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PAPER-1

PART-I (Physics)

1. Plane surface .........................................

Sol . (3)

x

1

x

1

=

1

2

3

300

11

300

1.

2

1

x

2 x = 1200 cm = 12m

so 3m

2. A plane mirror is .........................................Sol . (5)

Let 1

, 2

and 3

be the image formed by :(i) Refraction from ABC

(ii) Reflection from DEF and(iii) Again refraction from ABC

then B1= 5n = 7.5 cm

Now E1= 7.5 + 2.5 = 10 cm

Now B2= 10 + 2.5 = 12.5 cm

BI3=

n

5.12=

5.1

5.12= 8.33 cm.

3. (1)

4.

Sol . (5)

Initial velocity of centre of mass of (A + B) system is

ucm

=3

8522001 = 10m/s upwards

maximum height attained by centre of mass aboveits initial position is

H =g2

u2

cm =102

100

= 5 metres

Ans . 5 m

5.

Sol . (1)

Ans. 1

Vefflux

= gh2

time of fall t =g

2)h4(

HINTS & SOLUTIONS

DATE : 28-04-2013

COURSE : VIJETA (JP)

PART TEST-1 (PT-1)

TARGET : JEE (MAIN+ADVANCED)-2014

x = Veffloy

t = )h4(h2

the roots of x are (0,4) and the maximum of x is at h = 2.The permitted value of h is 0 to 1 clearly h = 1 will give themaximum value of x is this interval.

Al it er : If the column of water itself were from ground upto aheight of 4m, h = 2m would give the maximum range x.Farther the hole is from this midpoint, lower the range. Herethe nearest point possible to this midpoint is the base of the

container. Hence h = 1m.

6. 7

7. 2

8. In the given .........................................

Sol . (6)

for lens :20

1

30

1

v

1

v = 60 cm

height of image formed by mirror is

30

4

90

= 1.2 cm

9. In the figure shown .........................................Sol . (4)

The image will move as shown in the figure. It is very clear fromthe figure that the required velocity is20 m/s So x = 4

10. A loaded vertical .........................................So l. (2)

KE =2

1m2a2cos2 t

PE =2

1m2a2sin 2 t

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KE PE =2

1m2a2(cos2 t sin2 t)

=2

1m2a2cos2t

Angular frequency = 2

The time period =2

T= 2 s

11. 5

12. 1.

13. Figure shows a .........................................Sol. (25)

mV = m2 OmV

m = 2

1 VG = VMG(1 + m

2)

= 10 4

5=

2

25cm/sec

14. 86

15. A body of mass .........................................Sol. (10)

=m

K = 50

8

400

2= 50

V = 22 xAV2= (A2 x2)

1 = 50

100

1A2

A2=

100

1

50

1

A2=100

3

A =10

3

100

3

x = 10

16. 10

PART-II (Chemistry)

19. When a block of sodium was ................

Sol. (8)

bcc

fcc

fcc

bcc

V

V

d

d (m - PV = const)

20. How many of the above closed diagrams ...............Sol. 4

a, c, d, e

21. Number of rectangular faces .............Sol. (4)

a b c = = 90, 120 90 4 faces

22. An element A forms compounds ................Sol. (8)

AB Can't be CsCl strucutre as 4 formula units in one unit cell.A

2B Na

2O type structure.

AD2CaF

2type structure.

Coordination no. of A

AB (ZnS) AB (NaCl) A C2 AD24 6 4 8

3

8443 = 8

3

8463 = 10.

Since single digit so answer is 8.

23. x g. of a cubic crystal of NaOH ..................Sol. (9)

number of NaOH units dissolved = 3 1022 4

moles of NaOH dissolved = 23

22

106

1012

= 0.2 molarity of NaOH = 0.2 mol/L molarity of Na+= 0.2 mol/L = y

mass of NaOH = 0.2 40 = 8g = x

Hence (x + 5y) = 8 + 5 0.2 = 9.

24. Consider an element A. It crystallizes ...................

Sol. (6)

nearest

next nearest

farthest

p =3x

4

q =x

2

r = 3x

4 4

3x x3 2x

2= 6.

25. Elements A, B and C form ................

Sol. (6)Alternate corners = 4.Alternate edge centres = 4.

A =8

14 =

2

1

B =4

14 = 1

C = 8

1

4 + 2

1

2 = 2

3

2

1A , B,2

3C AB2C3

x = 1y = 2z = 3 x2+ y + z = 1 + 2 + 3

= 6.

26. 3

27. 4

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SOLJPPT12804913 - 3

28. 9

29. 13

30. 11

31. In hcp lattice, if a plane is .........

Sol. (10).

32. 'u'is the number of C3axis ..........Sol. 19

PART-III (Mathematics)

33. If equations ax2+ 2bx + c = 0......................Ans .2

Sol. Adding all three equation, we get(a + 2b + c)(x2+ x + 1) = 02+ + 1 = 0

+

1= 1 3+ 3

1

= 2

34. If the ratio of the roots..................................Ans .1Sol. ax2+ 2bx + c = 0 , k

px2+ 2qx + r = 0 , k

+ k= a

b2; k2 =

a

c

+ k= p

q2; k2 =

p

r

aq

bp

andar

pc2

2

pr

q

ac

b 22

35. Find the number of integral ............................

Ans .1

Sol. log5

15log

3

13 < 0

2xlog3xlog 222 0

log2

2x 3 log2x + 2 = 0

or (log2x 1) (log

2x 2) = 0

x = 2, 4

only x = 4 satisfies the inequation x x 2 0The given system of inequalities has only one integral solution.

36. Number of pair(s) of..................................

An s. 1

Sol. [(3, 2) only. Note that p

2 for p > 2, p2

1 = (p

1) (p + 1)

is even and divisible by 4 ]

37. If =

1sin1

cos31sin

1cos31

.......................

An s. 2Sol.= (3cos sin)2

now 10 3cos sin 100 10

5max

= 2

38. Number of real value ..................................Ans.1Sol. For non-trivial solution

=a02

2a1

32a2

= 0

a3 2a 4 = 0 (a 2)(a2+ 2a + 2) = 0 a = 2

39. If A = [aij]44

such..................................Ans. 2Sol. |adj(adjA)| = |adjA|3

= |A|9

= (24)9

b = 36 = 22.32

40. The greatest common ..................................Ans . 6Sol. x2+ ax + b = (x + 1)(x + b) b + 1 = a

x2+ bx + c = (x + 1)(x + c) c + 1 = b(x + 1)(x + b)(x + c) = x3 4x2+ x + 6

1 + b + c = 4 2b = 4 b = 2, c = 3 and a = 1

Alter : x3 4x2+ x + 6 = 0 = (x + 1)(x 2)(x 3)

So quadratic will be x2 x 2 = 0 ; x2 2x 3 = 0

After compairing

b = 2, c = 3 and a = 1

41. Ifclogalog2

1

bb

..................................

Ans. 1

Sol.clogalogblog

2/1

1

1

bbb

+blogalogclog

2/1

1

1

ccc

+

clogblogalog2/1

1

1

aaa

=clogalogblog

1

bbb

+blogalogclog

1

ccc

+

clogblogalog

1

aaa

=abclog

1

b

+abclog

1

c

+abclog

1

a

= logabc

b + logabc c + logabc a

= logabc

abc =2

1

42. Let x be the solution ..................................Ans. 2

Sol.A2= A.A = I

A4= A6= A8=....... = I x = 2,4,6,8,..... .

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x x(cos sin ) = (cos2+ sin2) + (cos4+ sin4) +

(cos6+ sin6) + .........= (cos2+ cos4 + cos6 +...) + (sin2 + sin4 + sin6+ .........)

=

2

2

cos1

cos+

2

2

sin1

sin

= cot2+ tan2minimum value = 2

43. Let A =

x61

x3 2

,..................................

An s. 7Sol. tr(AB) = tr(C)

3ax2+ b + 6cx = 6x2+ 6x + 4 x R a = 2, b = 4 and c = 1

44. If A is non-zero square ................................An s. 3

Sol. I= (I 3A)(I A)

I =I A 3A + 3A2

I =I+ ( 2 3)A 2 = 3

45. If and are roots of the...............................Ans .18Sol. A

n+1= n+1+ n +1

= n.+ .n+ n+ n n n = n(+ ) + n(+ ) (n 1 + n 1)

An+1

= aAn bA

n 1

46. If a, b, c are roots of the ................................Ans .13

Sol. = abc

1c1

c1

c1

b

11

b

1

b

1a

1

a

11

a

1

R1R1+ R2+ R3

= abc

c

1

b

1

a

11

1c

1

c

1

c

1b

11

b

1

b

1111

C1C1 C 3 & C2C2 C3

= (abc + ab + bc + ca)

1c

111

b

110

100

= 11 + 2 = 13

47. If a, b, c, d respectively ..................................

Ans .14

Sol. a, b, c, d are consecutive integer in decreasing order

a = b2 c2+ d2

= b2+ (d c) (d + c)

as d c = 1

a = b2+ d + c

b2+ d + c a = 0

b2+ d + 2 = 0 not possible

a, b, c, d are in decreasing order

a = b2 c2+ d2

= (b c) (b + c) + d 2

= b + c + d2

d2+ c = a b

d2+ c = 1

d = 0, c = 1, b = 2, a = 3

a2

+ b2

+ c2

d2

= 9 + 4 + 1 + 0 = 14

48. If [ . ] denotes the greatest...............................Ans . 02Sol. [x]2+ (x)2< 4

if x , then [x] = (x) = x

x2< 2 i.e. 2 < x < 2 i .e. x = 1, 0, 1if x , then (x) = 1 + [x] [x]2+ (1 + [x])2< 4i.e. 2[x]2+ 2 [x] 3 < 0

i.e.4

282< [x] 2

1

n

n sin i > isin

icos

sin2i > cos i1 cos2i > cos icos2i + cos i 1 < 0

cos i I

H

cm

x yC

5

A

B

3

IC= I

CM+ my2

= IB

1 mx2+ my2

= IB

1+ m (y2x2) = I

P + I

B+ m (y2 x2)

> IP

+ IB

> IPHere I

B1 is moment of inertia of the plate about an axis

perpendicular to it and passing through B. I

C> I

P > I

B > I

H

6. A particle is .........................................Sol . (AC)

For convex mirror]|m| < 1 for any real objectNow, V

image= m2 V

object

|Vimage

| < | Vobject

| always

7. White light .........................................Sol. (ACD)

(A) It is true when i = 0.

(C) By cauchy's formula n = a + 2

b

, medium has lowest 'n'for red colour (out of all the colours in white light)V = C/n V is max. for red.(D) = | i r | 1 sin i = n sin rfor violet, n is max r is min. in max.

8. A convex lens .........................................Sol. (BCD)

Height of the object h = 2I/I

h = 6 cm.

m1=

h

I1=

6

9=

2

3

m2=

h

2=

6

4=

3

2

|h|

|u|=

2

3and | u | + | h | = 90

| u | = 54 cm| h | = 36 cm

distance between two position

f =a4

da 22 =

904

)18()90( 22

= 21.6 cm

9. Which of the .........................................Ans. (AD)

10. A prism of r.........................................Sol. (ABC)

sin C =2

1c = 45

for A > 2C no. refractionC < A < 2C some refraction, some reflection

A < C all refraction.

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11. At t = 0 the .........................................Sol. (ABD)

for t = 0x = A sin ........... (1)v = Acos ...........(2)a = A2sin.........(3)

By (1) and (3)

=x

a =

m08.0

s/m32 2

= 20 rad/s

By (1) and (2)

vx

=

tan tan = 1 or =4 or

45

At t = 0 , x is negative hence =4

5

from (1) , A = sinx

= 0.08 2 m = 11.3 cm

12. (ABC)

13. (ABCD)

14. (ABC)

15. (A) No sliding pure rollingTherefore, acceleration of the tube = 2a (since COM ofcylinders are moving at 'a')

PA

= Patm

+ (2a) L (From horizontal limb)

Also ; PA

= Patm

+ g

H (From vertical limb)

a = L2

gHAns.

16. Initially from .........................................

Sol . (C)

happ=1

2h

when viewed from left side, Lapph =

2/3

= 2 cm

when viewed from right side, Rapph =

2/3

17 = 14/3 cm.

17. The slab in the .........................................

Sol . (C)

Here L= 2

When viewed from left, Lapph =

2/3

23= 4 cm

When viewed from right, Rapph =

2/3

27=

3

28cm.

18. Maximum kinetic .........................................So l. (B)

Maximum KE is acquired by the block when it passes themean position of SHM where F = 0 or mg = kx.

x = mg/k =200

10=

20

1m

Applying work energy theorem from position A to B on theblock K

f K

i= W

gravity+ W

spring

Kf 2J = 10N

m

20

1+

2

20

1200

2

1

kf= 2.25 J.Ans.

19. The amplitude .........................................So l. (C)

KEmax

=21

kA2

A =k

)KE(2 max=

200

)25.2(2=

400

9=

20

3m =

15 cm.

20. Sol . (A) As, dm = Wvdt

dt

dm= AAWv

dt

dm= VW

4

D2

where D is the diameter of stream.

21. Sol . (D) V1A1= V2A2

4

Dv2

00=

4

Dv 2

D =v

vD 00 .

22. (A)

Let u be the required minimum velocity. By momentum

conservation :mu = (m + m)v v = u/2.

Energy equation :

2

1mu2=

2

1(2m)v2+ mgH.

Substituting v = u/2 :

u = 2 gH

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23. (A)

When the block comes to rest, the wedge continues

to move at V =2

u= gH on the smooth surface.

(since, momentum of wedge-block system remains

conserved).

24. A small wooden .........................................Sol. (4)

25. Light travelling .........................................Sol. (1)

we can use sin x = x (where x is in radians)

sin2 =2

3sin r1

r1=

3

4; r2= A r1=

3

8

2

3sin r2= 3

4sine e =

8

9r2= 3

= i + e A = 2 + 3 4 = 1

26. 2

27. Ans.4

Sol. (m + m)5

x= mx + mO

m + m= 5 m ; m= 4 m

;m

m= 4

PART-II (Chemistry)

28. Crystal system in which ..............

Sol. (C)

29. How many Zn+2ions are ..........Sol. (A)

Zn+2 occupy alternate tetrahedral voids

31. (B)

32. C

33. The correct statement(s) ..............Sol. (ABD)

36. Which of the following crystal ..............

Sol. (AB)

cubic and orthorhombic both have simple, body centered, facecentered unit cells.

37. The distance between nearest ..............Sol. (BC)

38. Select the correct statement ............Sol. (ABC)

43. What % of anion sites ..........Sol. (B)

44. If edge length of unit cell ..........Sol. (D)

45. Volume of the cube is ..........

Sol. (D)

6a2= S,33

3

4ar

46. Ratio of the surface areas ..........

Sol. (A)

Ratio =2

2

4

6

r

a

,

47.Sol. (B)

(A) (B)

(C) (D)

48.

Sol. (B)

Sol.

4-Ethyl hepta-1,3,5-triene

51. Atoms from six body centered .............

Sol. (3)

Number of unit cell =4

26

52. In case of one unit cell of ..............Sol. (8)

In diamond above shaded plane pass through = 4 corner + 2

face center + 2 tetrahedral voids so 8 atom.

PART-III (Mathematics)

55. The number of integral...................................Sol. (D) (x + a)(x + 1991) = 1

x + a = 1 and x + 1991 = 1 or x + a = 1 and x + 1991= 1

a = 1993 or a = 1989

56. Let and be the roots of the ..............................Sol. (C) + = 1154 and = 1

2 = + + 2 = 1154 + 2 = 1156 = (34)2

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= 34

Again (1/4+ 1/4)2= + 2()1/4= 34 + 2 = 361/4+ 1/4= 6

57. If A and B are squares ...............................................Sol. (A) AB = A + B

B = AB A = A(B l) det(B) = det (A) det(B l) = 0 det(B) = 0

58. B

59. If adjB = A, |P| = |Q| ...................................Sol. (C)We know that adjA = |A| A1

adj(Q1BP1) = |Q1BP1|. (Q1BP1)1

= |P||Q|

|B|. PB1Q

= |B|. P. |B|

1adjB.Q

= PAQ

60. If 2576a 456b is ..............................................Sol. (AD) 2576a456b is divisible by 5 b = 0 or 5

2576a456b is divisible by 3 35 + a + b = 3, Ia = 1, 4, 7, 2, 5, 8.

61. The value of x ..................................................

Sol. (AB) Let log2x = A, log

3x = B, log

5x = C

ABC = AB + BC + CA

1C

1

B

1

A

1 , ABC = 0

A = 0, B = 0, C = 0

x = 1

& logx2 + log

x3 + log

x5 = 1

logx30 = 1

logx30 = 1

x = 30

62. If f(x) = |x + 1|...................................Sol. (ABC) f(x) = |x + 1| 2 |x 1|

(i) x < 1

f(x) = x 1 + 2x 2 = x 3

(ii) 1 x 1f(x) = x + 1 + 2x 2 = 3x 1

( iii) x > 1f(x) = x + 1 2x + 2 = x + 3

63. If log|sinx|

(x2 8x + 23)...................................Sol. (BCD) log

|sinx|(x2 8x + 23) 3 log

|sinx|2 > 0

log|sinx|

8

23x8x2

> 0

0 < |sin x| < 1 x R {n, (4n 1)2

} , n I

x 0,2

, ,

2

3......(i)

8

23x8x

2

< 1

x2 8x + 15 < 0 x (3, 5) ......(ii)

(i) and (ii) x (3, )

2

3,

5,

2

3

64. If a, b are non-zero real...................................Sol.(BCD)

(A) S = 2+ 2= a2 2bP = 22= b2

equation is x2 (a2 2b) x + b2= 0

(B) S =

1

+

1

= b

a

, P =

1

.

1

= b

1

x2+b

ax +

b

1= 0

bx2+ ax + 1 = 0

(C) S =

+

=

22=

b

b2a2

P =

.

= 1

x2b

b2a

2

x + 1 = 0 bx2 (a2 2b) x + b = 0

(D) S = + 2 = a 2P = ( 1) ( 1)= (+ ) + 1= b + a + 1

equation is x2+ (a + 2)x + (a + b + 1) = 0.65. The equation............................................

Sol. (BC) 2)x( 45

xlog)x(log4

32

22

taking log both sides with base 2

4

5xlogxlog

4

32

22 log

2x=

2

1

02xlog5xlog4xlog3 22

23

2

02xlog7xlog31xlog 2222

log2x = 1,

3

1 , 2 x = 2, 2 1/3 ,

4

1

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66. Consider the equation.................................

Sol. (ABD)25

125

log5

xx2525 =

2

1+ log

52

2

)25()25( xx = 5

Let t25 x ..........(i)

t +t

1= 52

t2 52 t + 1 = 0 t = 25 , 25Putting these values of t in (i) we get x = 1, 1

67. If f() =

sin cos sin

cos sin cos

cos sin sin

........................

Sol. (AC) f() = (sin+ cos)(1 sin2)nowf() = 0 tan= 1 or sin2= 1

=4

3,

4

Also

2sin1

)(f= sin+ cos ( 2,2 )

68. If A1=

3

100

130

211

.................................

Sol. (ABC) |A1| = 1 |A| = 1Now adj A = |A|. A1= A1

A = (A1)1

69. If A and B are symmetric...............................Sol. (ABC) Given BT= B and AT= A and AB = BA

ABA1= BAA1

B = ABA1

A1B = A1ABA1

A1B = BA1

Similarly AB1= B1A

Now(A) (A1B)T= BT(A1)T= BT(AT)1= BA1= A1B(B) (AB1)T= (B1)TAT= (BT)1AT= B1A = AB1

(C) (A1B1)T= (B1)T(A1)T= (BT)1(AT)1= B1A1= A1B1

70. C .................................Ans . (C)

71. The value of |B| .................................Ans. (A)

Sol.(70-71) Let A =

333231

232221

131211

aaa

aaa

aaa

Now AC = B

C = A1B = |A|

1.adjA.B

C = |A|

111 21 31

12 22 32

13 23 33

C C C

C C C

C C C

323133313332

222123212322

121113111312

aaaaaa

aaaaaa

aaaaaa

=|A|

1.

0|A||A|

|A|0|A|

|A||A|0

C =

011101

110

which is symmetric

|C| = 2 |A1B| = 2 |B| = 2|A|

72. If < 0, then the.....................................Ans. (D)73. If a, b, c are rational.......... ............ ......... .

Ans. (D)

Sol.=443322

3322

22

111

111

11111

=221

1

111

.

221

1

111

= {(2 2) (2 ) + 2}2

= ( )2(+ 1 ( +))2

=

a

c4

a

b2

2 2

a

b

a

c1

= 4

22

a

)cba)(ac4b(

(1) < 0 b2 4ac < 0

imaginary roots

(2) ax2+ bx + c = 0 21 , 21

a

b= 2 and

a

c= 1

= 32

74. Number of integral solutions...........................Ans . (C)

75. Solution set of ............................................Ans . (C)

Sol. 74 f(x) 0 )4x5x)(1x)(3x2)(1e( 2x2

0

)4x)(1x)(1x)(3x2)(1e(2x 0

Critical points are x = 0, 1, 3/2, 4

x [1, 1] [3/2, 4]

also g(x) 0 (x 1)2(x + 1)2 0 x = 1 Number of integral solutions = 2 (x = 1)

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75. 0)x(g

)x(f 22

2x

)1x()1x(

)4x5x)(1x)(3x2)(1e(2

> 0

)4x)(1x)(1x)(3x2)(1e(2x 0 x 1

x (, 1) (1, 3/2] [4, ) {0}

76. Ans. (A)

77. Ans. (C)

78. Ans. 0

79. The number of values of..........................

Ans .9

Sol.

2

x+

4

x+

6

x=

12

x11= integer number

Let x = 12k , then

2

x+

4

x+

6

x= 6k + 3k + 2k = 11k.1k.

Also12

x11= 11k

statement is true for all integeral value of k. x must be multiple of 12. x = 0, 12, 24, 36, 48, 60, 72, 84, 96 which are 9 in number

80. If A is a square .................................An s. 4

Sol. |(adjA1)1| = 1

1

| adjA |= 21 |A|

1= |A|2= 4

81. The entire graph ...............................An s. 6

Sol. Here for D < 0 , entire graph will be above x-axis

( a > 0)

(k 1)2 36 < 0

(k 7) (k + 5) < 0 5 < k < 7