally from B to B, the stone rises 5 m and thencomes down.

time taken t by the bird to fly from B to B =2h

2g

(i.e.,) t =2 5

2 2s10

= .

And hence u cos = speed of the bird = 5 ms1

.

u2 = 300 + 25 = 325 or u = 18 ms1

Distance covered by the bird = 5 2 = 10m

Now, u sin = 300 10 3=

u cos = 5

tan = 2 3 , = tan1

(2 3)

32. (A), (B), (C)

Let T be the tension in the string, T = 30 N.

Horizontal force on leg and foot by the device

FH

= T + T cos 30 .

= (30 + 15 3 ) N.

Vertical force FV

= T + T sin 30 = 45 N.

Net force by the device = 72 N.

horizontal force on the device = (30 + 15 3 ) N = 56 N

Effective weight of leg and foot = 45 N = 4.5 kg wt.

33. (B), (C), (D)

x = 1 + t + t2

t3, 2

dxv 1 2t 3t

dt= = +

v = 0 at t = 2 4 126

+ = 1s (t = 13

s not admissible)

v > 0 for 0 < t < 1s

v < 0 for t > 1s.

T

30 N

3 kg


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a =dv

dt= 2 6t, a = 0 at t =

1s

3.

a > 0 for 0 < t 1

s3

Thus,

v and a

are parallel during 0 < t 1s.

v and a

are anti-

parallel only during1

s3

< t < 1s.

Now, at t = 0, x = 1 m; t = 1 s, x = 2 m, t = 2s, x = 1 m. Therefore, x = 1 m att0, 1s < t

0< 2s.

34. (A), (B), (C), (D)

2 kg

A f

F T

2 g

2a1N

fT

4a 1N 2N

4 kg

B

4 g

B

AF

a22 ms =

N1

+ 2a 2g = 0, N1

= 2(g a)

T + f F = 0, F = T + N1

(1)

But f = T = N1

max. force F = 2 N1

= 4 (g a) = 4 0.4 (10 2)

F = 12.8 N

SECTION III

35. (A) Let F = hx

cy

z,

MLT2

= Mx

L2x

Tx

Ly

Ty

Lz x = 1

2x + y + z = 1

F =

2

hc (x + y) = 2

y = 1 and z = 1 3 = 2

one unit of F =34 8

262

6.6 10 3 1019.8 10

1m

=

= 1.98 1025

N.


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36. (A) In new system, energy E =

hc

one unit = 1.98 1025

J

Now, 13.6 eV = 13.6 1.6 1019

J

ionization energy in above system

196

25

13.6 1.6 10 13.6 1.610

1.981.98 10

= =

j 1.1 107

units

37. (B) Let ux

= u cos , uy

= u sin

ax

= a, ay

= 0

x = u cos t 21

at2

; y = u sin t

x = cot y 2

2 2

1 ay

2 u sin = py qy

2

p = cot , q = 2 22 2 2 2

a a acosec (1 cot )

2u sin 2u 2u= = +

q = 2 2

2

a a(1 p ) or u (1 p )

2q2u+ = +

38. (C) = + = +

dx dyv i j ai bx jdt dt

(given)

dx

adt

= x = at + c1. At t = 0, x = 0 c

1= 0.

x = at.

dybx abt

dt= = y =

2abt

2( At t = 0, y = 0)

=2

22

ab x b x2 2aa

=

path of the particle is y = 2b

x2a

parabolic.

Ox

y

u


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39. (C) x = a sin t, y = a(1 cos t)

(i.e.,) x = 2a sin t2

cos t2

, y = 2a sin2 t2

sint

2

=

y

2a,

cost

2

=

y1

2a

x =y y y

2a 1 2ay 12a 2a 2a

=

SECTION IV

40. (2)

+ =

=

= =

1 A 1

2 1 21 2 3 4

3 2 3

B 3 4

2x x C

2x x CC , C , C , C

2x x C

x x C

are constants.

Bx

3x

2x Ax

1x

A

B 10 kg

1 kg

Differentiating twice w.r.t. time t, A 1 2 3 Bx 2x 4x 8x 8x= = = =

Let a be the acceleration of block B downward. Then, the acceleration of A is 8a

upward.

FBD of A:

T

g

8a T g = 8a (1)


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FBD of B:

8T

10 g

a

10 kg 10g 8T = 10a (2)

From (1) and (2), 10g 8(8a + g) = 10a

2g = 74a

a =2g xg

74 74=

x = 2

41. (2) Let a be the acceleration of wedge Balong horizontal direction.

FBD of A:

a

mg

( )90 1N

ma

P

Q45 =

Perpendicular to PQ: N1

+ ma sin = mg cos

N1

= m(g cos a sin ) (1)

Parallel to PQ: mg sin + ma cos = ma (2)

FBD of B:

a

N

1N

Mg

1.6 kg

a

B

P

A

0.8 kg

Q


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N1

sin = Ma a =m

sin (g cos a sin )M

a (M + m sin2) = mg sin cos

a =2

mgsin cos 0.8 10

0.8M m sin 2 1.62

=

+ +

28 2ms2 2

= =

42. (3) Let a be the acceleration of each block.

For A,

Tmg

1N

60 60

O

A

B

a

T sin 60 + mg sin 30 = ma (1)

For B,

2N

T

amg

O

mg T sin 60 = ma (2)

T sin 60 + mg sin 30 = mg T sin 60

2T sin 60 = mg(1 sin 30)

T =

1mg 1

mg2

3 2 322

=

=5

N

3

.

=(n 2)

Nn

+, n = 3

A

B O

TT

( )1 kg

( )1 kg

60 =


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43. (5) T4

= 2T2

= 4T1

= 4T3, T

1= T

3

Let a1, a2, aA, aB, aC and aD be the accelerations of pulleys 1, 2 and blocks A, B,C, D respectively.

Bx

Ax

1x

2x

Cx

Dx

A

B

1T

2T

1T

1

2

2T

3T

3T

4T

C

D

Taking downward + ve,

xA

+ xB

2x1

= C1, a constant

aA

+ aB

= 2a1

(1)

xC

+ xD

2x2

= C2

aC

+ aD

= 2a2

(2)

And x1 + x2 = C3

a1

+ a2

= 0 (3)

For pulley 1,

1 g T1

= 1 aA

, 2 g T1

= 2 aB

4g 3T1

= 2(aA

+ aB

) = 4a1

(4)

Similarly for pulley 2,

1 C

1 D

3g T 3a

4g T 4a

=

= 24g 7T

1= 12(a

C+ a

D) = 24a

2 (5)

But a1

+ a2

= 0. Hence equation (4) and (5)

(4g 3T1

= 4a1) 6

24g 7T1

= 24a1

148g 25T 0 =


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19


148

T g25

=

Hence the tension in fixed string = 4T1

=1920

N25

=2

1920N

n

n = 5

44. (8) Let m be the mass per unit length. At

an instant t, let x be the length

hanging from the edge. Then FBD are:

N

( )m 1 x g

aT

a

T

mxg

T = m(1 x)a

mxg T = mxa

xg = a(x + 1 x) = a =dV dV

Vdt dx

=

By integration,

1.02 2

0.6

V gx

2 2

=

20V g(1 0.36)= , V0 = 0.8 10 =

8

10

n = 8

45. (3) Let P

be the compressive force (alongAO

)

A

DE

B C

O

0.4 m

0.6 m


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20


P = 4 1000 cos

= 4000 1824

= 3000 N

= 3 kN.

46. (3) Constraint equation:

2xB

+ xA

= constant

2aB

+ aA

= 0

For B

300 g 2T = 300 a2

3000 150 a = 2T

T = 1500 75a (1)

For A

200 g T F = 200 a

F = 2000 T 200 a

= 2000 1500 + 75a 200a

= 500 125a

= 375 N

power delivered

= FV = 375 2 =750

kW1000

=3

kW4

=n

n 1+kW,

n = 3

A2

Aa a 1 ms

= =

200 g

2T

300 g

BB

aa

2=

FT

Bx Ax

B A

M


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21


SECTION I

47. (C) Now1 1 1 1 4

1 .....14 16 64 5

14

+ + = =

1 1 1 1 21 ....

12 4 8 31

2

+ + = =

10 alog a log 104 2

20 35 3

=

( )1

t4 t2 2= where t = log10

a

4t =1

t t =

1

2

log10

a =1

2 a =

1

210 10=

48. (B) Sum to n terms

Sn

=n

[56 (n 1)( 2)]2

+

=n

[58 2n] n(29 n)2 =

= n2

+ 29n

= {n2

29n}

=

2 229 29

n2 2

=

2 229 29

n2 2

Sum is maximum when

2

29n2

is least.

But n is a natural number, n can be 15 or 14

S14

= S15

= 15 14 = 210.

PART C: MATHEMATICS


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22


49. (C) Given 2 sin cos sin = sin sin cos + sin cos sin

Dividing throughout by sin sin sin , we get2 cot = cot + cot

cot , cot , cot are in A.P.

tan , tan , tan are in H.P.

50. (A) Let S = 1 6 + 2 62

+ 3 63

+ + n 6n

6S = 1 62

+ 2 63

+ + (n 1) 6n

+ n 6n + 1

subtracting 5S = 6 + 62

+ . + 6n

n 6n + 1

=n

n 16(6 1)n 6

5

+

5S =n 1 n 16 6 5n 6

5

+ +

S =n 1(5n 1)6 6

25

+ +

a + b = n + 1 + 6

= n + 7

51. (C) Let E = 3 sec2 + 12 cosec

2

= 3 + 12 + 3 tan2 + 12 cot

2

= 15 + ( )2

3 tan 12 cot 2 3 12 +

= 15 + ( )2

3 tan 12 cot 12 +

= 27 + ( ) 2

3 tan 12 cot

E 27 Minimum value of E is 27.

It is attained when 3 tan 12 cot = .

52. (B)

xtan A tan B

x tan A y tan B y

xx y1

y

++

=+

+

cos A sin A sin B

cos B cos A cos BcosA

1cosB

+=

+


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sin A sin B

cos A cos B

+=

+

A B A B2sin cos

2 2A B A B

2cos cos2 2

+

=+

A Btan

2

+ =

53. (B) tan 2 =tan( ) tan( ) 4

1 tan( ) tan( ) 1 1

+ + =

+

2 =2

=

4

Now tan ( + ) + tan ( ) = 4

i.e.,1 tan 1 tan

41 tan 1 tan

+ + =

+

i.e.,2

2

2(1 tan )4

1 tan

+ =

i.e.,2

4cos 2

=

cos 2 =1

2

2 =3

=6

2 + 3 =2 2

+ =

SECTION II

54. (A), (D)

The first equation is sin + cos + 1 = x

x 1 = sin + cos

Similarly y + 1 = sin cos

(x 1)2

+ (y + 1)2

= 2


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x 1 cos sin 1 tan

y 1 cos sin 1 tan

+ + = = +

tan4

= +

cot2 4

= + +

3cot

4

= +

55. (A), (C)

The given equation is

2

2 2

2t 1 t

6 7 91 t 1 t

+ = + + where t = tan 2

16t2

12t + 2 = 0

8t2

6t + 1 = 0

1t

2= ,

1t

4=

But2

12tan 2

42 2tan1 3

1 tan 12 4

= = =

Also tan =

12 84

1 151

16

=

56. (A), (D)

Given2 2 2 2

22 2 2

2x z 2x zy

x z (x z) 2xz= =

+ +

2 2

2

2x z

(2y) 2xz=

{since x, y, z are in A.P.}

2 2

2

x z

2y zx

=

2y4

xy2z x

2z

2= 0

(2y2

+ xz) (y2

xz) = 0


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25


y2 zx since x, y, z are distinct terms in A.P.

2y2 + xz = 0 either x , y , z2

are in G.P.

orz

, y , x2

are in G.P.

57. (B), (C)

In an A.P. the sum of two terms which are equidistant from the first and fromthe last are the same.

Hence a2

+ a18

= a4

+ a16

= a8

+ a12

each =336

1123

=

a1 + a19 = 112

sum of all the terms = ( )1 1919

a a2

+

=19

1122

= 1064

Also a7

+ a13

= 112 and a5

+ a15

= 112

SECTION III

58. (A) tan7

tan cot16 2 16 16

= =

Similarly tan6

16

= cot

2

16

and tan5

16

= cot

3

16

given expression

= 2 2tan cot16 16

+

+ 2 2

2 2tan cot

16 16

+

+ 2 2

3 3tan cot

16 16

+

I bracket = = =

8 8 8 22 2 2

1 2 11 cos 14

2

III bracket =8 8 8 2

2 2 23 1 2 11 cos 14 2

= = + +


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II bracket =8

2 8 2 6

1 cos 2

= =

the whole expression = 2 + 8 2 2 2

= 34

59. (C) Given tan2 + cot

2 + 2 =

8

1 cos 4

i.e., (tan + cot )2

=8

1 cos4

2

8tan cot

12 121 cos

3

+ =

= 16

60. (B) Let A and R be the first term and common ratio of the G.P.

Now A Rp 1

= 64 (1)

A Rq 1

= 27 (2)

A Rr 1

= 36 (3)

(1)

(2)gives

3p q 4R

3

=

(3)

(2) givesr q 4

R 3

=

p q = 3r 3q

i.e., p + 2q = 3r

p 2q3

r

+=

61. (B) p = 8, q = 2

3r = 12

r = 4

This is possible.

A R7 = 64 and AR = 27

R =

1

24

3

R =2

3


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62. (A) f(0) = r = ve { r is + ve r cannot be 0}

f(1) = p + 2q r= 3r r = 2r = + ve

f(0) f(1) < 0

SECTION IV

63. (1) uk

= Sk

Sk 1

= { }1

k(k 1)(k 2)(k 3 k 1)16

+ + +

=1

k(k 1)(k 2)4

+ +

Let tk

=k

1 4

u k(k 1)(k 2)=

+ +

Let vk

= 4(k 1)(k 2)+ +

vk

vk 1

=4 4

(k 1)(k 2) k(k 1)

+ + +

= 2 tk

tk

= k k 11

(v v )2

[ ]n

k n 0k 1

1t v v

2== = [ ]0 n

1v v

2

=1 4 4

2 2 (n 1)(n 2)

+ +

kk 1

1t [2 0] 1

2

== =

64. (6) a + c = 2b

a3

+ c3

= 8b3

3ac 2b

a3

+ c3

+ 4b3

= 12b3

6abc

= 3b{4b2

2ac}

= 3b{(a + c)2

2ac}

= 3b(a2 + c2)

( )

3 3 3

2 2

2a 2c 8b6

b a c

+ +=

+


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65. (3) Given expression =

2 2

2 2

1 1

x w1 1

y z

=

+

+

1 1 1 1

x w x w

1 1 1 1

y z y z

=

1 1 1 13k 3k

x x x x

1 1 1 1k 2k k 2k

x x x x

+ +

+ + + +

where k is the common difference of the corresponding A.P.

=

23k ( 3k)

x3

23k ( k)

x

+

=

+

66. (9) Given2

cos x 3 3

4 41 sin x

3

=

2

3cosx 3 3

43 4sin x =

sin xcosx 3

sin 3x 4=

sin2x 3

2sin3x 4=

sin2x 3

sin3x 2=

2

2

sin 2x 3

4sin 3x

=

2

2

sin 2x12 9

sin 3x=


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67. (4) L.H.S. is (x 4)2

+ 5 which is 5

R.H.S. = 5 cos (y ) where tan =3

4

R.H.S. 5

Equality is possible.

L.H.S. = 5 when x = 4 (R.H.S. = 5 when y = )

68. (1) Now sin2

47 + sin2

13 + sin2

38 + sin2

22

= 1 cos2

47 + sin2

13 + 1 cos2

38 + sin2

22

= 2 {cos 60 cos 34 + cos 60 cos 16}

= 2 1

2{1 2 sin

217 + 1 2 sin

28}

= 1 + sin2

17 + sin2

8

given expression = 1 + sin2

17 + sin2

8 sin2

8 sin2

17

= 1

69. (6)sin(2 )

3sin

+ =

sin(2 ) sin 4

sin(2 ) sin 2

+ + =

+

2sin( ) cos2

2cos( )sin

+ =

+

tan( )2

tan

+ =

3tan( )6

tan

+ =

Iit 2012 Pti Pi Solns

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