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Page 1
PAGE # 1
PART : MATHEMATICS
1. Let A = [aij], B=[bij] are two 3 × 3 matrices such that
ij
2 j i
ij
a b
? ?
? ? & |B| = 81. Find |A| if ? = 3.
ekuk A = [aij], B=[bij], 3 × 3 Øe dh nks vkO;wg bl izdkj gS fd
ij
2 j i
ij
a b
? ?
? ? rFkk |B| = 81. ;fn ? = 3 gS rc
|A|dk eku gS&
(1)
9
1
(2) 3 (3)
81
1
(4)
27
1
Ans. (1)
Sol. |B|=
33 32 31
23 22 21
13 12 11
b b b
b b b
b b b
=
33
4
32
3
31
2
23
3
22
2
21
1
13
2
12
1
11
0
a 3 a 3 a 3
a 3 a 3 a 3
a 3 a 3 a 3
? 81 = 3
3
.3. 3
2
|A| ??3
4
= 3
6
|A| ??|A|=
9
1
2. From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of mid point of PQ.
js[kk x = 2y ij fLFkr fdlh fcUnq P ls js[kk y = x ij yEc [khapk tkrk gSA ekuk Q yEcikn gS rc PQ ds e/;fcUnq
dk fcUnqiFk Kkr djks&
(1) 2x = 3y (2) 5x = 7y (3) 3x = 2y (4) 7x = 5y
Ans. (2)
Sol.
y = x
Q
( ?, ?)
P(2 ?, ?)
x = 2y
(h,k)
slope of PQ dh izo.krk = 1
2 h
k
? ?
? ?
? ?
? ? ?k – ??= – h + 2?? ?
? ? ? ?=
3
k h ?
......(1)
Also rFkk 2h = 2 ? + ? ?
2k = ? + ?
? 2h = ? + 2k
? ? ? = 2h – 2k .....(2)
from (1) vkSj & (2) ls
) k h ( 2
3
k h
? ?
?
so locus is vr% fcUnqiFk 6x –6y = x + y ? ? ? ?5x = 7y ?
Page 2
PAGE # 1
PART : MATHEMATICS
1. Let A = [aij], B=[bij] are two 3 × 3 matrices such that
ij
2 j i
ij
a b
? ?
? ? & |B| = 81. Find |A| if ? = 3.
ekuk A = [aij], B=[bij], 3 × 3 Øe dh nks vkO;wg bl izdkj gS fd
ij
2 j i
ij
a b
? ?
? ? rFkk |B| = 81. ;fn ? = 3 gS rc
|A|dk eku gS&
(1)
9
1
(2) 3 (3)
81
1
(4)
27
1
Ans. (1)
Sol. |B|=
33 32 31
23 22 21
13 12 11
b b b
b b b
b b b
=
33
4
32
3
31
2
23
3
22
2
21
1
13
2
12
1
11
0
a 3 a 3 a 3
a 3 a 3 a 3
a 3 a 3 a 3
? 81 = 3
3
.3. 3
2
|A| ??3
4
= 3
6
|A| ??|A|=
9
1
2. From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of mid point of PQ.
js[kk x = 2y ij fLFkr fdlh fcUnq P ls js[kk y = x ij yEc [khapk tkrk gSA ekuk Q yEcikn gS rc PQ ds e/;fcUnq
dk fcUnqiFk Kkr djks&
(1) 2x = 3y (2) 5x = 7y (3) 3x = 2y (4) 7x = 5y
Ans. (2)
Sol.
y = x
Q
( ?, ?)
P(2 ?, ?)
x = 2y
(h,k)
slope of PQ dh izo.krk = 1
2 h
k
? ?
? ?
? ?
? ? ?k – ??= – h + 2?? ?
? ? ? ?=
3
k h ?
......(1)
Also rFkk 2h = 2 ? + ? ?
2k = ? + ?
? 2h = ? + 2k
? ? ? = 2h – 2k .....(2)
from (1) vkSj & (2) ls
) k h ( 2
3
k h
? ?
?
so locus is vr% fcUnqiFk 6x –6y = x + y ? ? ? ?5x = 7y ?
PAGE # 2
3. Pair of tangents are drawn from origin to the circle x
2
+ y
2
– 8x – 4y + 16 = 0 then square of length of
chord of contact is
ewyfcUnq ls o`Ùk x
2
+ y
2
– 8x – 4y + 16 = 0 ij Li'kZ js[kk,sa [khph tkrh gS rc Li'kZthok dh yEckbZ dk oxZ gS&
(1)
5
64
(2)
24
5
(3)
8
5
(4)
8
13
Ans. (1)
Sol. L = 4 16 S
1
? ?
2 16 4 16 R ? ? ? ?
Length of Chord of contact Li'kZthok dh yEckbZ =
20
16
4 16
2 4 2
R L
LR 2
2 2
?
?
? ?
?
?
square of length of chord of contact Li'kZthok dh yEckbZ dk oxZ =
5
64
4. Contrapositive of if A ? B and B ? C then C ? D
(1) C ? D or A ? B or B ? C (2) C ? D and A ? B or B ? C
(3) C ? D or A ? B and B ? C (4) C ? D or A ? B or B ? C
;fn A ? B vkSj B ? C rc C ? D dk izfrifjofrZr gS&
(1) C ? D ;k A ? B ;k B ? C (2) C ? D vkSj A ? B ;k B ? C
(3) C ? D;k A ? B vkSj B ? C (4) C ? D ;k A ? B ;k B ? C
Ans. (4)
Sol. Let P = A ? B, Q = B ? C, R = C ? A
Contrapositive of (P ? Q) ? R is ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
Sol. ekuk P = A ? B, Q = B ? C, R = C ? A
(P ? Q) ? R dk izfrifjofrZr ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
5. Let y(x) is solution of differential equation (y
2
– x)
dx
dy
= 1 and y(0) = 1, then find the value of x where
curve cuts the x-axis
ekuk y(x), vody lehdj.k (y
2
– x)
dx
dy
= 1 dk gy gS rFkk y(0) = 1, rc x dk eku gksxk] tgkaa oØ x v{k dks
izfrPNsn djrk gS&
(1) 2 – e (2) 2 +e (3) 2 (4) e
Ans. (1)
Sol. Sol.
2
y x
dy
dx
? ?
I.F . =
?
dy . l
e = e
y
x.e
y
=
?
dy . e . y
y 2
= y
2
.e
y
–
?
dy . e . y 2
y
Page 3
PAGE # 1
PART : MATHEMATICS
1. Let A = [aij], B=[bij] are two 3 × 3 matrices such that
ij
2 j i
ij
a b
? ?
? ? & |B| = 81. Find |A| if ? = 3.
ekuk A = [aij], B=[bij], 3 × 3 Øe dh nks vkO;wg bl izdkj gS fd
ij
2 j i
ij
a b
? ?
? ? rFkk |B| = 81. ;fn ? = 3 gS rc
|A|dk eku gS&
(1)
9
1
(2) 3 (3)
81
1
(4)
27
1
Ans. (1)
Sol. |B|=
33 32 31
23 22 21
13 12 11
b b b
b b b
b b b
=
33
4
32
3
31
2
23
3
22
2
21
1
13
2
12
1
11
0
a 3 a 3 a 3
a 3 a 3 a 3
a 3 a 3 a 3
? 81 = 3
3
.3. 3
2
|A| ??3
4
= 3
6
|A| ??|A|=
9
1
2. From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of mid point of PQ.
js[kk x = 2y ij fLFkr fdlh fcUnq P ls js[kk y = x ij yEc [khapk tkrk gSA ekuk Q yEcikn gS rc PQ ds e/;fcUnq
dk fcUnqiFk Kkr djks&
(1) 2x = 3y (2) 5x = 7y (3) 3x = 2y (4) 7x = 5y
Ans. (2)
Sol.
y = x
Q
( ?, ?)
P(2 ?, ?)
x = 2y
(h,k)
slope of PQ dh izo.krk = 1
2 h
k
? ?
? ?
? ?
? ? ?k – ??= – h + 2?? ?
? ? ? ?=
3
k h ?
......(1)
Also rFkk 2h = 2 ? + ? ?
2k = ? + ?
? 2h = ? + 2k
? ? ? = 2h – 2k .....(2)
from (1) vkSj & (2) ls
) k h ( 2
3
k h
? ?
?
so locus is vr% fcUnqiFk 6x –6y = x + y ? ? ? ?5x = 7y ?
PAGE # 2
3. Pair of tangents are drawn from origin to the circle x
2
+ y
2
– 8x – 4y + 16 = 0 then square of length of
chord of contact is
ewyfcUnq ls o`Ùk x
2
+ y
2
– 8x – 4y + 16 = 0 ij Li'kZ js[kk,sa [khph tkrh gS rc Li'kZthok dh yEckbZ dk oxZ gS&
(1)
5
64
(2)
24
5
(3)
8
5
(4)
8
13
Ans. (1)
Sol. L = 4 16 S
1
? ?
2 16 4 16 R ? ? ? ?
Length of Chord of contact Li'kZthok dh yEckbZ =
20
16
4 16
2 4 2
R L
LR 2
2 2
?
?
? ?
?
?
square of length of chord of contact Li'kZthok dh yEckbZ dk oxZ =
5
64
4. Contrapositive of if A ? B and B ? C then C ? D
(1) C ? D or A ? B or B ? C (2) C ? D and A ? B or B ? C
(3) C ? D or A ? B and B ? C (4) C ? D or A ? B or B ? C
;fn A ? B vkSj B ? C rc C ? D dk izfrifjofrZr gS&
(1) C ? D ;k A ? B ;k B ? C (2) C ? D vkSj A ? B ;k B ? C
(3) C ? D;k A ? B vkSj B ? C (4) C ? D ;k A ? B ;k B ? C
Ans. (4)
Sol. Let P = A ? B, Q = B ? C, R = C ? A
Contrapositive of (P ? Q) ? R is ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
Sol. ekuk P = A ? B, Q = B ? C, R = C ? A
(P ? Q) ? R dk izfrifjofrZr ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
5. Let y(x) is solution of differential equation (y
2
– x)
dx
dy
= 1 and y(0) = 1, then find the value of x where
curve cuts the x-axis
ekuk y(x), vody lehdj.k (y
2
– x)
dx
dy
= 1 dk gy gS rFkk y(0) = 1, rc x dk eku gksxk] tgkaa oØ x v{k dks
izfrPNsn djrk gS&
(1) 2 – e (2) 2 +e (3) 2 (4) e
Ans. (1)
Sol. Sol.
2
y x
dy
dx
? ?
I.F . =
?
dy . l
e = e
y
x.e
y
=
?
dy . e . y
y 2
= y
2
.e
y
–
?
dy . e . y 2
y
PAGE # 3
? y
2
e
y
– 2(y.e
y
–e
y
) + c
x.e
y
= y
2
e
y
–2ye
y
+ 2e
y
+ C
x = y
2
–2y + 2 + c. e
–y
x = 0, y = 1
0 = 1 – 2 + 2 +
e
c
c = –e
y = 0, x = 0 – 0 + 2 + (–e) (e
–0
)
x = 2 –e
6. Let ?1 and ?2 (where ?1 < ?2)are two solutions of 2cot
2
? –
? sin
5
+ 4 = 0, ?? ? [0, 2 ?) then
?
?
?
? ?
2
1
d 3 cos
2
is
equal to
ekuk ?1 vkSj ?2 (tgka ?1 < ?2) lehdj.k 2cot
2
? –
? sin
5
+ 4 = 0, ?? ? [0, 2 ?) ds nks gy gS rc
?
?
?
? ?
2
1
d 3 cos
2
dk
eku gS&
(1)
3
?
(2)
3
2 ?
(3)
9
?
(4)
6
1
3
?
?
Ans. (1)
Sol. 0 4
sin
5
cot 2
2
? ?
?
? ?
0 4
sin
5
sin
cos 2
2
2
? ?
?
?
?
?
0 sin 4 sin 5 cos 2
2 2
? ? ? ? ? ? , 0 sin ? ?
0 2 sin 5 sin 2
2
? ? ? ? ?
0 ) 2 )(sin 1 sin 2 ( ? ? ? ? ?
2
1
sin ? ?
6
5
,
6
? ?
? ?
? ?
?
?
?
?
?
? ?
? ? ? ?
6 / 5
6 /
6 / 5
6 /
2
d
2
6 cos 1
d 3 cos
6 / 5
6 /
6
6 sin
2
1
?
?
?
?
?
?
?
? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
? ) 0 0 (
6
1
6 6
5
2
1
3 6
4
.
2
1 ?
?
?
?
7. Let 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 terms = S. If S = (102)m then m =
ekuk 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 in rd = S. ;fn S = (102)m rc m =
(1) 20 (2) 25 (3) 10 (4) 5
Ans. (1)
Page 4
PAGE # 1
PART : MATHEMATICS
1. Let A = [aij], B=[bij] are two 3 × 3 matrices such that
ij
2 j i
ij
a b
? ?
? ? & |B| = 81. Find |A| if ? = 3.
ekuk A = [aij], B=[bij], 3 × 3 Øe dh nks vkO;wg bl izdkj gS fd
ij
2 j i
ij
a b
? ?
? ? rFkk |B| = 81. ;fn ? = 3 gS rc
|A|dk eku gS&
(1)
9
1
(2) 3 (3)
81
1
(4)
27
1
Ans. (1)
Sol. |B|=
33 32 31
23 22 21
13 12 11
b b b
b b b
b b b
=
33
4
32
3
31
2
23
3
22
2
21
1
13
2
12
1
11
0
a 3 a 3 a 3
a 3 a 3 a 3
a 3 a 3 a 3
? 81 = 3
3
.3. 3
2
|A| ??3
4
= 3
6
|A| ??|A|=
9
1
2. From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of mid point of PQ.
js[kk x = 2y ij fLFkr fdlh fcUnq P ls js[kk y = x ij yEc [khapk tkrk gSA ekuk Q yEcikn gS rc PQ ds e/;fcUnq
dk fcUnqiFk Kkr djks&
(1) 2x = 3y (2) 5x = 7y (3) 3x = 2y (4) 7x = 5y
Ans. (2)
Sol.
y = x
Q
( ?, ?)
P(2 ?, ?)
x = 2y
(h,k)
slope of PQ dh izo.krk = 1
2 h
k
? ?
? ?
? ?
? ? ?k – ??= – h + 2?? ?
? ? ? ?=
3
k h ?
......(1)
Also rFkk 2h = 2 ? + ? ?
2k = ? + ?
? 2h = ? + 2k
? ? ? = 2h – 2k .....(2)
from (1) vkSj & (2) ls
) k h ( 2
3
k h
? ?
?
so locus is vr% fcUnqiFk 6x –6y = x + y ? ? ? ?5x = 7y ?
PAGE # 2
3. Pair of tangents are drawn from origin to the circle x
2
+ y
2
– 8x – 4y + 16 = 0 then square of length of
chord of contact is
ewyfcUnq ls o`Ùk x
2
+ y
2
– 8x – 4y + 16 = 0 ij Li'kZ js[kk,sa [khph tkrh gS rc Li'kZthok dh yEckbZ dk oxZ gS&
(1)
5
64
(2)
24
5
(3)
8
5
(4)
8
13
Ans. (1)
Sol. L = 4 16 S
1
? ?
2 16 4 16 R ? ? ? ?
Length of Chord of contact Li'kZthok dh yEckbZ =
20
16
4 16
2 4 2
R L
LR 2
2 2
?
?
? ?
?
?
square of length of chord of contact Li'kZthok dh yEckbZ dk oxZ =
5
64
4. Contrapositive of if A ? B and B ? C then C ? D
(1) C ? D or A ? B or B ? C (2) C ? D and A ? B or B ? C
(3) C ? D or A ? B and B ? C (4) C ? D or A ? B or B ? C
;fn A ? B vkSj B ? C rc C ? D dk izfrifjofrZr gS&
(1) C ? D ;k A ? B ;k B ? C (2) C ? D vkSj A ? B ;k B ? C
(3) C ? D;k A ? B vkSj B ? C (4) C ? D ;k A ? B ;k B ? C
Ans. (4)
Sol. Let P = A ? B, Q = B ? C, R = C ? A
Contrapositive of (P ? Q) ? R is ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
Sol. ekuk P = A ? B, Q = B ? C, R = C ? A
(P ? Q) ? R dk izfrifjofrZr ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
5. Let y(x) is solution of differential equation (y
2
– x)
dx
dy
= 1 and y(0) = 1, then find the value of x where
curve cuts the x-axis
ekuk y(x), vody lehdj.k (y
2
– x)
dx
dy
= 1 dk gy gS rFkk y(0) = 1, rc x dk eku gksxk] tgkaa oØ x v{k dks
izfrPNsn djrk gS&
(1) 2 – e (2) 2 +e (3) 2 (4) e
Ans. (1)
Sol. Sol.
2
y x
dy
dx
? ?
I.F . =
?
dy . l
e = e
y
x.e
y
=
?
dy . e . y
y 2
= y
2
.e
y
–
?
dy . e . y 2
y
PAGE # 3
? y
2
e
y
– 2(y.e
y
–e
y
) + c
x.e
y
= y
2
e
y
–2ye
y
+ 2e
y
+ C
x = y
2
–2y + 2 + c. e
–y
x = 0, y = 1
0 = 1 – 2 + 2 +
e
c
c = –e
y = 0, x = 0 – 0 + 2 + (–e) (e
–0
)
x = 2 –e
6. Let ?1 and ?2 (where ?1 < ?2)are two solutions of 2cot
2
? –
? sin
5
+ 4 = 0, ?? ? [0, 2 ?) then
?
?
?
? ?
2
1
d 3 cos
2
is
equal to
ekuk ?1 vkSj ?2 (tgka ?1 < ?2) lehdj.k 2cot
2
? –
? sin
5
+ 4 = 0, ?? ? [0, 2 ?) ds nks gy gS rc
?
?
?
? ?
2
1
d 3 cos
2
dk
eku gS&
(1)
3
?
(2)
3
2 ?
(3)
9
?
(4)
6
1
3
?
?
Ans. (1)
Sol. 0 4
sin
5
cot 2
2
? ?
?
? ?
0 4
sin
5
sin
cos 2
2
2
? ?
?
?
?
?
0 sin 4 sin 5 cos 2
2 2
? ? ? ? ? ? , 0 sin ? ?
0 2 sin 5 sin 2
2
? ? ? ? ?
0 ) 2 )(sin 1 sin 2 ( ? ? ? ? ?
2
1
sin ? ?
6
5
,
6
? ?
? ?
? ?
?
?
?
?
?
? ?
? ? ? ?
6 / 5
6 /
6 / 5
6 /
2
d
2
6 cos 1
d 3 cos
6 / 5
6 /
6
6 sin
2
1
?
?
?
?
?
?
?
? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
? ) 0 0 (
6
1
6 6
5
2
1
3 6
4
.
2
1 ?
?
?
?
7. Let 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 terms = S. If S = (102)m then m =
ekuk 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 in rd = S. ;fn S = (102)m rc m =
(1) 20 (2) 25 (3) 10 (4) 5
Ans. (1)
PAGE # 4
Sol. S =
?
.40 … … … 19 + 18 + 14 + 13 + 9 + 8 + 4 + 3
? ? ? ? ? ? ? ? ? ? ? ? ?
terms in rd
S = 7 + 17 + 27 + 37 + 47 +.........20 terms in rd
? ? 10 ) 19 ( 7 2
2
20
S
40
? ? ? = 10[14 + 190] = 10[2040] = (102) (20)
? m = 20
8. If ? ? ? ?
r
35 2
1 r
36
C 3 k C ? ? ?
?
6 , then number of ordered pairs (r, k) are –(where k ?I).
;fn ? ? ? ?
r
35 2
1 r
36
C 3 k C ? ? ?
?
6 , rks Øfer ;qXeksa (r, k) dh la[;k gS –(tgk¡ k ?I).
(1) 6 (2) 2 (3) 3 (4) 4
Ans. (4)
Sol. ? ?
r
2
r
C
35
3 k
C
35
1 r
36
? ? ?
?
k
2
–3 =
6
1 r ?
??k
2
= 3 +
6
1 r ?
r can be 5, 35
for r = 5, k = ±2
r = 35, k = ±3
Hence number of order pair = 4
Hindi. ? ?
r
2
r
C
35
3 k
C
35
1 r
36
? ? ?
?
k
2
–3 =
6
1 r ?
??k
2
= 3 +
6
1 r ?
r 5, 35 gks ldrk gSA
r = 5, ds fy, k = ±2
r = 35, ds fy,k = ±3
vr% Øfer ;qXeksa dh la[;k = 4
9. Let 4 ? dx e
2
1 –
| x | –
?
?
= 5 then ? =
ekuk fd 4 ? dx e
2
1 –
| x | –
?
?
= 5 rks ? =
(1) ?n2 (2) ?n 2 (3) ?n
4
3
(4) ?n
3
4
Ans. (1)
Sol. 5 dx e dx e 4
2
0
x
0
1
x
?
?
?
?
?
?
?
? ?
? ?
? ?
?
?
? 5
e e
4
2
0
x
0
1
x
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
Page 5
PAGE # 1
PART : MATHEMATICS
1. Let A = [aij], B=[bij] are two 3 × 3 matrices such that
ij
2 j i
ij
a b
? ?
? ? & |B| = 81. Find |A| if ? = 3.
ekuk A = [aij], B=[bij], 3 × 3 Øe dh nks vkO;wg bl izdkj gS fd
ij
2 j i
ij
a b
? ?
? ? rFkk |B| = 81. ;fn ? = 3 gS rc
|A|dk eku gS&
(1)
9
1
(2) 3 (3)
81
1
(4)
27
1
Ans. (1)
Sol. |B|=
33 32 31
23 22 21
13 12 11
b b b
b b b
b b b
=
33
4
32
3
31
2
23
3
22
2
21
1
13
2
12
1
11
0
a 3 a 3 a 3
a 3 a 3 a 3
a 3 a 3 a 3
? 81 = 3
3
.3. 3
2
|A| ??3
4
= 3
6
|A| ??|A|=
9
1
2. From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of mid point of PQ.
js[kk x = 2y ij fLFkr fdlh fcUnq P ls js[kk y = x ij yEc [khapk tkrk gSA ekuk Q yEcikn gS rc PQ ds e/;fcUnq
dk fcUnqiFk Kkr djks&
(1) 2x = 3y (2) 5x = 7y (3) 3x = 2y (4) 7x = 5y
Ans. (2)
Sol.
y = x
Q
( ?, ?)
P(2 ?, ?)
x = 2y
(h,k)
slope of PQ dh izo.krk = 1
2 h
k
? ?
? ?
? ?
? ? ?k – ??= – h + 2?? ?
? ? ? ?=
3
k h ?
......(1)
Also rFkk 2h = 2 ? + ? ?
2k = ? + ?
? 2h = ? + 2k
? ? ? = 2h – 2k .....(2)
from (1) vkSj & (2) ls
) k h ( 2
3
k h
? ?
?
so locus is vr% fcUnqiFk 6x –6y = x + y ? ? ? ?5x = 7y ?
PAGE # 2
3. Pair of tangents are drawn from origin to the circle x
2
+ y
2
– 8x – 4y + 16 = 0 then square of length of
chord of contact is
ewyfcUnq ls o`Ùk x
2
+ y
2
– 8x – 4y + 16 = 0 ij Li'kZ js[kk,sa [khph tkrh gS rc Li'kZthok dh yEckbZ dk oxZ gS&
(1)
5
64
(2)
24
5
(3)
8
5
(4)
8
13
Ans. (1)
Sol. L = 4 16 S
1
? ?
2 16 4 16 R ? ? ? ?
Length of Chord of contact Li'kZthok dh yEckbZ =
20
16
4 16
2 4 2
R L
LR 2
2 2
?
?
? ?
?
?
square of length of chord of contact Li'kZthok dh yEckbZ dk oxZ =
5
64
4. Contrapositive of if A ? B and B ? C then C ? D
(1) C ? D or A ? B or B ? C (2) C ? D and A ? B or B ? C
(3) C ? D or A ? B and B ? C (4) C ? D or A ? B or B ? C
;fn A ? B vkSj B ? C rc C ? D dk izfrifjofrZr gS&
(1) C ? D ;k A ? B ;k B ? C (2) C ? D vkSj A ? B ;k B ? C
(3) C ? D;k A ? B vkSj B ? C (4) C ? D ;k A ? B ;k B ? C
Ans. (4)
Sol. Let P = A ? B, Q = B ? C, R = C ? A
Contrapositive of (P ? Q) ? R is ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
Sol. ekuk P = A ? B, Q = B ? C, R = C ? A
(P ? Q) ? R dk izfrifjofrZr ~ R ? ~ (P ? Q)
R ? (~ P ? ~ Q)
5. Let y(x) is solution of differential equation (y
2
– x)
dx
dy
= 1 and y(0) = 1, then find the value of x where
curve cuts the x-axis
ekuk y(x), vody lehdj.k (y
2
– x)
dx
dy
= 1 dk gy gS rFkk y(0) = 1, rc x dk eku gksxk] tgkaa oØ x v{k dks
izfrPNsn djrk gS&
(1) 2 – e (2) 2 +e (3) 2 (4) e
Ans. (1)
Sol. Sol.
2
y x
dy
dx
? ?
I.F . =
?
dy . l
e = e
y
x.e
y
=
?
dy . e . y
y 2
= y
2
.e
y
–
?
dy . e . y 2
y
PAGE # 3
? y
2
e
y
– 2(y.e
y
–e
y
) + c
x.e
y
= y
2
e
y
–2ye
y
+ 2e
y
+ C
x = y
2
–2y + 2 + c. e
–y
x = 0, y = 1
0 = 1 – 2 + 2 +
e
c
c = –e
y = 0, x = 0 – 0 + 2 + (–e) (e
–0
)
x = 2 –e
6. Let ?1 and ?2 (where ?1 < ?2)are two solutions of 2cot
2
? –
? sin
5
+ 4 = 0, ?? ? [0, 2 ?) then
?
?
?
? ?
2
1
d 3 cos
2
is
equal to
ekuk ?1 vkSj ?2 (tgka ?1 < ?2) lehdj.k 2cot
2
? –
? sin
5
+ 4 = 0, ?? ? [0, 2 ?) ds nks gy gS rc
?
?
?
? ?
2
1
d 3 cos
2
dk
eku gS&
(1)
3
?
(2)
3
2 ?
(3)
9
?
(4)
6
1
3
?
?
Ans. (1)
Sol. 0 4
sin
5
cot 2
2
? ?
?
? ?
0 4
sin
5
sin
cos 2
2
2
? ?
?
?
?
?
0 sin 4 sin 5 cos 2
2 2
? ? ? ? ? ? , 0 sin ? ?
0 2 sin 5 sin 2
2
? ? ? ? ?
0 ) 2 )(sin 1 sin 2 ( ? ? ? ? ?
2
1
sin ? ?
6
5
,
6
? ?
? ?
? ?
?
?
?
?
?
? ?
? ? ? ?
6 / 5
6 /
6 / 5
6 /
2
d
2
6 cos 1
d 3 cos
6 / 5
6 /
6
6 sin
2
1
?
?
?
?
?
?
?
? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
? ) 0 0 (
6
1
6 6
5
2
1
3 6
4
.
2
1 ?
?
?
?
7. Let 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 terms = S. If S = (102)m then m =
ekuk 3 + 4 + 8 + 9 + 13 + 14 + 18 +……….40 in rd = S. ;fn S = (102)m rc m =
(1) 20 (2) 25 (3) 10 (4) 5
Ans. (1)
PAGE # 4
Sol. S =
?
.40 … … … 19 + 18 + 14 + 13 + 9 + 8 + 4 + 3
? ? ? ? ? ? ? ? ? ? ? ? ?
terms in rd
S = 7 + 17 + 27 + 37 + 47 +.........20 terms in rd
? ? 10 ) 19 ( 7 2
2
20
S
40
? ? ? = 10[14 + 190] = 10[2040] = (102) (20)
? m = 20
8. If ? ? ? ?
r
35 2
1 r
36
C 3 k C ? ? ?
?
6 , then number of ordered pairs (r, k) are –(where k ?I).
;fn ? ? ? ?
r
35 2
1 r
36
C 3 k C ? ? ?
?
6 , rks Øfer ;qXeksa (r, k) dh la[;k gS –(tgk¡ k ?I).
(1) 6 (2) 2 (3) 3 (4) 4
Ans. (4)
Sol. ? ?
r
2
r
C
35
3 k
C
35
1 r
36
? ? ?
?
k
2
–3 =
6
1 r ?
??k
2
= 3 +
6
1 r ?
r can be 5, 35
for r = 5, k = ±2
r = 35, k = ±3
Hence number of order pair = 4
Hindi. ? ?
r
2
r
C
35
3 k
C
35
1 r
36
? ? ?
?
k
2
–3 =
6
1 r ?
??k
2
= 3 +
6
1 r ?
r 5, 35 gks ldrk gSA
r = 5, ds fy, k = ±2
r = 35, ds fy,k = ±3
vr% Øfer ;qXeksa dh la[;k = 4
9. Let 4 ? dx e
2
1 –
| x | –
?
?
= 5 then ? =
ekuk fd 4 ? dx e
2
1 –
| x | –
?
?
= 5 rks ? =
(1) ?n2 (2) ?n 2 (3) ?n
4
3
(4) ?n
3
4
Ans. (1)
Sol. 5 dx e dx e 4
2
0
x
0
1
x
?
?
?
?
?
?
?
? ?
? ?
? ?
?
?
? 5
e e
4
2
0
x
0
1
x
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
PAGE # 5
? 5
1 e e 1
4
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
??4(2 – e
–??
– e
–2 ?
) = 5 Put e
– ?
= t
? 4t
2
+ 4t – 3 = 0 ??(2t + 3) (2t – 1) = 0
??e
– ?
=
2
1
? ? ? = ln2
10. Let f(x) is a five degree polynomial which has critical points x = ±1 and 4
x
) x ( f
2 lim
3
0 x
? ?
?
?
?
?
?
?
?
then which one
is incorrect.
(1) f(x) has minima at x = 1 & maxima at x = –1
(2) f(1) –4f(–1) = 4
(3) f(x) is maxima at x = 1 and minima at x = –1
(4) f(x) is odd
ekuk fd f(x) ,d 5 ?kkr dk cgqin gS ftlds Økafrd fcUnq x = ±1 gS rFkk 4
x
) x ( f
2 lim
3
0 x
? ?
?
?
?
?
?
?
?
rks fuEu esa ls dkSulk
xyr gSA
(1) f(x) dk x = 1 ij fuEufu"B rFkk x = –1 ij mfPPk"B gSA
(2) f(1) –4f(–1) = 4
(3) f(x) dk x = 1 mfPp"B rFkk x = –1 ij fuEuf"B gSA
(4) f(x) fo"ke ?kkr gSA
Ans. (1)
Sol. f(x) = ax
5
+bx
4
+ cx
3
4
x
cx bx ax
2 lim
3
3 4 5
0 x
?
?
?
?
?
?
?
?
?
? ?
?
?
?? 2 + c = 4 ? c = 2
f'(x) = 5ax
4
+ 4bx
3
+ 6x
2
= x
2
? ? 6 bx 4 ax 5
2
? ?
f'(1) = 0 ? 5a + 4b + 6 = 0
f'(–1) = 0 ? 5a – 4b + 6 = 0
b = 0
a = –
5
6
f(x) =
3 5
x 2 x
5
6
?
?
f'(x) = –6x
4
+ 6x
2
= 6x
2
(–x
2
+1)
= –6x
2
(x+1) (x–1)
–1 + 1–
1– 1
Minimal at x = –1
Maxima at x = 1
x = –1 ij fuEufu"B
x = 1 ij mfPp"B
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