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RESONANCE SOLJP*CT1260513 - 1
PAPER-1PART-I (Physics)
1. Two large vertical .................................
Sol. (C)
qE = mg
X = Ed
= mg
dq
X = 2�
19�
27�
101106.1
10106.1
X = 10�9
Volt
2. (A)Work done by external agent = q [Vfinal � Vinitial ]
Wext =
22 r
º135coskp�
r
º45coskpq
= 2r4
qp
2
1��
2
1 Wext = 2
0r4
qp2
3. (C)
4. (C)
2
2
1
1
rGm
rGm
= 43
; 21
1
r4
m
= 2
2
2
r4
m
m1 + m2 = m 2R4
m
= 2
1
1
r4
m
or =
1
1
rGm
R
Gm
= 35
Ans.
5. In an electric .................................
Sol. (B)
Rate of change of potential is increasing
v1 � v2/x1 < v2 � v3/x2
x1 > x2
6. The diagram .................................
Sol. (A)
HINTS & SOLUTIONS
DATE : 26-05-2013COURSE : VIJETA (JP)
CUMULATIVE TEST-1 (CT-1)
TARGET : JEE (MAIN+ADVANCED)-2014
(Easy) It can be seen from the diagram that only the sphere B
and sphere C repel. Hence they both must be of same type.
According to the fact that at least two spheres are positively
charged, therefore both spheres should be positively charged.
Since attraction occurs for two remaining pairs it can be
concluded that the sphere A is negatively charged.
7. (C)
m = uf
f
m1 = 1xf
f
m2 = 2xf
f
m1 = �m2
2f = (x1 + x2) f = 2
xx 21 .
8. An electric .................................
Sol. (A)
Work done on 2 C charge = rd.Eq
= q
3
1
Edr
[ r for (1, 1, 0) = 2 & r for (3,0, 0) = 3]
= 2 × area of E-r graph from r = 2 m to r = 3
= 2 ×
1(3 2)20
2
= 20 (3 2) J.
9. A thin prism .................................
Sol. (D)
a =
1
23
× A = 2A
W = A13/42/3
=
8A
water
air
=
14
10. A converging .................................
Sol. (A)
401
� 601
=
f1
....(1)
v1
� 601
=
f21
....(2)
RESONANCE SOLJP*CT1260513 - 2
v1
+ 601
= 21
.
601
401
v1
+ 601
= 21
. 6040
100
�v1
= 481
� 601
v = 240 cm
11. (A)The inner sphere is grounded, hence its potential is zero.The net charge on isolated outer sphere is zero. Let thecharge on inner sphere be q�.
Potential at centre of inner sphere is
= 0a'q
41
o
+ a4q2
41
0 = 0 q� =
4q2
= �2q
12. (B)
The electric field at B is = 20 x2
q.
41
towards left.
VC = VC � VA = dxx2
q4
12
a
a2 0
= a16q
o
13. The distance .................................
Sol. (C)
The image distance of A is vA = � xf
)xf( ( x < f)
The image distance of B is vB = x)xf2(]f)xf2[(
x x
A F B Cpole
Solving | vA | + | vB | = 4f we get x = 2f
required time t = u2f
ux
14. The magnitude .................................
Sol. (B)
At time t = 0, velocity of image of A and B are i�u and i�urespectively. Therefore magnitude of their relative velocity
is 2u.
15. The magnitude .................................
Ans. (C)
16. The square .................................
Sol. (D)
The direction of electric field is in x-y plane as shown in figure
E
60°
N
O y
x
The magnitude of electric field is
E = 2y
2x E E = 13 = 2V/m.
The direction of electric field is given by
= tan�1 x
y
E
E= tan�1
3
1 = 30°
Hence electric field is normal to square frame LMNO as shown
in figure.
electric flux = E
. A
= E A cos0 = 2 × 1 = 2 V/m
C is correct option of Q. 15.
Since flux is maximum at = 60° , rotation by 30° either way
would lead to decrease in flux.
D is correct option of Q.16.
Lines ON and NM are both normal to uniform electric field E
.
Hence work done by electric field as a point charge 1 C is
taken from O to M is zero.
17. (8)
� R
GMm �
R9GMm
+ 21
mv02 =
21
mv2 � 2R5
GMm
� R9
]10[GMm +
R5GMm2
= 21
mv2 � 21
mv02
R45
]1850[GMm =
21
mv2 � 21
mv02
� R4532GMm
+ 21
mv02 0
21
v02
R45Gm32
v0 R45
Gm64 n = 8
18. (6)
E =
22
2
1R
)2(k2
2
1Rk2
= 52
1.
Rk2
= R
k 10 = 9 × 109 × 3 × 10�6 = 18 × 103 = 1.8 × 104 .
Alternate Solution:
Enet = Rk
102
= 10
103109102
69
= 54 × 103 N/C.Ans. 6
RESONANCE SOLJP*CT1260513 - 3
19. (1)
1 2kq kqV
3R 3R
1 2k q q9000
3R
99 10 3 c 9000 3R
R 1m
20. A positive .................................
Sol. (4)
� q2q1
x
4 7
1 2q qx 4 4
= 01 2q q
x 7 7
= 0
1
2
= x 4
4 1
2
qq =
x 77
x 44
= x 7
7
7x � 28 = 4x + 28
3x = 56
x = 563
1
2
=
567
37
= 113
| q2 | = + 1211
c q1 = 1211
× 113
= 4 c
21. A particle is .................................
Ans. 9
22. The side of the .................................
Sol. (3)Only effective pairs will be body diagonals for this interaction
electrostatic potential energy is
= 3
kq4 2
.
23. A ray of light .................................
Sol. (3)Apply Snell's law on various surfaces one by one :
1 sin 90° = 1 sin r1 sin r1 = 2
1 r1 = 45°
1 cos r1 = 2 sin r2 sin r2 = 2
1
2 cos r2 = 3 sin r3 sin r3 = 3
rsin1 22
2
3 cos r3 = 1 = 3
122
sin2 r3 + cos2 r3 = 1
3
122
+ 31
= 1 22 = 3 2 = 3
PART-II (Chemistry)
24. The number of carbon ....................
Sol. (A)Diamond structure is like ZnS (Zinc blende).
Carbon forming CCP(FCC) and also occupying half of tetrahedral
voids.
Total no. of carbon atoms per unit cell
=
)TV()FC()Corners(
421
681
8 = 8
25. How many grams of concentrated ...................
Sol. (D)
2501000
63Mass
2
mass = 2
63gm
mass of acid × 2
6310070
mass of acid = 45 gm
26. A metal has a fcc lattice ..................
Sol. (B)
3AaN
ZMd
31023 )10404(1002.6
M472.2
7
3
104
)404(02.672.2M
= 26.99 = 27 gm mole�1
27. (A)
P × 1 =
60001
20001
× 0.082 × 300
P = 0.0164 atm.
28. Which of the following represents correctly ...................
Sol. (D)
Ideal solution Gmix
= � ve,
TSmix
= + ve, Hmix
= 0.
RESONANCE SOLJP*CT1260513 - 4
29. (B)
30. (A)
31. (B)
32. (B)Cyclic unsaturated structural isomers are :
33. (D)
Seniority of F.G is �COOR > � COX > �CN >
O||
R�C�
34. Vant Hoff factor ....................
Sol. (A)
i = 1 � 2
i = 1 � 23.0
= 0.85
35. For 1 M solution of HA .............................
Sol. (D)
HA H+ + A¯
i = 1 +
= (i � 1)
Now Ka =
1C 2
Ka = )1i(1
)1i( 2
Ka = i2)1i( 2
36. An aqueous solution of 0.1 ....................
Sol. (D)Tf = Kfm = 1.86 × 0.1 = 0.186 ; Freezing point
= 0 � Tf = � 0.186ºC.
37. When 250 mg of eugenol is ......................
Sol. (A)
0.62 = 100M10250 3
× 1000 × 39.7 × 1
M = 160 or 1.6 × 102 g/mol
38. (A)
39. (C)
40. 32 g of hydrated magnesium sulphate MgSO4.x H2O ...................
Sol. (7)molar mass of salt = 120 + 18x
mass of water present in the salt = g32x18120
x18
Molality of the solution =
x18120x3218
84
1000x18120
32
= x26112604000
Tf = 4.836 = 2 1.86 x26112604000
x = 6.9 7
41. The difference in height of the mercury column in two.................
Sol. (4)
2CaClM = 111 g
2CaCln = 111222
= 2 mole O2Hn = 18324
= 18 mole
Relative lowering in vapour pressure
R.L.V.P. = 0
S0
PPP
= 21
1
ninin
= 100
80100 =
182i2i
or 0.2 = 18i2i2
or 0.4i + 3.6 = 2i
i = 2.25.
So i = 2.25
For CaCl2 i = 1 + (n �1)
2.25 = 1 + (3 � 1)
= 225.1
= 0.625.
6.4 × 0.625 = 4.
42. 5
XN2 (gas) = 2520
= 54
PN2 =
54
50 = 40 atm
40 = 20108
K3
H
KH = 3108
2040
= 105 atm.
43. For the reaction : A + 2B C .........................
Sol. (4)A + 2B C
5 8
15
28
(B is L.R)
From mole�mole analysis
28
= 1
nC
44. 5
45. 4
46. 6
C�C�C�C�OCH3, , ,
, C�C�C�OC2H5,
RESONANCE SOLJP*CT1260513 - 5
PART-III (Mathematics)
47. If the roots of the equation..............
Sol. (D)Product of roots = 1
)cb(a)ba(c
= 1 b = )ca(
ac2
ab
= ca
c2
=
1ca
2
� (+) =
11
2
2= � 1 �
48. If x2 + ax + b is an integer...............
Sol. (A)Let f(x) = x2 + ax + b
Now it is integer for every integral value of x a, b I
Now roots of f(x) = 0 are
2a a 4b2
..........(1)
roots are rational a2 � 4b = k2where k I
Now k is even or odd according as a is even or odd
(1) roots must be integers.
49. The least value of |a|..............
Sol. (A)sin + cosec = � a
|sin + cosec |= |a|
|sin|1sin2
= |a|
|sin|1
+ |sin |= |a|
|a| 2
50. If p, q R and p2 + q2 � pq................
Sol. (C)p2 + q2 � pq � p � q + 1 0
2
)1q()1p()qp( 222
0
p = q = 1
Now =
1coscos
cos1cos
coscos1
= sin2 � cos2 + 2 cos cos cos � cos2
= 0
= 0
cos () = cos
or (cos cos � cos )2 = sin2 sin2
= (1 � cos2 ) (1 � cos2 )
51. The set of values of 'a' for ..............
Sol. (D)Let y = x2 � 4x + 5
y = (x � 2)2 + 1
sin�1 y, cos�1 y defined only at y = 1
at which x = 2
So given equation
4 + 2a + 2
= 0
8a
4
52. The domain of definition ..............
Sol. (A)for f(x) to be defined
025284 )1x(2)2x(
32
x
and 01xx
x2
52]221[2 24x2
and 0x
642 x2 3x
Hence domain is ,3
53. tan�1 1 + tan�1 .............
Sol. (C)tan�1 1 + (tan�1 2 + tan�1 3)
= tan�1 1 + + tan�1
6132
=
54. The period of sin [x]4
+............
Sol. (D)
period of sin [x]4
+ cos [x]2
+ cot 3
]x[ are 8, 4, 3 respec-
tively.
Hence L.C.M. of 8, 4, 3 i.e.,24 is required period.
55. The domain of the...........
Sol. (C)
)xx3x2(
)1x2(23
0 x (� , �1)
0,
21
1,
2
for log2 x, x > 0
for )x(logsin 21 , 0 log2 x 1
x [1, 2] ........(2)
So by (1) and (2) we get x [1, 2]
RESONANCE SOLJP*CT1260513 - 6
56. If the function f : [1, ) ...........
Sol. (A)y = f(x) = 2x(x � 1)
x(x � 1) = log2y
x2 � x � log2 y = 0
x = 21 1 4log y
2
x 1
x = 21 1 4log y
2
y = 21
2(1 1 4log x )
57. Minimum value of y = g(x)..........
Ans. (C)
58. The value of b2 .........
Ans. (B)
Sol. 57, 58( � ) = (( + ) � ( + ))
( + )2 � 4 = [( + )+ ( + )]2 � 4( + ) ( + )
(�b1)2 � 4c1 = (�b2)
2 � 4c2
D1 = D2 ..........(i)
minimum of y = f(x) is � 4
D1 = �
41
D1 = 1 D2 = 1
minimum of y = g(x) is � 4
D2 = �
41
minimum of y = g(x) occur at x = � 2
b2 =
27 b2 = �7
b22 � 4c2 = D2
49 � 4c2 = 1 4
48 = c2 c2 = 12
59. Which of the following .................
Ans. (D)
60. Which of the following.............
Ans. (D)
Sol. cos�1
2x
+ cos�1
3y
=
cos = 6xy
� 4x
12
9y
12
4x
12
9y
12
=
cos
6xy
4x
12
9y
12
=
2
cos6xy
1 � 4x2
� 9y2
+ 36
yx 22
= 36
yx 22
+ cos2 � 3cos xy
1 � cos2 = 4x2
+ 9y2
� 3cos xy
9x2 + 4y2 � 12 xycos = 36 sin2
N = 36
and (cos�1 x)2 � (sin�1 x)2 > 0
2
xsinxcos 11 > 0
x
2
1,1 [p, q)
p = � 1 , q = 2
1
59. N � 6 = 36 � 6 = 0
2
1,1
60. sec�1x is not defined at x = 0
61. If D is the set of .............
Sol. (B)
1 � 1�
x1
e > 0i.e. 1�
x1
e < 1
i.e.x1
� 1 < 0 i.e. x
1x > 0 i.e. x (� , 0) (1, )
62. f(x) is a/an..............
Sol. (A)
Here f(x) = 2
1�x1
x
e > 0 ;
f(x) is one-one
63. Sum of all elements...........
Ans. 5
Sol. For defining f(x)
(16 � x), (20 � 3x) N ; (2x � 1), (4x � 5) W
(16 � x) (2x � 1) ; (20 � 3x) (4x � 5)
Hence x = 2, 3
when x = 2, f(x) = 14C3 + 14C3 = 728
when x = 3, f(x) = 13C5 + 11C4 = 1617
Hence sum 728 + 1617 = 2345
64. If 1 lies between the roots.............
Ans. 3
Sol. 1 lies between the rootss
hence f (1) < 0
3 � 3 sin � 2 cos2 < 0
2 sin2 � 3 sin + 1 < 0
21
< sin < 1
RESONANCE SOLJP*CT1260513 - 7
2n + 6
< < 6
5+ 2n, n I
In [0, 2],
65
,6 �
2
= 1, 2
Required sum 1 + 2 = 3
65. If f(x) =
0x0
n,nex
21
eex
)1xxlog(x 2
I
.........
Ans.0
Sol. Let f(x) = )x(h
)x(g.)x(p
Function (M)where p(x) = x odd function
g(x) = log (x + 1x2 )
g(x) + g(�x) = log (x + 1x2 ) + log (�x + 1x2
)
= log (�x2 + x2 + 1)
= log (1)
= 0 g(x) is an odd function.
h(x) = 21
eex
=
ex
+ 21
h(x) + h(�x) =
ex
+ 21
+
ex
+ 21
=
ex
+
ex
+ 1
= �1 + 1 [x] + [�x] = �1 if x R � I
= 0
h(x) is an odd function
f(x) is an odd function f(1947) + f(� 1947) + f(0) = 0.
66. (7)
67. If sum of the series .............
Ans.0Sol. Tr = cot�1 (2r2)
S =
1r2
1�
1r
21�
r2
1tan)r2(cot
=
1r
1�
)1�r2)(1r2(1)1�r2(�)1r2(
tan
=
1r
1�1� )1�r2(tan�)1r2(tan
= (tan�13 � tan�11) + (tan�15 � tan�13) + (tan�17 � tan�15) +.... + tan�1
S = �4
+ 2
= 4
Given S = 4
= k k = 41
[k] = 0
68. Given equation |x2 � 5x + 4 + sinx|...............
Ans. 5Sol. |a + b| = |a| + |b|
ab 0 (x2 � 5x + 4) sinx 0 x [0, 1] [,
4] {2}
sum of all integral values is 0 + 1 + 4 = 5
69. If f(x) =
)�cos()�cos()�cos(
)xcos()xcos()xcos(
)xsin()xsin()xsin(
.............
Ans. 9
Sol. Using the idea of the differentiation of determinant, we get
f(x) =
)�cos()�cos()�cos(
)xcos()xcos()xcos(
)xcos()xcos()xcos(
+
)�cos()�cos()�cos(
)xsin(�)xsin(�)xsin(�
)xsin()xsin()xsin(
+
000
)xcos()xcos()xcos(
)xsin()xsin()xsin(
= 0 + 0 + 0
f(x) = 0 for all x
f(x) = a constant
But f(9) =
f(x) = for all x.
9
k 1
f(k)
= 9
RESONANCE SOLJP*CT1260513 - 8
PAPER-2PART-I (Physics)
1. (D)
2. A positively ....................................
Sol. (A)Electric field on surface of a uniformly charged sphere is given
by 0
30 3
R
R4
Q
Electric field at outside point is given by
E = 20
3
20 r3
R
r4
Q
|E| r =
0
0
3
r
�
20
0
30
2r3
3
2r
= 0
0
54
r17
leftt
3. (C)
Given 8 = 21
2
=
21 T2
T2
2
, TT1 = 24 hours for
earth. T2 = 12 hours (T2 being the time period ofsatellite, it will remain same as the distancefrom the centre of the earth remains constant).
T = 12
2
=
12 T2
T2
2
= 24 hours.
4. (B)
5. In the figure ....................................
Sol. (C)
F = 2
2
r
qk K =
41
C
2
2
2
Rmv
r
qk RC = 2
22
qk
rmv;
RC = 2
22
q
mrv4
6. If uniform electric ....................................
Sol. (B)A = (O, O), B = (xo, O)
B
A
nd.E = VA � VB = O � VB
Eo Xo = � VB
VB = � Eo xo
7. A triangular medium ....................................
Sol. (B)
Clearly, PM = 23
cm
37º > sin�1 )2/3(an1
0
53
>
2a3
n
1
0
3n0 + 2a9
> 5
2a9
> 1 a > 92
8. A uniform electric ....................................
Sol. (D)
j�Ei�EE yx
, V = �Ex x � Ey y
for A and B
16 � 4 = � Ex (� 2 � 2) � Ey (2 � 2)
Ex = 3 V/m
for B and C
12 � 16 = �Ex {2� (� 2)} � Ey (4 � 2)
Ey = �4 V/m.
m/V)j�4�i�3(E
9. A cubical region ....................................
Sol. (ACD)
RESONANCE SOLJP*CT1260513 - 9
Net flux through x = a2
& x = a
�2
will be equal
Net flux through y = a2
& y = a
�2
will be equal
y = a2
y = a
�2
innet
0 0 0
q �q 3q � q q
.
Net flux through Z = +a
�2
& x = +a
�2
will be equal.
Ans. A, C & D
10. (A,B,C,D)
11. Two identical ....................................
Sol. (BCD)
Potential r1 < r2
V+ = 1r
kq2
+ � + � x
r2
r1
y
V� =
2r
kq2
Vnet = V+ + V� = 2kq
21 r
1
r
1 0
Electric field :
+ � + � x
r2r1
E�E�
EEz
E = 21r
kq
E' = 22r
kq
E' < E, hence electric field for z > 0 is directed along +ve z-axis.
Similarly for z < 0, electric field is directed along �ve z-axis.
Hence electric field at origin is zero.
+ � + � x
z
E� E�
E E
12. (BCD)
13. A luminous point ....................................
Sol. (AC)
(A)
(C)
(D) Image is inverted It should be real
14. S1 : When a concave ....................................
Sol. (C)S1 : The focal length of a concave mirror depends only on its
radius of curvature.
S2 : f1
= (nrel � 1)
21 R1
�R1
nrel = gsurroundin
rel
n
n
nsurrounding fnrel
S3 : Since E = drdv�
: if E = 0
V = constant not necessarily equal to zero
RESONANCE SOLJP*CT1260513 - 10
15. Four particles are ....................................
Sol. (B)The potential of the centre is positive. From symmetry of the
problem B and D will move away on the line BD and A and C will
move away on the line AC.
Since masses are different, their speeds will be different.
16. (A)
17. (C)
18. In each figure ....................................
Ans. (A) � r ; (B) � p,q,r ; (C) � q ; (D) � s,t
19. In each situation ....................................
Ans. (A) � p,r, t ; (B) � p,q,r,s,t ; (C) � p,q,r,t ; (D) � p,s,tSol. The electric field due to one dipole at centre of other dipole
is parallel to that dipole. Hence torque on all given dipoles
is zero.
In case B and C the electric field at second dipole due to
first is along the second dipole. hence electrostatic potential
energy of second dipole is negative.
In case A and B x-axis is line of zero potential. In case B
and C electric field at origin is zero.
PART-II (Chemistry)
20. The olivine series of minerals consists of crystal in which Fe2+
and Mg2+ .........................
Sol. (C)According to given data :
O2� = 8 × 81
+ 6 × 21
= 4
Si4+ = 41
× octahedral void = 41
× 4 = 1
M++ = 41
× tetrahedral void = 41
× 8 = 2.
Forsterite = Mg2SiO4 Fayalite = Fe2SiO4
Let the forsterite is x% and fayalite is (100�x) % then
10034.4)x100(21.3x
= 3.88
x = 40.71% (Forsterite) and 59.29% (Fayalite).
21. What is the concentration of nitrate ions .........................
Sol. (C)
[NO3�] = V2
0V1.0 = 2
1.0 = 0.05 M
22. (C)For immiscible solution = PT = PA
0 + PB0
= 400 + 200 = 600
23. In the closest packing of atoms.........................
Sol. (B)
24. (C)�COOH is the prior F,G and there are four carboxylic acid
are possible.
25. (B)
26. (A)
27. (C)
28. For a binary ideal liquid solution ...............
Sol. (B)In (B) option component B is more volatile where as in (D) option
A is more volatile.
29. (D)
30. (A,C,D)
31. (C,D)Esters do not show chain and position isomerism.
32. (A,C)(a) and are structural
isomers.(b) These compounds are identicle(c) These are chain isomers(d) These are homologs.
33. (B)
34. S1 : The lowering of vapour pressure ...............
Sol. (D)S1 : Relative lowering = Xsolute
S2 : NaCl is electrolyte while sugar is not
S3 : Eqilibrium constant of an endothermic reaction decrease
with decrease in temperature.
S4 : PP
is a colligative property..
35. (B)
36. (B)
37. (A) 1 : 0.2 M KCl, 2 :0.2 ...............
Sol. (A) � r , s ; (B) � p ; (C) � r, s ; (D) � q, t
38. (A) - p ; (B) - p,q ; (C) - p,r ; (D) - s
PART-III (Mathematics)
39. The value of where...............
Sol. (D)
= sin�1
813
2
+ cos�1
23
+ sec�1 ( 2 )
= sin�1
22
13 + cos�1
23
+ sec�1 2
RESONANCE SOLJP*CT1260513 - 11
= 12
+ 6
+ 4
= 12
32 =
2
.
40. 2 cot�1 7 + cos�1 53
..............
Sol. (C)
2 cot�1 = 2 tan�1 71
= cos�1
49/1149/11
= cos�1 2524
Hence cos�1 2524
+ cos�1 53
= cos�1
54
.257
53
.2524
= cos�1 12544
= cot�1
11744
= cosec�1 117125
41. Let f : R R defined...........
Sol. (C) f ' (x) = 3x2 + 2x + 100 + 5 cos x
= 3x2 + 2x + 94 + (6 + 5 cos x) > 0
f is strictly increasing hence one one also
when x � , f(x) �
x , f(x) onto
42. The principle argument of z = x....................
Ans.(B)
43. The graph of |y| = | |x| � 1|....................
Sol. (C)
|x|
|x| � 1
||x| � 1|
|y| = ||x| � 1|
44. (B)
As |sin 4x| � |cos 4x| has period 4
But on taking x4cosx4sin as g(x)
we get
8xg =
x4
2cosx4
2sin
= x4sinx4cos = g(x)
period of g(x) is 8
. Now h(x) = cos (cos 6x)
then x6coscos6
xh
= cos (� cos 6x) = cos (cos 6x) Period is 6
Taking L. C. M of 6
, 8
we get 2
45. Period of f(x) = sin ...............
Sol. (D)
L.C.M.
3
2,
2
2
= For period L.C.M. is not possible
46. If A =
43
21.............
Sol. (D)
AB =
43
21
dc
ba =
d4b3c4a3
d2bc2a
BA =
dc
ba
43
21 =
d4c2d3c
b4a2b3a
if AB = BA, then a + 2c = a + 3b
2c = 3b
b 0
b + 2d = 2a + 4b
2a � 2d = � 3b
cb3d3a3
=
b23
b3
b29
= � 1
RESONANCE SOLJP*CT1260513 - 12
47. The value(s) of x...........
Sol. (A,B,C)
1 � 2
3)1x(log 2 =
21
3x5x
log 2/13
1 � 22
log3 |x + 1| = 22
log3
3x5x
log3
|1x|3
= log3
3x5x
|1x|3
= 3x5x
Case - I x + 1 > 0 x > � 1
3(x + 3) = (x + 1)(x + 5)
x2 + 3x � 4 = 0
x = � 4 or x = 1
x = � 4 rejected ( x > � 1)
x = 1
Case - II x + 1 < 0 x < � 1
3(x + 3) = � (x + 1)(x + 5)
x2 + 9x + 14 = 0
x = � 2 or x = �7
Set of value of x = {�7, �2, 1}
48. For the function f(x)...............
Sol. (A,C)
f(x) = ex (x2 + bx + c) > 0 iff D = b2 � 4c < 0
g(x) = ex (x2 + (b + 2)x + (b + c)) > 0 iff
D = b2 � 4c + 4 = D + 4 < 0
49. If A and B are two................
Sol. (B)(A + B)2 = (A + B). (A + B)
= A2 + AB + BA + B2 .........(1)
Since B = � A�1 BA
AB = A (�A�1 BA)
A(BC) = (AB) C
= �(AA�1) BA
AB = � BA
put in (1), we get
(A + B)2 = A2 � BA + BA + B2
(A + B)2 = A2 + B2
50. (B,C)
51. The maximum number...............
Sol. (B)The maximum number of distinct elements
= no. of positions in diagonal & above
= 50 + 49 + ...... + 1 = 1275
52. Consider the following statements....................
Sol. (D)
S1 : x2 + 3x + 3 = x2 + 3x + 43
49 =
2
23
x
+
43
least value of 3x3x2 is
23
3
sin�1 3x3x2
2
3x3xsin 21 = 1
n
3x3xsin 21 = 0
S2 : f(x) = tan�1
2�xtanxcot2
1 11
= tan�1
xcot1 1
Range of f(x) is
43
tan,41
tan 11
no integral value in the range
S3 : 2x
sin 1 < 2
xcos 1
i.e. 2x
sin 1 < cos�1
2x
which is true for �2 x < 2
integral value are
�2, �1, 0, 1
53. Consider the following statements.................
Sol. (B)S1 : f(x) = sin�1{x}
f(x + 1) = sin�1 {x + 1} = sin�1 {x} = f(x)
1 is a period and f(x) is periodic
S2 : Let h(x) = f(x) + |g(x)|
h(�x) = f(�x) + |g(�x)| = f(x) + |�g(x)| = f(x) +
|g(x)| = h(x)
h(x) is an even function
S3 : sin (sin�1x) = x and cos (cos�1x) = x for all x [�1, 1]
domains are equals
both are identical
54. Consider the following statements.............
Sol. (D)Obvious
55. (C)
RESONANCE SOLJP*CT1260513 - 13
56. Match the inequality in column-I with................
Ans. (A) - (r), (B) - s, (C) - q, (D) - pSol. (A) logsinx (log3 (log0.2 x)) < 0 = logsinx1
log3 (log0.2x) > 1 log0.2x > 3 = log0.2(0.2)3
0 < x < (0.2)3 0 < x < 125
1
(B) )1x(x)2x(sin)2xx)(3x2)(1e( 2x
0
)1x(x
)2/3x)(1e( x
0 x < � 1 or x
23
x (�, �1)
,
23
(C) |2 � | [x] � 1| | 2
||[x] � 1| � 2| 2 0 |[x] � 1| 4
� 3 [x] 5 x [�3, 6)
(D) |sin�1 (3x � 4x3)| 2
�2
sin�1 (3x � 4x3) 2
� 1 3x � 4x3 1 � 1 x 1
57. If f(x) = 2�x1�x2
.............
Ans. (A) - (r), (B) - (t), (C) - (q), (D) - (p)
Sol. (A) x 0 and g(x) 2 x 1
Domain is x (�, 0) (0, 1) (1, )
(B) x 2 and f(x) 0 x 1/2
Domain is x (�, 1/2) (1/2, 2) (2, )
(C) f(f(x)) = x and x 2
Range is (�, 2) (2, )
(D) g(g(x)) =1x1x2
and x 0
Range is (�, 1) (1, )
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