Solutions CT-2 JP

8/11/2019 Solutions CT-2 JP

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SOLJPCT2180813 - 1

PAPER-A : (MATHEMATICS)

1. At the x ! 5

6

", ......................................

Sol. (4)

x3sin213 – $ + 3cos3x $ 13

f(x) % ]13,13[ –

f ' (

)*+

, "

6

5= 2sin '

(

)*+

, "

2 + 3cos '

(

)*+

, "

2= 2

at x =

6

5", neither maximum nor minimum and not zero

value occurs.

2. If the v ectors ...........................................Sol. (1)

c11

1b1

11a

= 0

c11

c –11 –b0

0b –11 –a

= 0

- (a – 1) (c(b – 1) – (1 – c)) + ((1 – b)(1 – c)) = 0

- (a –

1)(bc –

1) + (1 –

b)(1 –

c) = 0- (1 – b)(1 – c) = (1 – a)(bc – 1) ...........(1)Required value

a –1

1+

b –1

1+

c –1

1=

)c –1)(b –1)(a –1(

]b –1c –11 –bc)[a –1( ##= 1

3. Let R be a relation........................................Sol. (3)

If a is negative, then |a| $ a is not true. R is not reflexiveIf a = 1, b = 2, then |a| $ b but |b| $ a is not true . R is notsymmetriclet a, b % R and b, c % R - |a| $ b and |b| $ c, then c / 0

! c / |b| - c / b . c / |a| or |a| $ c. (a, c) % R

Hence R is transitive.

4. If f(x) =01

02

3

4

!

5#

1x2 –x5

1xk

1x1x2

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

Sol. (2)

)h –1(flim0h6 = f(1) = )h1(flim

0h#

63 = k = 3 - k = 3

5. (2 )

6. (3)

HINTS & SOLUTIONS

CUMULATIVE TEST-2 (CT-2)

(JEE MAIN)TARGET : JEE (MAIN+ADVANCED) 2014

COURSE NAME : VIJETA (JP)

7. The number of solutions of.....................

Sol. (4)

L.H.S. % [ –1, 1] but R.H.S. / 2. Hence cannot be equal, no

solution.

8. (1) 9. (2) 10. (3)

11. If z is a complex.................................Sol. (1)

z = |z| 6

5i

e

"

= 78

9:;

< "#

"

6

5sini

6

5cos4

=778

9

::;

<

# 2

i

2

3 –4 = i232 – #

12. If A = {1, 3, 5, 7, 9, 11......................................Sol. (3)

(A = B) > B? = AA? = A = N

13. (1)

14. (2)

15. The value of.......................................Sol. (3)

–1x 0

sin(2tan3x)lim

sin 4x6×

x3

x3tan2×

x4

x3×

x4sin

x41 –

= 1 × 2 ×4

3 × 1 =

2

3

16. The chord of contact.............................Sol. (1)

1

2

1

2

2 2 2 2a b c b a b – –!

- !b ac

!A M G M. . . ./

17. (1) 18. (1)

19. (4) 20. (2)

21. The reciprocal of the ...................................

Sol. (3)

The shortest distance (SD)

!

2 1 4 2 5 3

2 3 4

3 2 5

2 3 4

3 2 5

– – –

" " "i j k 78

1!

DATE : 18-08-2013

CODE : 0 & 3



SOLJPCT2180813 - 2

22. The distance between .............................

Sol. (2)

Normal vector :

543

432

k ji

= –" "

– "i j k# 2

Equation of plane is –1(x – 1) +2(y – 2) –1(z – 3)=0

- # !x y z – 2 0

So, required distance ! # # !| |6

1 4 16

23. If the equation z2 + z + @ = 0........................

Sol. (4)

Let z i R! %A A( ) be a root, then

–A A @2 0# # !i - !@ A A2 – i

Now as | |@ ! 1 - # !A A4 2 1 - ! #

A2 5

2

–1

24. If |z| = 1 and z B C1 ................................

Sol. (3)

arg(z) – arg(z + 1) – arg(z – 1)

! F

H G I

K J ! F

H G I

K J arg –

arg –

z

z

z

z zz2 21

! F

H G I

K J !arg –

arg1

z z (purely imaginary no.

25. (2)

26. (3)

27. Equation of circle..................................

Sol. (2)

Radius of the required circle is12

and centre is (a, b)

Hence equation is ( – ) ( – )x a y b2 2 1

2# !

28. Let# #b and c are......................................

Statement 1 : # # #a b c. # !e j 4 ..........................

Sol. (2) Statement-1

c)b.a( –b)c.a(######

= (x2 – 2x + 6) b#

+ siny c#

- c.a ##

= x2 – 2x + 6 and siny = – (

b.a

##)

Given b.a##

+ c.a##

= 4

- – siny + x2 – 2x + 6 = 4

- x2 – 2x + 2 = siny

- x2 – 2x + 1 = siny – 1

(x – 1)2 = siny – 1

now –1 $ siny $ 1 - –2 $ siny – 1 $ 0

Hence x – 1 = 0 - x = 1 and siny = 1

Hence locus is straight line x = 1.

Statement-2 is standard result.

29. Statement 1 : If sin(2 cos –1(cot(2 tan –1x))).......

Sol. (1)

0x2

x –1cos2sin

21 – !

''

(

)

**

+

,

'' (

)**+

,

- ''

(

)**

+

,

x2

x –1cos2

21 –

= n"

- ''

(

)**+

,

x2

x –1cos

21 –

=2

n"

- ''

(

)**+

,

x2

x –1 2

= 0 or – 1 or 1

- x = ±1 or x2 ± 2x – 1 = 0

- x =

2

222 CC= ± (1 ± 2 )

Hence x = ±1, ±(1 ± 2 )

30. Let# # #a b c, , are ...........................................

Statement 1 : [ – – – ]2 2 2 0# # # # # #a b b c c a ! .........

Sol. (1) Obvious.

PAPER-B : (PHYSICS)

31. Consider a .............................Sol. (2)

2

1 2

V 1R R !# ...(1)

2

2 31

2 3

V2

R RR

R R

!#

#...(2)

2

2 3

V3

R R!

# ...(3)

221 2

1 2 1 3 2 3 1 23

1 2

V (R R )VP

R R R R R R R RR

R R

#! !

# ###

21 2

2

2 3

V (R R )P 6 watt

V(R R )

2

#! !

#

32. ( 2)

33. In the given .............................Sol. (2)

0 04B [sin 45 ]

4 (0.2) 4(0.1)

D E D E! F #

"



SOLJPCT2180813 - 3

=0 5 1 5

0.2 0.42

D "< 9#: 7" ; 8

, =050

[ 2 ]4

D# "

". Ans. (2)

34. Particle A ................. .......... ..Sol. (4)

1st separation will decrease then sep will increaseFor slope :- 1st there will be some V

app, then V

app will decrease

slowly and become zero, after Vsep

will increase gradually.

Hence slope will decrease till it become zero afterward it will

increase.

35. Three rings .............................Sol. (1)

Bz = B

y =

R2

0ED

Bnet

=2z

2y

2x BBB ##

= 3 R20ED

36. The distance .............................Sol. (2)

Potential in conductor is constantPotential in dielectric decreases at slower rate.

37. If x, y and .............................Sol. (2)

Resistance decreases - EG - xG

38. A reflecting .............................Sol. (2)

dx

dy = 2cos '

(

)*+

, L

x2 = –1

cosL

x2= –

2

1

x =3L ,

3L2

39. A particles with .............................Sol. (1)

U + k = Ewhen U = EK = 0

40. The equivalent .............................Sol. (1)

Circuit can be reduced to :

A B R R

2 R /3

2 R /3

2 R /32 R /3

2 R /3 2 R /3

R R

R R

R eq =H

35

22

41. A rod of length .............................Sol. (2)

Ldx e 1R e dx

A A A e

IJ JJ I, )! ! ! * '

+ (K K /0 0

0 0

..

L L

x L

' (

)*+

, IJ

!!.1e

e

L

AV

R

V I

0

00

42. The connection .............................

Sol. (1)

16µC 16µC

16µC 16µC

I + I+

+II+

24V

0V

12µC 24µC

12µC24µC

24V

0V

I + + I

I+ +I

Initially finally

. q through switch = 12 µC

43. Consider a long .............................Sol. (3)

q inner is equal to q outer in magnitude. E inner is greater than E outer always.. Pressure on A greater than Pressure on B . Force of interac-tion is equal and opposite. . Net force = 0.

44. The speed of .............................Sol. (2)

speed of boat in upstream = 15 km/hrDistance = 1.5 km

time t1 =

15

5.1 = 6 min

Dist. moved by float in this time

boat

1.5 km

floatvk

= 5 ×10

1 =

2

1 km

speed of boat w.r.t. float = 20 km/hr, t2 =

20km2 = 6 min.

45. The speed of .............................

Sol. (2)

Distance = v0 × 1 +

2

v0 × 1 +

4

v0 × 1 + ............

= v0

' (

)*+

, ### ..............

4

1

2

11

= 2v0

46. The focal length .............................

Sol. (2)Given f

0 = 2 cm, f

e = 5 cm

uo = -3, vo = 6 cm mo = -2for normal adjustment

ue = -5 cm, v

e = infinity m

e = 5

47. A bead is released .............................

Ans. (4)

48. The graph shows .............................

Sol. (1)At t = 3

Slope =

1d

V

dt

, )* '+ ( = –1



SOLJPCT2180813 - 4

- – 2V

1

dt

dV = –1

-

2

3

1''

(

)**+

,

dt

dV = 1 - a =

dt

dV = 3 m/s2

49. Two large .............................Sol. (4)

For 0 $ x $ d ; magnetic induction is negative.

For d $ x $ 2d ; magnetic fields due to the two planes canceleach other, hence becomes zero.For 2d $ x $ 3d ; magnetic field B is positive.

50. A magnet of .............................

Ans. (4)

51. A small element ............. ............ ....Ans. (2)

52. Four students .............................

Sol. (1)From given graph highest precision that means sharpness ismaximum in graph (I)

53. A point source .............................Sol. (1)

54. There is layer .............................Sol. (1)

cos

' cos

glass

air

LMN!

MN L putting the values we have MN’ =3

2MN

55. (2)

56. The current .............................

Sol. (3)

B =)r2(2

io D

"

"

2 =

r8

io D

57. Distance between .............................

Sol. (1)

Lf

1 = '

(

)*+

, I1

2

3' (

)*+

, I

20

1

20

1- F

L = 20 cm

f

1 =

10

1

I –

20

2 = –

5

1-

v

1 +

30

1

I =

5

1

I

- v

1 =

30

1 –

5

1

for lensv

1 –

30

1

I =

20

1

58. STATEMENT –1 : If current .............................Sol. (4)

Statement-1 is false but two is true.

59. STATEMENT –1 : When .............................

Ans. (1)

60. STATEMENT –1: When .............................Ans. (1)

PAPER-C : (CHEMISTRY)

61. Alternate tetrahedral void..............

Sol. (2)

In ZnS, S2 – occupy FCC lattice points and Zn2+ alternate tetrahedral

voids.

62. A compound contains two..............Sol. (2)

Zx = 1

8

18 !O . Z

y = 1 × 1 = 1 - XY

63. A solution contains 0.05 (M)..............

Sol. (1)

Ksp

(AgCl) = [Ag+] [Cl –]

[Ag+] required for AgCl ppt

= M103405.0

107.1

]Cl[

)AgCl(K 1010

sp II

I O!

O!

Ksp (Ag2CrO4) = [Ag+]2 [CrO42 –

][Ag+] required for Ag

2CrO

4 ppt

=

M103805.0

109.1

]CrO[

)CrOAg(K12

12

24

42sp II

I O!

O!

[Ag+] = 6.16 × 10 –6 M

[Ag+] required for AgCl is less than that of required for Ag2CrO

4.

Hence Cl – ion precipitated first.

64. Which of the following constitutes..............

Sol. (3)

H2O , I32POH , I24HPO are amphoteric in nature.

65. When the depression in freezing..............

Sol. (1)

Depresing in freezing point at equilibrium exist between Liquid

solvent and solid solvent

66. An aqueous solution of methanol..............

Sol. (3)

Methanol + water

PT = P

Aº X

A + P

BºX

B

Vapour pressure will lie between both of than. It will greater

than OH2P .

67. The donor atoms in EDTA –4 are..............

Sol. (2)



SOLJPCT2180813 - 5

In EDTA –4 , 4 oxygen atom and 2 nitrogen atom are donor in

nature.

68. The spin magnetic moment of..............

Sol. (3)

Hg[Co(SCN)4] P6 P Hg2+ + [Co(SCN)

4]2 –

Oxidation state of cobalt x + 4 × ( –1) = –2

x = 4 – 2 = +2

In +2 oxidation state cobalt have 3 unpaired electron (d7 elec-

tronic configurations).

So 15)23(3s !#!D B.M.

Hg[Co(SCN)4] P6 P Hg2+ + [Co(SCN)

4]2 –

69. (1)

70. (4)

71. (2)

72. What volume of 3 molar HNO3 is..............

Sol. (2)

Eq(HNO3) = Eq (Fe2+

)

(3 × 3) × V =1 / 56

8 V = 0.01588 L Q 16 ml

73. (3)

74. (3)

75. Which of the following form..............

Sol. (1)

76. Which of the following is optically..............

Sol. (4)

77. Normality of 1 M H3PO3 against..............Sol. (1)

78. (1) 79. (1)

80. (3) 81. (1)

82. (4) 83. (4)

84. (3) 85. (1)

86. (1) 87. (1)

88. (3)

89. (2)

90. Statement-1 : Nitrating mixture used for ..............

Statement-2 : In presence of H2SO

4, HNO

3..............

Ans. (1)

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