Solution 3 (Class XII)

HINTS & SOLUTIONS (PRACTICE PAPER-3) ANSWER KEY Que s. Ans. Que s.

1 A 16

2

3

B 17

4

C 18

5

B 19

6

D 20

B 21

7 D 22

8 A 23

9 A 24

10

11

12

13

14

15

C 25

B

B

D

C

A

26

27

28

29

30

Ans.

B

C

B

C

D

B

B

D

D

C

C

Que s.

31

32

33

34

35

36

37

38

39

40

41

D 42

C 43

A 44

D 45

Ans.

D 46

C 47

A 48

B 49

B 50

C 51

A 52

A 53

A 54

B 55

C 56

D 57

C 58

C 59

D 60

A 61

C 62

B 63

A 64

B 65

A 66

D 67

C 68

D 69

B 70

D 71

A 72

D 73

B 74

B 75

A 76

C 77

C 78

C 79

B 80

B 81

C 82

D 83

A 84

D 85

D 86

D 87

B 88

A 89

B 90

Ans. B Que s. 91 D Ans. Que s. 106 C Ans.

B 92 A

D 93 B

C 94 C

A 95 D

C 96 A

B 97 C

C 98 C

B 99 B

B 100 B

D 101 B

C 102 D

D 103 A

B 104 A

C 105 B

107 A

108 C

109 D

110 B

111 C

112 B

113 A

114 B

115 A

116 D

117 B

118 A

119 C

120 C

Que s. Ans. Que s. Ans. Que s.

PART-I (1 Mark) MATHEMATICS 1.

Given : a1, a2, a3 .........AP and a1, a2, a4, a8 ......GP. Let common difference of A.P. = d a2 = a1 + d a4 = a1 + 3d a8 = a1 + 7d 

a8 a2 a4 = = a1 a2 a4 = r a1  d a1  3d a1  7d a1 = a1  d = a1  3d = r (a1 + d)2 = a1(a1 + 3d) a12 + d2 + 2 a1d = a12 + 3a1 d d2 = a1 d

(d  0)

d = a1

....(i)

a2 a1  d Hence, a = r ; a =r 1 1 a1  a1 =r a1

(using (i))

r = 2.

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2.

Tk 101  till k = 10 Tk –1 k Tk > Tk–1 Let k = 11 T11 < T10  T10 is maximum at k = 10.

3.

x=

2+ 3+ 6 2

2

x  2  =  3  6 

x2 + 2 – 2 2 = 9 + 6 2 x2 – 7 = 8 2 (x2 – 7)2 = 64 × 2 So, smallest possible value of n is 4. 4.

Let the three players are A, B, C. Now, each player get 0 score after playing 9 games. It happened only when each player wins 3 games and loss 6. So, A win 3 games out of 9  9C3 B win 3 games out of remaining 6  6C3 C win 3 games out of remaining 3  3C3 So, required way = 9C3 × 6C3 × 3C3 =

9  8  7  6! 6  5  4  3! × ×1 3  2  1 6! 3  2  1 3!

= 1680. A (2, 3)

5.

O (2, z) B (4, 0)

C (x, y)

O is circumcenter  OA = OB = OC = circumradius (2 – 2)2 + (z– 3)2 = (4 – 2)2 + (0 – z)2 z2 + 9 – 6z = 4 + z2 9 – 6z = 4 5 = 6z 5 =z 6

 Circumcenter =

5 13 . ( z  3 ) 2  ( 2  2) 2 = | z – 3 | = 6  3 = 6

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6. (5, 15) L(21, 15) P

P'

A

 5  21 15  15  ,  Mid point of PP’ =  2   2 L = (13, 15)  Point A will be (13, 0) By property PA + PA’ = 2a

PA =

(13  5)2  (0  15 )2 =

64  225

=

289 = 17 cm

PA’ = =

(13  21) 2  (0  15) 2 64  225

=

289 = 17 cm  2a = PA + PA’ 2a = 17 + 17 2a = 34 cm So, length of major axis = 2a = 34 cm. 7.

P (10, 10)

B (0, 6)

2x + 3y = 18

C (a, b)

PB = PC (10 – 0)2 + (10 – 6)2 = (a – 10)2 + (b – 10)2 100 + 16 = a2 + 100 – 20a + b2 + 100 – 20b a2 + b2 – 20a – 20b + 84 = 0 ....(i) Also (a, b) i.e. on 2x + 3y = 18 2a + 3b = 18

3b 2 Using equation (i) a=9–

2

3b   3b    – 20b + 84 = 0 9   + b2 – 20  9  2  2   

81 +

9b2 – 27b + b2 – 180 + 30b – 20b + 84 = 0 4

13b 2 – 17b – 15 = 0 4 13b2 – 68b – 60 = 0

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13b2 – 78b + 10b – 60 = 0 13b(b – 6) + 10 (b – 6) = 0 10 13

b = 6 or b =



When b = 6, then a = 9 –

When b =

10 3  10 30 132 , then a = 9 + =9+ = 13 2  13 26 13

132 10 +2× 13 13

 8a + 2b = 8 ×

1056  20  79 . 13

=

8.

36 =0 2

cosec2( + ) – sin2( – ) + sin2(2 – ) = cos2( – ) cosec2( + ) + sin2(2 – ) =

cos 2 (  )  sin 2 (   ) 1

cosec2( + ) = 1 – sin2(2 – ) cosec2( + ) = cos2(2 – ) Minimum value of cosec2( + ) is 1 and maximum value of cos2(2 – ) is 1.  They will be equal for the value 1.

 2

.....(i)

2 –  = 0 By adding (i) & (ii)

.....(ii)

+=

 2

3 =



9.

=

 6

=

 3

sin( – ) = sin (

sinx + siny =

7 5

cos x + cosy =

   1 – ) = – sin ( ) = . 6 3 6 2

....(1) 1 5

....(2)

By (1)2 + (2)2 we get 2 + 2sinx siny + 2 cosx cosy = 2

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sinx siny + cosx cosy = 0 cos(x – y) = 0  x – y = 90º By (1) × (2) we get sinx cosx + sinx cosy + siny cosx + siny cosy =

7 25

sin(90 + y)cosx + sin(x + y) + sin(x – 90) cos y =

7 25

cosy cosx + sin(x + y) – cosx cosy = sin(x + y) =

7 25

7 . 25

10.

6 x

y=

(1, 1) 1 0 y = sin x Clearly, curve meet each other twice in

 11.

2 – 3 4 – 5 6 – 7 8 – 9 10 – 11

Total 10 Times.

f(x) is differentiable on R. So, it will be contincous on R. Continuity at x = 0 LHL 2 lim– sin x x 0 x Put x = 0 – h, then h  0 2 lim sin(0  h) h0 h

lim sin h  h = 0 h h RHL h0

lim x2 + ax + b

x 0 

Put x = 0 + h, then h  0

lim h2 + ah + b = b

h0

Value of f(x) at x = 0 f(0) = b.

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f(x) is contineous at x = 0 LHL = RHL = f(0) 0=b=b b=0 Differentiability at x = 0 LHD  

lim

f ( 0  h )  f ( 0) h

lim

sinh 2 b h h

h0

h0

2 lim sin h = 1 h0 h2 RHD

lim f (0  h)  f (0) h

h0

2 lim h  ah  b – b h0 h

lim h(h  a ) = a. h  f(x)is differentiable at x = 0, LHD = RHD a = 1. h0

12.

Let point p(x1, y1) is on the curve y2 = 4x. 2



y12 = 4x1  x1 =

y1 4

( x1  0) 2  ( y1  3 ) 2

PA =

AP2 = x12 + y12 – 6y1 + 9 AP2 = x12 + y12 – 6y1 + 9 Let AP = z z2 = x12 + y12 – 6y1 + 9 2

 y12    2 z =  4  + y12 – 6y1 + 9   4

z2 =

y1 + y12 – 6y1 + 9 16

Diff. w.r.t. y1 3 dz 4 y1 2z dy = + 2y1 – 6 16 1

3 dz y 2z dy = 1 + 2y1 – 6 1 4 3

=

y1  8 y1  24 4

dz 2z dy = (y1 – 2) (y12 + 2y1 + 12) 1

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For the critical points

dz dy 1 = 0 (y1 – 2)(y12 + 2y1 + 12) = 0 y1 = 2 y12 = 4x1 (2)2 = 4x1 x1 = 1.

  

 dz   2   dy 1 

2

d2 z

+ 2z

2

dy 1

= (y12 + 2y1 + 12) + (y1 – 2) ( 2y1 + 2) = y12 + 2y1 + 12 + 2y12 – 4y1 + 2y1 – 4 = 3y12 + 8.

dz when y1 = 2 and dy = 0 1 d2 z 2

dy 1

>0

z is min at (1, 2)



Minimum distance = (1  0)2  (2  3)2 = 13.

1 1 =

2.

We can find the answer through option as the sum of weight of packet taken from trucks is 1022870 gm and its unit digit is 0. The truck that have heavier bags have unit digit 0. So, the truck have lighter bags in which the sum of weight of bags must have unit digit 0. So, according to option D. i.e. truck no. 2, 8 Track 2 have 21 bags and total weight = 21 × 999 gm = .......8 gm Truck have 27 bags and total weight = 27 × 999 = 128 × 999 gm = ......2 gm So, the unit digit of the weight contain by truck 2, 8 together is 0. 1

14.

 cos( x ) cos([2x]) dx 0

1/ 2

1

cos(  x ) cos 0 dx +

=



=



=

 sin x      

0 1/ 2

0



1/ 2

cos(  x ) cos  dx

1

cos(  x ) dx – 1/ 2



1/ 2

cos(  x ) dx 1

– 0

 sin x      

1/ 2

1  1  =   – 0     

=

2 . 

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77

1

cos(nx )x 9 dx   cos(nx)x10  10   + n n   0 1

15.

IN =



0

=

= 

  10  0+ n   

1

  9 sin(nx)x dx    n n 0  0 

9

 sin(nx)x   n  

1

8



1  10  9   sin(nx )x 8 dx   n 2  0 



1  10 !   sin(nx )dx  = 10   n 0 



= 0 as Denom  

16.

y = x2 & y = 1 – x2 Point of intersections of graphs x2 = 1 – x2 2x2 = 1

1 x=±

2

 1 1  1 1 ,  and  ,  .  Point of intersections =   2 2  2 2 Area under graph : 1/ 2

x

=

2

 (1  x )2

1 / 2

1/ 2

=



1 /

1/ 2

 3  2x 2  1 =  2x  x   1/  3 2

2 =2×

=2 17.

6 2

26 6 2



1 2 4

=

2

3 2

=

2 2 . 3

      a  3 i  4k and b = 5 j  12k

  2 2 2 a  (3)2  4   5 and b  (5)  12   13

       Therefore, a vector which bisects the angle is 13 3 i  4k + 5 ( 5 j  12k ) = 39 i  25 j  8k .



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

88

19.

Let M  2 x1 .3 x 2 .5 x 3 ...... , N  2 y1 .3 y 2 .5 y 3 ...... xi & yi  w x1

 5 x3 – 1  2 y1 – 1  3 y 2 – 1        5 – 1  ....... =  2 – 1  3 – 1  .........     

 3 x2 – 1    3 –1    

 d =  d   22 ––11

d/m

d/N



x2   1  x1      – 1   1  – 1  2   3      ....  1 / 2 – 1  1 / 3 – 1        1/ d   1  y1   1  y 2  d/M    1/ d =   2  – 1   3  – 1         .... d/N  1 / 2 – 1  1 / 3 – 1       

 

(2 x1 – 1)(3 x 2 – 1).....



2 x1 3 x 2 .....

20.

m C0 

(2

y1

m C1 n 

– 1)(3

y2

– 1)......

=

N  1. M

2 y1 3 y 2

 m C2 n 2    mC m n m  (1 + n)m

only one element from A

PHYSICS 27.

Sphere is hollow so potential inside sphere will be same as that on surface.

28.

Heat supplied Q = du + W PV = RT PdV = RdT

(at contat pressure)

PdV R Q = CVdT + PdV dT =

PdV + PdV R Work done at constant pressue, W W = PdV Q = CV

Q  W

CV

PdV  PdV R PdV

Q CV  1 W R

(For diatomic gos, CV =

5 R) 2

Q 5R  1 W 2R Q 7  W 2



Pre-foundation Career Care Programmes (PCCP) Division

W 2  Q 7

99

29.

1 1  1  R 2  2  For lyman series  2  1 1 1   1  R 2  2  For balmer series  3  2 





(

1 1 94   ) 4 9 36

3/4 3 9 27 =   5 / 36 1 5 5

 5    27

A

B

q

q

30.

q q and 2 2

charge divides

Than, on touching

q sphere to q 2

Charge divides

q / 2  q 3q  2 4

force between

3q q R  4 2

f=

f=

unchanged

K 3q2 8R 2 3 kq2 3  F × 8 R2 8

31.

Intially block enters in the magnetic filed rate of change in flux will be constant so costant current will produce, when it mass in side the magnetic field there is no change in magnetic flux, current I = 0, when it use the filed the rate of the change in flux will be again constant between in decrecaing order so contant current will induced on opposite.

32.

No change in moment of inertia

34. E

–q

A

O

+q

Electric field at each point of OA obtained  to it and opposite to direction of dipole moment.

Pre-foundation Career Care Programmes (PCCP) Division

10 10

38.

a

m

Total force in upward direction m × (g + a) because mass m is stationary on inclined plane and whole system is accelerated with acceleatration a in upward. 39.

Force of positve charge = Electric force + Magnetic force F = (qE + qVB) This force is in upward direction so no any particle will pass through the hole.

40.

Potential energy at H height = Kinetic energy at the lowest point of circular path. mgH =

1 mv2 2

To complete the circular motion minmum velocity at lowest point will be V = mgH =

H=

5gR

1 m (5gR) 2

5 R 2

CHEMISTRY 41.

According to Graham’s law

1 Rate of diffusion 

Molar mass

due to highest molar mass of CO2 rate of diffusion is slowest.

42.

Moles of H2 =

3 4 , Moles of O2 = 2 32

Kinetic energy of n moles of gas =

3 nRT 2

3 n1RT 2 Kinetic energy of hydrogen so, = 3 Kinetic energy of oxygen n 2RT 2 n1 = n 2 =

3/2 4 / 32

= 12 : 1 Pre-foundation Career Care Programmes (PCCP) Division

11 11

44.

ClF3  sp3d hybridisation, but due to presence of two lp on central atom Cl, according to VSEPR theory shape is ‘T’

45.

HCO3– + H+  H2CO3 Bronsted base HCO3–  H+ + CO32– Bronsted acid

47.

Isoelectronic means same no. of electrons CO has 6 + 8 = 14 electrons CN– has 6 + 7 + 1 = 14 electrons

48.

CO2, due to sp hybridisation bond angle = 180º

49.

Diethyl ether, because it is inert towards the Grignard reagent

50.

CH3 – CH2 – CH2 – CHO + CH3 – CH2 – CH2 – MgBr

H

H +

CH3 – CH2 – CH2 – C – CH2 – CH2 – CH3

H3O

CH3 – CH2 – CH2 – C – CH2 – CH2 – CH3

OMgBr 51.

[Ni (PPh3)2 Cl2]  [NiCl4]2–

52.



OH Achiral Secondary alcohol

dsp2 hybridisation, because PPh3 is strong ligand hence pairing of electrons takes place sp3 hybridisation, because Cl– is weak ligand hence pairing of electrons is not takes place

16H++ 2MnO4– + 5COO– COO–

53.

Suppose equilibrium constant for the following reaction is K1 N2 + 3H2

2NH3

; K1 =

[NH3 ]2

-------- (i)

[N2 ][H2 ]3

and equilibrium constant for the following reaction is K2 [NH3 ] 1 3 N2 + H2 NH3 ; K2 = [N ]1/ 2 [H ]3 / 2 2 2 2 2 square the both side of equation (ii)

K22 =

[NH3 ]2 [N2 ][H2 ]3

K22 = k1 K2 =

-------- (ii)

[by equation (i)

k1

K2 = 41 K2 = 6.4

Pre-foundation Career Care Programmes (PCCP) Division

[ K1 = 41]

12 12

54.

Suppose reaction is 2A  Product according to rate law Rate R = k [A]2

[ A ]12 R1 or R = [ A ]22 2 R1 [A]12 according to question R = 2 2  [A]1     2 

[ A ]1    [ A]2   2  

R1  R =4 2 R2 =

55.

R1 4

HCO3–  H+ + CO32– Conjugate base NH3  H+ + NH2– Conjugate base

56.

57.

(II) & (IV) Because both have close system of conjugated double bond and follow Huckel’s (4n+2)  e– rule.

••

N H p of N takes part in resonance with conjugated double bonds, so it is not easily available on N for the protonation.

N ••

p is not taken part in resonance so easily available for the protonation.

O ••

N H due to high E.N. of O availability of p on N decreases.

••

N H No extra effect, so availability of p on N increases.

Pre-foundation Career Care Programmes (PCCP) Division

13 13

58.

Gauche conformer. because angle between same groups is 60º

59.

Suppose initial quantity = No after 75% completion of the reaction remaining quantity N = No ×

T=

 No  2.303  log   N  K

T=

 No  2.303 log  N / 4  K  o 

T=

1.386 -------- (i) K

 T1/2 =

K=

25 N = o 100 4

0.693 K

0.693 -------- (ii) 30

so by equation (i) and (ii) T=

1.386 0.693 / 30

T = 60 min.

60.

Concentration of H+ ions in H2SO4 solution = 2 × 0.1 = 0.2 M So no. of moles of H+ ions in 10 ml H2SO4 solution =

0.2  10 = 0.002 1000

concentration of OH– ions in 0.1 KOH solution = 1 × 0.1 = 0.1 M So no. of moles of OH ions in 10 ml KOH –

solution =

0.1 10 1000

= 0.001 after mixing remaining moles of H+ ions

= 0.002 – 0.001 = 0.001

+

so concentration of H ions in mixture of solutions =

0.001 × 1000 = 0.05 M 10  10

Pre-foundation Career Care Programmes (PCCP) Division

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PART-II (2 Mark) MATHEMATICS 81.

p(x) = a0 + a1x + ........+ anxn p(0) = 7 a0 = 7 p(1) = a0 + a1 + a2 + ............+ an = 9 p(– 1) = a0 – a1 + a2 ........... = 1 p(2) = a0 + 2a1 + 4a2 + .......... = 13 p(– 2) = a0 – 2a1 + 4a2 ........... = – 15 p(1) + p(– 1) = 2[a0 + a2 + .........] = 10 a0 + a2 + a4 = 5 .....(1) 7 + a2 + a4 = 5 a2 + a4 = – 2 .....(2) p(2) + p(– 2) = 13 – 15 2(a0 + 4a2 + .........) = – 2 a0 + 4a2 + 16a4 = – 1 4a2 + 16a4 = – 8 .....(2) p(3) = 25 a0 + 3a1 + 9a2 + ........ = 25 a0 + 3a1 + 9a2 + 27a3 + 81a4 + 243a5 = 25 From (1) and (2)

.....(3)

4a2 + 4a4 = – 8 4a2 + 16a4 = – 8 – – + a4 = 0 and a2 = – 2  Smallest possible value of n is 3. 2

82.

 (a – b )

2

0 

abc

a  1  ab

|a – b| < c ...... (1) |b – c| < a ...... (2) |c – a| < b ...... (3) Squaring and adding a2 + b2 + c2 < 2ab + 2bc + 2ca

[Triangle inequalities]

 a2 2  ab So, b  [1,2). 83.

y = | x 3 | 4 – 5 When, x < – 1 y=|3–x–4|–5 y=–x–1–5 y=–x–6 When –1  x < 3 y=|3–x–4|–5 = | – x – 1| – 5 =x+1–5 =x–4 When, 3  x < 7 y = |x – 7| – 5 y=7–x–5 y=2–x

Pre-foundation Career Care Programmes (PCCP) Division

12 –6

–

3

–1

7

O

x

–

6 x

2– x

–4

x

–

12

–5

15 15

When, x  7 y = | x – 7| – 5 =x–7–5 = x – 12. Area bounded region 1

=



3

(  x  6 )dx



+

6

7

( x  4 )dx

+

1



12

( 2  x )dx +

3

1

3

 ( x  12)dx 7

7

12

  x2   x2    x2  x2  2 x  6 x  4 x   12 x         = + + + 2   2   6  2  1   2  7 3

49  9  1    36  9  1     144   49   6 –   36  +   12  –   4  + 14   144  –   84   – 6   +  =  2  2  2   2  2  2     2   2 

=

11 15 9 21 3 119 – 18 – – – – – 72 + 2 2 2 2 2 2

=

130 48 – – 90 2 2

= 65 – 24 – 90 = 49 sq. unit.

A 84.

b

a

b

C

a

D

cos =

B

b

b a2  b2  a2 = 2 a 2ab

cos(180º – ) =

– cos =

cos =

....(i)

b2  b2  a2 2b 2

2b 2  a 2 2b 2

a 2  2b 2

....(ii)

2b 2

From (i) & (ii) a 2  2b 2 2b

2

=

b 2a

3

a a   –2   –1=0 b b

Pre-foundation Career Care Programmes (PCCP) Division

16 16

Let

a =x b

x3 – 2x – 1 = 0 (x + 1)(x2 – x –1) = 0 x = – 1 or x =

1  1  4( 1)(1) 2



1 5 2 x cannot be negative



x=

=

85.

1 ( 5  1) 2

an =

1  a n1 2

a1 =

1  a0 2

a1 =

1  cos  2

2 cos 2 a1 =

2

a1 = cos

 2

1  a1 2

a2 =

=

 2

1  cos

 2

2

2 cos 2 =

 4

2 = cos

 22

an = cos

 2n

lim 4n(1 – a ) n

n 

   lim 4n 1  cos  2n  

n 

Pre-foundation Career Care Programmes (PCCP) Division

17 17



lim 4n × 2sin2

2 n 1

n 



2 2n  2 sin2

2 

lim

n 

 2n 1



 ×

2

n 1



 2

n 1

2n 1

sin2

2

n 1

 n 1

2 lim  ×   2  2n 1 2n 1

n 

2 . 2

= 86.

f(x) = (sin x)sinx f(x) = e sin x log sin x Minimum value of sinxlog(sinx) is 0. Maximum value of (sin x)sinx is e0 = 1. Maximum value of sinxlog(sinx) is – Minimum value of (sin x)sinx is 10

87.

1 . e

1 e

e .

1

 x dx 1

=

10

log x

1

= log10 = 2.303

1 1 + ....+ 9 2 = 1 + 0.5 + 0.33 + 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11  2.81

B=1+

1 1 1 + ....+ +  2.81 – 1 + 0.1 =1.91 9 10 2 So, C < A < B and B – A  0.51, A – C = 2.303 – 1.91  0.40. So, B – A > A – C.

C=

r

A

88.

D

B

r

60º

r

r

60º

r

r

C

As we want the distance between two point is at least r. Now when the point A, B are at distance r.Then the angle made arc BA is 60º. Now as chord AB come closer to centre the length of chord AB is increased that is it is greater than r and the angle is also increases i.e. from 60º to 180º and now when chord AB move way from centre then the length of chord AB decreases , when chord AB reach CD the length of AB equal to r and the angle chang from 180º to 60º Pre-foundation Career Care Programmes (PCCP) Division

18 18

So, the angle required for desired conditions = 2(180 – 60) = 240 Total angle for all around the circle = 360º So, required propability = 89.

240 2 = 360 3

Let aN be pth digit no.

4(5P –1) 4(5P – 1)