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Page 1
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 21 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 21 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Let a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? and c
?
is nonzero vector and c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 find c . b
?
?
ekuk a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? rFkk c
?
,d v'kwU; lfn'k gS rFkk c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 rks c . b
?
?
dk eku
gksxk%
(1)
2
1
(2)
3
1
(3)
2
1
? (4)
3
1
?
Ans. (3)
Sol. ) c b ( a
?
?
?
? ? = ) a b ( a
?
?
?
? ?
– ) b . a (
?
?
c
?
= ( a
?
. a
?
) b
?
– ( b . a
?
?
) a
?
– c 4
?
= 6( k
ˆ
j
ˆ
i
ˆ
? ? ) – 4( k
ˆ
j
ˆ
2 i
ˆ
? ? )
– c 4
?
= k
ˆ
2 j
ˆ
2 i
ˆ
2 ? ?
c
?
=
2
1
? ( k
ˆ
j
ˆ
i
ˆ
? ? )
c . b
?
?
=
2
1
?
2. Let coefficient of x
4
and x
2
in the expansion of
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? is ? and ? then ? – ? is
equal to
ekuk
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? ds izlkj esa x
4
rFkk x
2
ds xq.kkad Øe'k% ? rFkk ? gS rks ? – ? dk eku gS %
(1) 48 (2) 60 (3) –132 (4) –60
Ans. (3)
2[
6
C0 x
6
+
6
C2 x
4
(x
2
– 1) +
6
C4 x
2
(x
2
– 1)
2
+
6
C6 (x
2
– 1)
3
]
= 2[x
6
+ 15(x
6
– x
4
) + 15x
2
(x
4
– 2x
2
+ 1) + (–1 + 3x
2
– 3x
4
+ x
6
)]
= 2(32x
6
– 48x
4
+ 18x
2
– 1)
? ? = – 96 and vkSj ? = 36 ? ? ? – ? = –132
Page 2
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 21 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 21 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Let a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? and c
?
is nonzero vector and c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 find c . b
?
?
ekuk a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? rFkk c
?
,d v'kwU; lfn'k gS rFkk c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 rks c . b
?
?
dk eku
gksxk%
(1)
2
1
(2)
3
1
(3)
2
1
? (4)
3
1
?
Ans. (3)
Sol. ) c b ( a
?
?
?
? ? = ) a b ( a
?
?
?
? ?
– ) b . a (
?
?
c
?
= ( a
?
. a
?
) b
?
– ( b . a
?
?
) a
?
– c 4
?
= 6( k
ˆ
j
ˆ
i
ˆ
? ? ) – 4( k
ˆ
j
ˆ
2 i
ˆ
? ? )
– c 4
?
= k
ˆ
2 j
ˆ
2 i
ˆ
2 ? ?
c
?
=
2
1
? ( k
ˆ
j
ˆ
i
ˆ
? ? )
c . b
?
?
=
2
1
?
2. Let coefficient of x
4
and x
2
in the expansion of
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? is ? and ? then ? – ? is
equal to
ekuk
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? ds izlkj esa x
4
rFkk x
2
ds xq.kkad Øe'k% ? rFkk ? gS rks ? – ? dk eku gS %
(1) 48 (2) 60 (3) –132 (4) –60
Ans. (3)
2[
6
C0 x
6
+
6
C2 x
4
(x
2
– 1) +
6
C4 x
2
(x
2
– 1)
2
+
6
C6 (x
2
– 1)
3
]
= 2[x
6
+ 15(x
6
– x
4
) + 15x
2
(x
4
– 2x
2
+ 1) + (–1 + 3x
2
– 3x
4
+ x
6
)]
= 2(32x
6
– 48x
4
+ 18x
2
– 1)
? ? = – 96 and vkSj ? = 36 ? ? ? – ? = –132
3. Differential equation of x
2
= 4b(y + b), where b is a parameter, is
x
2
= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk %
(1)
2
2
x
dx
dy
y 2
dx
dy
x ? ? ?
?
?
?
?
?
(2) x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
(3)
2
2
x
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
(4)
2
2
x 2
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
Ans. (2)
Sol. 2x = 4by
'
? ?
'
y 2
x
b ?
So. differential equation is
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ?
vr% vody lehdj.k
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ? ? ? x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
4. Image of (1, 2, 3) w.r.t a plane is ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
then which of the following points lie on the plane
(1, 2, 3) dk fdlh lery esa izfrfcEc ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
rks fuEu esa ls dkSulk fcUnq lery ij fLFkr ugha gS %
(1) (–1, 1,–1) (2) (–1, –1, –1) (3) (–1, –1,1) (4) (1, 1, –1)
Ans. (4)
Sol. d.r of normal to the plane
lery ds yEcor~ f}dvuqikr
3
10
,
3
10
,
3
10
1, 1, 1
midpoint of P and Q is ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
P rFkk Q dk e/; fcUnq ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
equation of plane x + y + z = 1
lery dk lehdj.k x + y + z = 1
5.
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
is equal to
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
dk eku gS %
(1) 1 (2) 10 (3) 5 (4) 0
Page 3
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 21 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 21 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Let a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? and c
?
is nonzero vector and c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 find c . b
?
?
ekuk a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? rFkk c
?
,d v'kwU; lfn'k gS rFkk c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 rks c . b
?
?
dk eku
gksxk%
(1)
2
1
(2)
3
1
(3)
2
1
? (4)
3
1
?
Ans. (3)
Sol. ) c b ( a
?
?
?
? ? = ) a b ( a
?
?
?
? ?
– ) b . a (
?
?
c
?
= ( a
?
. a
?
) b
?
– ( b . a
?
?
) a
?
– c 4
?
= 6( k
ˆ
j
ˆ
i
ˆ
? ? ) – 4( k
ˆ
j
ˆ
2 i
ˆ
? ? )
– c 4
?
= k
ˆ
2 j
ˆ
2 i
ˆ
2 ? ?
c
?
=
2
1
? ( k
ˆ
j
ˆ
i
ˆ
? ? )
c . b
?
?
=
2
1
?
2. Let coefficient of x
4
and x
2
in the expansion of
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? is ? and ? then ? – ? is
equal to
ekuk
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? ds izlkj esa x
4
rFkk x
2
ds xq.kkad Øe'k% ? rFkk ? gS rks ? – ? dk eku gS %
(1) 48 (2) 60 (3) –132 (4) –60
Ans. (3)
2[
6
C0 x
6
+
6
C2 x
4
(x
2
– 1) +
6
C4 x
2
(x
2
– 1)
2
+
6
C6 (x
2
– 1)
3
]
= 2[x
6
+ 15(x
6
– x
4
) + 15x
2
(x
4
– 2x
2
+ 1) + (–1 + 3x
2
– 3x
4
+ x
6
)]
= 2(32x
6
– 48x
4
+ 18x
2
– 1)
? ? = – 96 and vkSj ? = 36 ? ? ? – ? = –132
3. Differential equation of x
2
= 4b(y + b), where b is a parameter, is
x
2
= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk %
(1)
2
2
x
dx
dy
y 2
dx
dy
x ? ? ?
?
?
?
?
?
(2) x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
(3)
2
2
x
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
(4)
2
2
x 2
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
Ans. (2)
Sol. 2x = 4by
'
? ?
'
y 2
x
b ?
So. differential equation is
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ?
vr% vody lehdj.k
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ? ? ? x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
4. Image of (1, 2, 3) w.r.t a plane is ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
then which of the following points lie on the plane
(1, 2, 3) dk fdlh lery esa izfrfcEc ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
rks fuEu esa ls dkSulk fcUnq lery ij fLFkr ugha gS %
(1) (–1, 1,–1) (2) (–1, –1, –1) (3) (–1, –1,1) (4) (1, 1, –1)
Ans. (4)
Sol. d.r of normal to the plane
lery ds yEcor~ f}dvuqikr
3
10
,
3
10
,
3
10
1, 1, 1
midpoint of P and Q is ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
P rFkk Q dk e/; fcUnq ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
equation of plane x + y + z = 1
lery dk lehdj.k x + y + z = 1
5.
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
is equal to
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
dk eku gS %
(1) 1 (2) 10 (3) 5 (4) 0
Ans. (4)
Sol. Using L’Hospital
L gkWfLiVy ls
0 x
lim
? 1
) x 0 1 sin( x
= 0
6. Let P be the set of points (x, y) such that x
2
? y ? – 2x + 3. Then area of region bounded by points in set
P is
ekuk P fcUnqvksa (x, y) dk leqPp; bl izdkj gS fd x
2
? y ? – 2x + 3. rks P ds lHkh fcUnqvksa }kjk lEc) {ks=kQy
gksxk%
(1)
3
16
(2)
3
32
(3)
3
29
(4)
3
20
Ans. (2)
Sol. Point of intersection of y = x
2
& y = – 2x + 3 is
y = x
2
rFkk y = – 2x + 3 dk izfrPNsnu fcUnq
obtained by izkIr gksrk gS] x
2
+ 2x – 3 = 0
? x = – 3, 1
So, Area vr% {ks=kQy =
?
?
? ?
1
3
2
dx ) x x 2 3 ( = 3(4) – 2
?
?
?
?
?
?
?
?
?
2
3 1
2 2
–
?
?
?
?
?
?
?
?
?
3
3 1
3 3
= 12 + 8 –
3
28
=
3
32
–3 1
7. Let f(x) =
? ?
1 x
x x
2
?
: (1, 3) ? R then range of f(x) is (where [ . ] denotes greatest integer function)
ekuk f(x) =
? ?
1 x
x x
2
?
: (1, 3) ? R rks f(x) dk ifjlj gS (tgk¡ [ . ] egÙke iw.kkZad Qyu gS)
(1) ?
?
?
?
?
?
2
1
, 0 ?
?
?
?
?
?
?
5
7
,
5
3
(2) ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
5
3
(3) ?
?
?
?
?
?
1 ,
5
2
?
?
?
?
?
?
?
5
4
, 1 (4) ?
?
?
?
?
?
3
1
, 0 ?
?
?
?
?
?
?
5
4
,
5
2
Ans. (2)
Page 4
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 21 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 21 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Let a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? and c
?
is nonzero vector and c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 find c . b
?
?
ekuk a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? rFkk c
?
,d v'kwU; lfn'k gS rFkk c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 rks c . b
?
?
dk eku
gksxk%
(1)
2
1
(2)
3
1
(3)
2
1
? (4)
3
1
?
Ans. (3)
Sol. ) c b ( a
?
?
?
? ? = ) a b ( a
?
?
?
? ?
– ) b . a (
?
?
c
?
= ( a
?
. a
?
) b
?
– ( b . a
?
?
) a
?
– c 4
?
= 6( k
ˆ
j
ˆ
i
ˆ
? ? ) – 4( k
ˆ
j
ˆ
2 i
ˆ
? ? )
– c 4
?
= k
ˆ
2 j
ˆ
2 i
ˆ
2 ? ?
c
?
=
2
1
? ( k
ˆ
j
ˆ
i
ˆ
? ? )
c . b
?
?
=
2
1
?
2. Let coefficient of x
4
and x
2
in the expansion of
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? is ? and ? then ? – ? is
equal to
ekuk
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? ds izlkj esa x
4
rFkk x
2
ds xq.kkad Øe'k% ? rFkk ? gS rks ? – ? dk eku gS %
(1) 48 (2) 60 (3) –132 (4) –60
Ans. (3)
2[
6
C0 x
6
+
6
C2 x
4
(x
2
– 1) +
6
C4 x
2
(x
2
– 1)
2
+
6
C6 (x
2
– 1)
3
]
= 2[x
6
+ 15(x
6
– x
4
) + 15x
2
(x
4
– 2x
2
+ 1) + (–1 + 3x
2
– 3x
4
+ x
6
)]
= 2(32x
6
– 48x
4
+ 18x
2
– 1)
? ? = – 96 and vkSj ? = 36 ? ? ? – ? = –132
3. Differential equation of x
2
= 4b(y + b), where b is a parameter, is
x
2
= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk %
(1)
2
2
x
dx
dy
y 2
dx
dy
x ? ? ?
?
?
?
?
?
(2) x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
(3)
2
2
x
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
(4)
2
2
x 2
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
Ans. (2)
Sol. 2x = 4by
'
? ?
'
y 2
x
b ?
So. differential equation is
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ?
vr% vody lehdj.k
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ? ? ? x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
4. Image of (1, 2, 3) w.r.t a plane is ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
then which of the following points lie on the plane
(1, 2, 3) dk fdlh lery esa izfrfcEc ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
rks fuEu esa ls dkSulk fcUnq lery ij fLFkr ugha gS %
(1) (–1, 1,–1) (2) (–1, –1, –1) (3) (–1, –1,1) (4) (1, 1, –1)
Ans. (4)
Sol. d.r of normal to the plane
lery ds yEcor~ f}dvuqikr
3
10
,
3
10
,
3
10
1, 1, 1
midpoint of P and Q is ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
P rFkk Q dk e/; fcUnq ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
equation of plane x + y + z = 1
lery dk lehdj.k x + y + z = 1
5.
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
is equal to
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
dk eku gS %
(1) 1 (2) 10 (3) 5 (4) 0
Ans. (4)
Sol. Using L’Hospital
L gkWfLiVy ls
0 x
lim
? 1
) x 0 1 sin( x
= 0
6. Let P be the set of points (x, y) such that x
2
? y ? – 2x + 3. Then area of region bounded by points in set
P is
ekuk P fcUnqvksa (x, y) dk leqPp; bl izdkj gS fd x
2
? y ? – 2x + 3. rks P ds lHkh fcUnqvksa }kjk lEc) {ks=kQy
gksxk%
(1)
3
16
(2)
3
32
(3)
3
29
(4)
3
20
Ans. (2)
Sol. Point of intersection of y = x
2
& y = – 2x + 3 is
y = x
2
rFkk y = – 2x + 3 dk izfrPNsnu fcUnq
obtained by izkIr gksrk gS] x
2
+ 2x – 3 = 0
? x = – 3, 1
So, Area vr% {ks=kQy =
?
?
? ?
1
3
2
dx ) x x 2 3 ( = 3(4) – 2
?
?
?
?
?
?
?
?
?
2
3 1
2 2
–
?
?
?
?
?
?
?
?
?
3
3 1
3 3
= 12 + 8 –
3
28
=
3
32
–3 1
7. Let f(x) =
? ?
1 x
x x
2
?
: (1, 3) ? R then range of f(x) is (where [ . ] denotes greatest integer function)
ekuk f(x) =
? ?
1 x
x x
2
?
: (1, 3) ? R rks f(x) dk ifjlj gS (tgk¡ [ . ] egÙke iw.kkZad Qyu gS)
(1) ?
?
?
?
?
?
2
1
, 0 ?
?
?
?
?
?
?
5
7
,
5
3
(2) ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
5
3
(3) ?
?
?
?
?
?
1 ,
5
2
?
?
?
?
?
?
?
5
4
, 1 (4) ?
?
?
?
?
?
3
1
, 0 ?
?
?
?
?
?
?
5
4
,
5
2
Ans. (2)
Sol f(x)
?
?
?
?
?
?
?
?
?
?
?
) 3 , 2 [ x ;
1 x
x 2
) 2 , 1 ( x ;
1 x
x
2
2
? f(x) is a decreasing function
? f(x) âkleku Qyu gSA
? y ? ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
10
6
? y ? ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
5
3
8. Let A =
?
?
?
?
?
?
4 9
2 2
and I =
?
?
?
?
?
?
1 0
0 1
then value of 10 A
–1
is –
ekuk A =
?
?
?
?
?
?
4 9
2 2
rFkk I =
?
?
?
?
?
?
1 0
0 1
rks 10 A
–1
dk eku gksxk %
(1) 4I – A (2) 6 I – A (3) A – 4 I (4) A – 6 I
Ans. (4)
Sol. Characteristics equation of matrix ‘A’ is
vkO;wg ‘A’ dh vfHkyk{kf.kd lehdj.k gS &
0
x – 4 9
2 x – 2
? ? ? x
2
– 6x – 10 = 0
? A
2
– 6A – 10 I = 0
? ? 10A
–1
= A – 6I
9. Solution set of 3
x
(3
x
–1) + 2 = |3
x
–1| + |3
x
– 2| contains
(1) singleton set (2) two elements
(3) at least four elements (4) infinite elements
3
x
(3
x
–1) + 2 = |3
x
–1| + |3
x
– 2| dk gy leqPPk; j[krk gSA
(1) ,d gy (2) nks gy (3) de ls de pkj gy (4) vUur gy
Ans. (1)
Sol. Let 3
x
= t
t(t –1) + 2 = |t –1| + |t –2|
t
2
– t + 2 = |t –1| + |t –2|
7/4
2
1
a 1 2
are positive solution
t = a
3
x
= a
x = a log
3
so singleton set
Page 5
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 21 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 21 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Let a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? and c
?
is nonzero vector and c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 find c . b
?
?
ekuk a
?
= k
ˆ
j
ˆ
2 i
ˆ
? ? , b
?
= k
ˆ
j
ˆ
i
ˆ
? ? rFkk c
?
,d v'kwU; lfn'k gS rFkk c b
?
?
? = a b
?
?
? , c . a
? ?
= 0 rks c . b
?
?
dk eku
gksxk%
(1)
2
1
(2)
3
1
(3)
2
1
? (4)
3
1
?
Ans. (3)
Sol. ) c b ( a
?
?
?
? ? = ) a b ( a
?
?
?
? ?
– ) b . a (
?
?
c
?
= ( a
?
. a
?
) b
?
– ( b . a
?
?
) a
?
– c 4
?
= 6( k
ˆ
j
ˆ
i
ˆ
? ? ) – 4( k
ˆ
j
ˆ
2 i
ˆ
? ? )
– c 4
?
= k
ˆ
2 j
ˆ
2 i
ˆ
2 ? ?
c
?
=
2
1
? ( k
ˆ
j
ˆ
i
ˆ
? ? )
c . b
?
?
=
2
1
?
2. Let coefficient of x
4
and x
2
in the expansion of
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? is ? and ? then ? – ? is
equal to
ekuk
6
2
6
2
1 x x 1 x x
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
? ? ds izlkj esa x
4
rFkk x
2
ds xq.kkad Øe'k% ? rFkk ? gS rks ? – ? dk eku gS %
(1) 48 (2) 60 (3) –132 (4) –60
Ans. (3)
2[
6
C0 x
6
+
6
C2 x
4
(x
2
– 1) +
6
C4 x
2
(x
2
– 1)
2
+
6
C6 (x
2
– 1)
3
]
= 2[x
6
+ 15(x
6
– x
4
) + 15x
2
(x
4
– 2x
2
+ 1) + (–1 + 3x
2
– 3x
4
+ x
6
)]
= 2(32x
6
– 48x
4
+ 18x
2
– 1)
? ? = – 96 and vkSj ? = 36 ? ? ? – ? = –132
3. Differential equation of x
2
= 4b(y + b), where b is a parameter, is
x
2
= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk %
(1)
2
2
x
dx
dy
y 2
dx
dy
x ? ? ?
?
?
?
?
?
(2) x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
(3)
2
2
x
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
(4)
2
2
x 2
dx
dy
y
dx
dy
x ? ? ?
?
?
?
?
?
Ans. (2)
Sol. 2x = 4by
'
? ?
'
y 2
x
b ?
So. differential equation is
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ?
vr% vody lehdj.k
2
' '
2
y
x
y .
y
x 2
x
?
?
?
?
?
?
?
?
? ? ? ? x
dx
dy
y 2
dx
dy
x
2
? ? ?
?
?
?
?
?
4. Image of (1, 2, 3) w.r.t a plane is ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
then which of the following points lie on the plane
(1, 2, 3) dk fdlh lery esa izfrfcEc ?
?
?
?
?
?
3
1 –
,
3
4 –
,
3
7 –
rks fuEu esa ls dkSulk fcUnq lery ij fLFkr ugha gS %
(1) (–1, 1,–1) (2) (–1, –1, –1) (3) (–1, –1,1) (4) (1, 1, –1)
Ans. (4)
Sol. d.r of normal to the plane
lery ds yEcor~ f}dvuqikr
3
10
,
3
10
,
3
10
1, 1, 1
midpoint of P and Q is ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
P rFkk Q dk e/; fcUnq ?
?
?
?
?
?
3
4
,
3
1
,
3
2 –
equation of plane x + y + z = 1
lery dk lehdj.k x + y + z = 1
5.
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
is equal to
0 x
lim
? x
dt ) t 10 sin( t
x
0
?
dk eku gS %
(1) 1 (2) 10 (3) 5 (4) 0
Ans. (4)
Sol. Using L’Hospital
L gkWfLiVy ls
0 x
lim
? 1
) x 0 1 sin( x
= 0
6. Let P be the set of points (x, y) such that x
2
? y ? – 2x + 3. Then area of region bounded by points in set
P is
ekuk P fcUnqvksa (x, y) dk leqPp; bl izdkj gS fd x
2
? y ? – 2x + 3. rks P ds lHkh fcUnqvksa }kjk lEc) {ks=kQy
gksxk%
(1)
3
16
(2)
3
32
(3)
3
29
(4)
3
20
Ans. (2)
Sol. Point of intersection of y = x
2
& y = – 2x + 3 is
y = x
2
rFkk y = – 2x + 3 dk izfrPNsnu fcUnq
obtained by izkIr gksrk gS] x
2
+ 2x – 3 = 0
? x = – 3, 1
So, Area vr% {ks=kQy =
?
?
? ?
1
3
2
dx ) x x 2 3 ( = 3(4) – 2
?
?
?
?
?
?
?
?
?
2
3 1
2 2
–
?
?
?
?
?
?
?
?
?
3
3 1
3 3
= 12 + 8 –
3
28
=
3
32
–3 1
7. Let f(x) =
? ?
1 x
x x
2
?
: (1, 3) ? R then range of f(x) is (where [ . ] denotes greatest integer function)
ekuk f(x) =
? ?
1 x
x x
2
?
: (1, 3) ? R rks f(x) dk ifjlj gS (tgk¡ [ . ] egÙke iw.kkZad Qyu gS)
(1) ?
?
?
?
?
?
2
1
, 0 ?
?
?
?
?
?
?
5
7
,
5
3
(2) ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
5
3
(3) ?
?
?
?
?
?
1 ,
5
2
?
?
?
?
?
?
?
5
4
, 1 (4) ?
?
?
?
?
?
3
1
, 0 ?
?
?
?
?
?
?
5
4
,
5
2
Ans. (2)
Sol f(x)
?
?
?
?
?
?
?
?
?
?
?
) 3 , 2 [ x ;
1 x
x 2
) 2 , 1 ( x ;
1 x
x
2
2
? f(x) is a decreasing function
? f(x) âkleku Qyu gSA
? y ? ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
10
6
? y ? ?
?
?
?
?
?
2
1
,
5
2
?
?
?
?
?
?
?
5
4
,
5
3
8. Let A =
?
?
?
?
?
?
4 9
2 2
and I =
?
?
?
?
?
?
1 0
0 1
then value of 10 A
–1
is –
ekuk A =
?
?
?
?
?
?
4 9
2 2
rFkk I =
?
?
?
?
?
?
1 0
0 1
rks 10 A
–1
dk eku gksxk %
(1) 4I – A (2) 6 I – A (3) A – 4 I (4) A – 6 I
Ans. (4)
Sol. Characteristics equation of matrix ‘A’ is
vkO;wg ‘A’ dh vfHkyk{kf.kd lehdj.k gS &
0
x – 4 9
2 x – 2
? ? ? x
2
– 6x – 10 = 0
? A
2
– 6A – 10 I = 0
? ? 10A
–1
= A – 6I
9. Solution set of 3
x
(3
x
–1) + 2 = |3
x
–1| + |3
x
– 2| contains
(1) singleton set (2) two elements
(3) at least four elements (4) infinite elements
3
x
(3
x
–1) + 2 = |3
x
–1| + |3
x
– 2| dk gy leqPPk; j[krk gSA
(1) ,d gy (2) nks gy (3) de ls de pkj gy (4) vUur gy
Ans. (1)
Sol. Let 3
x
= t
t(t –1) + 2 = |t –1| + |t –2|
t
2
– t + 2 = |t –1| + |t –2|
7/4
2
1
a 1 2
are positive solution
t = a
3
x
= a
x = a log
3
so singleton set
Hindi. ekuk fd 3
x
= t
t(t –1) + 2 = |t –1| + |t –2|
t
2
– t + 2 = |t –1| + |t –2|
7/4
2
1
a 1 2
/kukRed gy gSA
t = a
3
x
= a
x = a log
3
,dy leqPp;
10. Mean and variance of 20 observation are 10 and 4. It was found, that in place of 11, 9 was taken by
mistake find correct variance.
20 vakdM+ksa dk ek/; rFkk pfjrk Øe'k% 10 rFkk 4 gSA ;fn 11 ds LFkku ij xyrh ls 9 fy;k x;k gS rks lgh pfjrk
gS&
(1) 3.99 (2) 3.98 (3) 4.01 (4) 4.02
Ans. (1)
Sol.
20
x
i
?
= 10 …….(i)
100 –
20
x
2
i
?
= 4 …….(ii)
2
i
x ? = 104 × 20 = 2080
Actual mean lgh ek/; =
20
11 9 – 200 ?
=
20
202
Variance pfjrk =
20
121 81 – 2080 ?
–
2
20
202
?
?
?
?
?
?
=
20
2120
– (10.1)
2
= 106 – 102.01 = 3.99
11. ?x + 2y + 2z = 5
2 ?x + 3y + 5z = 8
? 4x + ?y + 6z = 10
for the system of equation check the correct option.
(1) Infinite solutions when ? = 8 (2) Infinite solutions when ? = 2
(3) no solutions when ? = 8 (4) no solutions when ? = 2
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JEE Main 2020 Mathematics January 8 Shift 2 Paper & Solutions JEE Questions
The "JEE Main 2020 Mathematics January 8 Shift 2 Paper & Solutions JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
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The content of JEE Main 2020 Mathematics January 8 Shift 2 Paper & Solutions is prepared as per the latest JEE syllabus.
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