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Page 1
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 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 20 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. Find the number of solution of log1/2 |sinx| = 2 – log1/2 |cosx| , x ? [0,2 ?]
x ?[0,2 ?] esa log1/2 |sinx| = 2 – log1/2 |cosx| ds gyksas dh la[;k gS&
(1) 2 (2) 4 (3) 6 (4) 8
Ans. (4)
Sol. log1/2 |sinx| = 2 – log1/2 |cosx|
log1/2 |sinx cosx| = 2
|sinx cosx| =
4
1
sin2x = ±
2
1
? ? 2 ? ?
2
1
2
1
–
Number of solution gyksa dh la[;k = 8.
2. If e1 and e2 are eccentricities of 1
4
y
18
x
2 2
? ? and 1
4
y
9
x
2 2
? ? , respectively and if the point (e1, e2) lies
on ellipse 15x
2
+ 3y
2
= k. Then find value of k
;fn e1 rFkk e2 Øe’’'k% 1
4
y
18
x
2 2
? ? rFkk 1
4
y
9
x
2 2
? ? dh mRdsUnzrk,sa gS rFkk fcUnq (e1, e2) nh?kZo`Ùk 15x
2
+ 3y
2
= k
ij fLFkr gS rks k dk eku gS&
(1) 14 (2) 15 (3) 16 (4) 17
Ans. (3)
Sol. e1 =
18
4
1 ? =
9
7
=
3
7
e2 =
9
4
1 ? =
9
13
=
3
13
k e 3 e 15
2
2
2
1
? ? ? k = 15 ?
?
?
?
?
?
? ?
?
?
?
?
?
9
13
3
9
7
? k = 16
Page 2
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 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 20 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. Find the number of solution of log1/2 |sinx| = 2 – log1/2 |cosx| , x ? [0,2 ?]
x ?[0,2 ?] esa log1/2 |sinx| = 2 – log1/2 |cosx| ds gyksas dh la[;k gS&
(1) 2 (2) 4 (3) 6 (4) 8
Ans. (4)
Sol. log1/2 |sinx| = 2 – log1/2 |cosx|
log1/2 |sinx cosx| = 2
|sinx cosx| =
4
1
sin2x = ±
2
1
? ? 2 ? ?
2
1
2
1
–
Number of solution gyksa dh la[;k = 8.
2. If e1 and e2 are eccentricities of 1
4
y
18
x
2 2
? ? and 1
4
y
9
x
2 2
? ? , respectively and if the point (e1, e2) lies
on ellipse 15x
2
+ 3y
2
= k. Then find value of k
;fn e1 rFkk e2 Øe’’'k% 1
4
y
18
x
2 2
? ? rFkk 1
4
y
9
x
2 2
? ? dh mRdsUnzrk,sa gS rFkk fcUnq (e1, e2) nh?kZo`Ùk 15x
2
+ 3y
2
= k
ij fLFkr gS rks k dk eku gS&
(1) 14 (2) 15 (3) 16 (4) 17
Ans. (3)
Sol. e1 =
18
4
1 ? =
9
7
=
3
7
e2 =
9
4
1 ? =
9
13
=
3
13
k e 3 e 15
2
2
2
1
? ? ? k = 15 ?
?
?
?
?
?
? ?
?
?
?
?
?
9
13
3
9
7
? k = 16
3. Find integration
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
lekdyu
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
=
(1) c
4 x
3 x 7
1
? ?
?
?
?
?
?
?
?
(2) c
4 x
3 x
7
7
1
? ?
?
?
?
?
?
?
?
(3) c
4 x
3 x
7
7
6
? ?
?
?
?
?
?
?
?
(4) c
3 x
4 x
7
7
6
? ?
?
?
?
?
?
?
?
Ans. (1)
Sol.
? ?
dx
4 x
1
4 x
3 x
2
7
6 –
?
?
?
?
?
?
?
?
?
?
Let
7
t
4 x
3 x
?
?
?
,
? ?
dt t 7 dx
4 x
7
6
2
?
?
?
? ? c t dt t t
6 6 –
4. If
i 2 z
i – z
?
= 1, |z| =
2
5
then value of |z + 3i| is
;fn
i 2 z
i – z
?
= 1, |z| =
2
5
rks |z + 3i| dk eku gS&
(1)
2
7
(2) 10 (3) 5 (4) 3
Ans. (1)
Sol. x
2
+ (y–1)
2
= x
2
+ (y+2)
2
–2y + 1 = 4y + 4
6y = –3 ? y = –
2
1
x
2
+ y
2
=
4
25
? x
2
=
4
24
= 6
? z = ± 6 –
2
i
|z + 3i| =
4
25
6 ? =
4
49
|z + 3i | =
2
7
5.
48
1
16
1
4
1
8 4 2 ? ? ………….. 8 =
(1) 2 (2) 2 (3)
4
1
2 (4) 1
Ans. (1)
Page 3
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 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 20 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. Find the number of solution of log1/2 |sinx| = 2 – log1/2 |cosx| , x ? [0,2 ?]
x ?[0,2 ?] esa log1/2 |sinx| = 2 – log1/2 |cosx| ds gyksas dh la[;k gS&
(1) 2 (2) 4 (3) 6 (4) 8
Ans. (4)
Sol. log1/2 |sinx| = 2 – log1/2 |cosx|
log1/2 |sinx cosx| = 2
|sinx cosx| =
4
1
sin2x = ±
2
1
? ? 2 ? ?
2
1
2
1
–
Number of solution gyksa dh la[;k = 8.
2. If e1 and e2 are eccentricities of 1
4
y
18
x
2 2
? ? and 1
4
y
9
x
2 2
? ? , respectively and if the point (e1, e2) lies
on ellipse 15x
2
+ 3y
2
= k. Then find value of k
;fn e1 rFkk e2 Øe’’'k% 1
4
y
18
x
2 2
? ? rFkk 1
4
y
9
x
2 2
? ? dh mRdsUnzrk,sa gS rFkk fcUnq (e1, e2) nh?kZo`Ùk 15x
2
+ 3y
2
= k
ij fLFkr gS rks k dk eku gS&
(1) 14 (2) 15 (3) 16 (4) 17
Ans. (3)
Sol. e1 =
18
4
1 ? =
9
7
=
3
7
e2 =
9
4
1 ? =
9
13
=
3
13
k e 3 e 15
2
2
2
1
? ? ? k = 15 ?
?
?
?
?
?
? ?
?
?
?
?
?
9
13
3
9
7
? k = 16
3. Find integration
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
lekdyu
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
=
(1) c
4 x
3 x 7
1
? ?
?
?
?
?
?
?
?
(2) c
4 x
3 x
7
7
1
? ?
?
?
?
?
?
?
?
(3) c
4 x
3 x
7
7
6
? ?
?
?
?
?
?
?
?
(4) c
3 x
4 x
7
7
6
? ?
?
?
?
?
?
?
?
Ans. (1)
Sol.
? ?
dx
4 x
1
4 x
3 x
2
7
6 –
?
?
?
?
?
?
?
?
?
?
Let
7
t
4 x
3 x
?
?
?
,
? ?
dt t 7 dx
4 x
7
6
2
?
?
?
? ? c t dt t t
6 6 –
4. If
i 2 z
i – z
?
= 1, |z| =
2
5
then value of |z + 3i| is
;fn
i 2 z
i – z
?
= 1, |z| =
2
5
rks |z + 3i| dk eku gS&
(1)
2
7
(2) 10 (3) 5 (4) 3
Ans. (1)
Sol. x
2
+ (y–1)
2
= x
2
+ (y+2)
2
–2y + 1 = 4y + 4
6y = –3 ? y = –
2
1
x
2
+ y
2
=
4
25
? x
2
=
4
24
= 6
? z = ± 6 –
2
i
|z + 3i| =
4
25
6 ? =
4
49
|z + 3i | =
2
7
5.
48
1
16
1
4
1
8 4 2 ? ? ………….. 8 =
(1) 2 (2) 2 (3)
4
1
2 (4) 1
Ans. (1)
Sol.
? ? ? ? .. ..........
48
3
16
2
4
1
2
=
? ? ? ? .. ..........
16
1
8
1
4
1
2 = 2
6. Value of
8
3
cos
8
cos
3
? ?
+ sin
3
8
?
sin
8
3 ?
is
8
3
cos
8
cos
3
? ?
+ sin
3
8
?
sin
8
3 ?
dk eku gS&
(1)
2 2
1
(2)
2
1
(3)
2
1
(4)
2
1
–
Ans. (1)
Sol. cos
3
8
?
?
?
?
?
?
? ? ?
8
cos 3 –
8
cos 4
3
+ sin
3
8
?
?
?
?
?
?
? ? ?
8
sin 4 –
8
sin 3
3
= 4cos
6
8
?
– 4sin
6
8
?
– 3cos
4
8
?
+ 3sin
4
8
?
= ?
?
?
?
?
?
?
?
?
?
?
? ? ?
8
sin –
8
cos 4
2 2
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
8
cos
8
sin
8
cos
8
sin
2 2 4 4
– 3 ?
?
?
?
?
?
?
?
?
?
?
? ? ?
8
sin –
8
cos
2 2
= cos
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
3 –
8
cos
8
sin – 1 4
4
2 2
=
2
1
?
?
?
?
?
?
2
1
– 1 =
2 2
1
7. Find the value of
?
?
?
2
0
8 8
8
dx
x cos x sin
x sin x
?
?
?
2
0
8 8
8
dx
x cos x sin
x sin x
dk eku gS&
(1) ?
2
(2) 2 ?
2
(3) 3 ?
2
(4) 4 ?
2
Ans. (1)
Sol.
?
?
?
?
?
?
0
8 8
8
8 8
8
dx
x cos x sin
x sin ) x – 2 (
x cos x sin
x sin x
=
?
?
?
?
0
8 8
8
dx
x cos x sin
x sin 2
= dx
x cos x sin
x cos
x cos x sin
x sin
2
2 /
0
8 8
8
8 8
8
?
?
?
?
?
?
= dx 1 2
2 /
0
?
?
? = 2 ? ?×
2
?
= ?
2
Page 4
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 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 20 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. Find the number of solution of log1/2 |sinx| = 2 – log1/2 |cosx| , x ? [0,2 ?]
x ?[0,2 ?] esa log1/2 |sinx| = 2 – log1/2 |cosx| ds gyksas dh la[;k gS&
(1) 2 (2) 4 (3) 6 (4) 8
Ans. (4)
Sol. log1/2 |sinx| = 2 – log1/2 |cosx|
log1/2 |sinx cosx| = 2
|sinx cosx| =
4
1
sin2x = ±
2
1
? ? 2 ? ?
2
1
2
1
–
Number of solution gyksa dh la[;k = 8.
2. If e1 and e2 are eccentricities of 1
4
y
18
x
2 2
? ? and 1
4
y
9
x
2 2
? ? , respectively and if the point (e1, e2) lies
on ellipse 15x
2
+ 3y
2
= k. Then find value of k
;fn e1 rFkk e2 Øe’’'k% 1
4
y
18
x
2 2
? ? rFkk 1
4
y
9
x
2 2
? ? dh mRdsUnzrk,sa gS rFkk fcUnq (e1, e2) nh?kZo`Ùk 15x
2
+ 3y
2
= k
ij fLFkr gS rks k dk eku gS&
(1) 14 (2) 15 (3) 16 (4) 17
Ans. (3)
Sol. e1 =
18
4
1 ? =
9
7
=
3
7
e2 =
9
4
1 ? =
9
13
=
3
13
k e 3 e 15
2
2
2
1
? ? ? k = 15 ?
?
?
?
?
?
? ?
?
?
?
?
?
9
13
3
9
7
? k = 16
3. Find integration
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
lekdyu
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
=
(1) c
4 x
3 x 7
1
? ?
?
?
?
?
?
?
?
(2) c
4 x
3 x
7
7
1
? ?
?
?
?
?
?
?
?
(3) c
4 x
3 x
7
7
6
? ?
?
?
?
?
?
?
?
(4) c
3 x
4 x
7
7
6
? ?
?
?
?
?
?
?
?
Ans. (1)
Sol.
? ?
dx
4 x
1
4 x
3 x
2
7
6 –
?
?
?
?
?
?
?
?
?
?
Let
7
t
4 x
3 x
?
?
?
,
? ?
dt t 7 dx
4 x
7
6
2
?
?
?
? ? c t dt t t
6 6 –
4. If
i 2 z
i – z
?
= 1, |z| =
2
5
then value of |z + 3i| is
;fn
i 2 z
i – z
?
= 1, |z| =
2
5
rks |z + 3i| dk eku gS&
(1)
2
7
(2) 10 (3) 5 (4) 3
Ans. (1)
Sol. x
2
+ (y–1)
2
= x
2
+ (y+2)
2
–2y + 1 = 4y + 4
6y = –3 ? y = –
2
1
x
2
+ y
2
=
4
25
? x
2
=
4
24
= 6
? z = ± 6 –
2
i
|z + 3i| =
4
25
6 ? =
4
49
|z + 3i | =
2
7
5.
48
1
16
1
4
1
8 4 2 ? ? ………….. 8 =
(1) 2 (2) 2 (3)
4
1
2 (4) 1
Ans. (1)
Sol.
? ? ? ? .. ..........
48
3
16
2
4
1
2
=
? ? ? ? .. ..........
16
1
8
1
4
1
2 = 2
6. Value of
8
3
cos
8
cos
3
? ?
+ sin
3
8
?
sin
8
3 ?
is
8
3
cos
8
cos
3
? ?
+ sin
3
8
?
sin
8
3 ?
dk eku gS&
(1)
2 2
1
(2)
2
1
(3)
2
1
(4)
2
1
–
Ans. (1)
Sol. cos
3
8
?
?
?
?
?
?
? ? ?
8
cos 3 –
8
cos 4
3
+ sin
3
8
?
?
?
?
?
?
? ? ?
8
sin 4 –
8
sin 3
3
= 4cos
6
8
?
– 4sin
6
8
?
– 3cos
4
8
?
+ 3sin
4
8
?
= ?
?
?
?
?
?
?
?
?
?
?
? ? ?
8
sin –
8
cos 4
2 2
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
8
cos
8
sin
8
cos
8
sin
2 2 4 4
– 3 ?
?
?
?
?
?
?
?
?
?
?
? ? ?
8
sin –
8
cos
2 2
= cos
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
3 –
8
cos
8
sin – 1 4
4
2 2
=
2
1
?
?
?
?
?
?
2
1
– 1 =
2 2
1
7. Find the value of
?
?
?
2
0
8 8
8
dx
x cos x sin
x sin x
?
?
?
2
0
8 8
8
dx
x cos x sin
x sin x
dk eku gS&
(1) ?
2
(2) 2 ?
2
(3) 3 ?
2
(4) 4 ?
2
Ans. (1)
Sol.
?
?
?
?
?
?
0
8 8
8
8 8
8
dx
x cos x sin
x sin ) x – 2 (
x cos x sin
x sin x
=
?
?
?
?
0
8 8
8
dx
x cos x sin
x sin 2
= dx
x cos x sin
x cos
x cos x sin
x sin
2
2 /
0
8 8
8
8 8
8
?
?
?
?
?
?
= dx 1 2
2 /
0
?
?
? = 2 ? ?×
2
?
= ?
2
8. If f(x) = a + bx + cx
2
where a, b, c ?R then
?
1
0
dx ) x ( f is
;fn f(x) = a + bx + cx
2
tgkaa a, b, c ?R rc
?
1
0
dx ) x ( f cjkcj gS
(1)
3
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
f 2 0 f 1 f
(2)
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
f 4 0 f 1 f
(3)
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
f 4 – 0 f 1 f
(4)
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
f 4 – 0 f – 1 f
Ans. (2)
Sol. ? ?
?
? ?
1
0
2
dx cx bx a = ax +
2
bx
2
+
1
0
3
3
cx
= a +
2
b
+
3
c
f(1) = a + b + c
f(0) = a
f ?
?
?
?
?
?
2
1
= a +
2
b
+
4
C
Now
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
f 4 0 f 1 f
=
6
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
4
c
2
b
a 4 a c b a
=
6
1
? ? c 2 b 3 a 6 ? ? = a +
2
b
+
3
c
9. If number of 5 digit numbers which can be formed without repeating any digit while tenth place of all of
the numbers must be 2 is 336 k find value of k
fcuk fdlh vad dh iqujko`fÙk ds cuus okys 5 vadks dh la[;kvksas dh la[;k tcfd ngkbZ ds LFkku ij lHkh la[;kvksa esa
2 vkrk gks 336 k gS rks k dk eku gS&
(1) 8 (2) 7 (3) 6 (4) 5
Ans. (1)
Sol.
2
Number of numbers la[;kvksa dh la[;k = 8 ? ?8 ? ?7 ? ?6 = 2688 = 336k ? ?k = 8
Page 5
PART : MATHEMATICS
SECTION – 1
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 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 20 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. Find the number of solution of log1/2 |sinx| = 2 – log1/2 |cosx| , x ? [0,2 ?]
x ?[0,2 ?] esa log1/2 |sinx| = 2 – log1/2 |cosx| ds gyksas dh la[;k gS&
(1) 2 (2) 4 (3) 6 (4) 8
Ans. (4)
Sol. log1/2 |sinx| = 2 – log1/2 |cosx|
log1/2 |sinx cosx| = 2
|sinx cosx| =
4
1
sin2x = ±
2
1
? ? 2 ? ?
2
1
2
1
–
Number of solution gyksa dh la[;k = 8.
2. If e1 and e2 are eccentricities of 1
4
y
18
x
2 2
? ? and 1
4
y
9
x
2 2
? ? , respectively and if the point (e1, e2) lies
on ellipse 15x
2
+ 3y
2
= k. Then find value of k
;fn e1 rFkk e2 Øe’’'k% 1
4
y
18
x
2 2
? ? rFkk 1
4
y
9
x
2 2
? ? dh mRdsUnzrk,sa gS rFkk fcUnq (e1, e2) nh?kZo`Ùk 15x
2
+ 3y
2
= k
ij fLFkr gS rks k dk eku gS&
(1) 14 (2) 15 (3) 16 (4) 17
Ans. (3)
Sol. e1 =
18
4
1 ? =
9
7
=
3
7
e2 =
9
4
1 ? =
9
13
=
3
13
k e 3 e 15
2
2
2
1
? ? ? k = 15 ?
?
?
?
?
?
? ?
?
?
?
?
?
9
13
3
9
7
? k = 16
3. Find integration
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
lekdyu
? ? ? ?
?
? ?
7 / 8 7 / 6
4 x . 3 x
dx
=
(1) c
4 x
3 x 7
1
? ?
?
?
?
?
?
?
?
(2) c
4 x
3 x
7
7
1
? ?
?
?
?
?
?
?
?
(3) c
4 x
3 x
7
7
6
? ?
?
?
?
?
?
?
?
(4) c
3 x
4 x
7
7
6
? ?
?
?
?
?
?
?
?
Ans. (1)
Sol.
? ?
dx
4 x
1
4 x
3 x
2
7
6 –
?
?
?
?
?
?
?
?
?
?
Let
7
t
4 x
3 x
?
?
?
,
? ?
dt t 7 dx
4 x
7
6
2
?
?
?
? ? c t dt t t
6 6 –
4. If
i 2 z
i – z
?
= 1, |z| =
2
5
then value of |z + 3i| is
;fn
i 2 z
i – z
?
= 1, |z| =
2
5
rks |z + 3i| dk eku gS&
(1)
2
7
(2) 10 (3) 5 (4) 3
Ans. (1)
Sol. x
2
+ (y–1)
2
= x
2
+ (y+2)
2
–2y + 1 = 4y + 4
6y = –3 ? y = –
2
1
x
2
+ y
2
=
4
25
? x
2
=
4
24
= 6
? z = ± 6 –
2
i
|z + 3i| =
4
25
6 ? =
4
49
|z + 3i | =
2
7
5.
48
1
16
1
4
1
8 4 2 ? ? ………….. 8 =
(1) 2 (2) 2 (3)
4
1
2 (4) 1
Ans. (1)
Sol.
? ? ? ? .. ..........
48
3
16
2
4
1
2
=
? ? ? ? .. ..........
16
1
8
1
4
1
2 = 2
6. Value of
8
3
cos
8
cos
3
? ?
+ sin
3
8
?
sin
8
3 ?
is
8
3
cos
8
cos
3
? ?
+ sin
3
8
?
sin
8
3 ?
dk eku gS&
(1)
2 2
1
(2)
2
1
(3)
2
1
(4)
2
1
–
Ans. (1)
Sol. cos
3
8
?
?
?
?
?
?
? ? ?
8
cos 3 –
8
cos 4
3
+ sin
3
8
?
?
?
?
?
?
? ? ?
8
sin 4 –
8
sin 3
3
= 4cos
6
8
?
– 4sin
6
8
?
– 3cos
4
8
?
+ 3sin
4
8
?
= ?
?
?
?
?
?
?
?
?
?
?
? ? ?
8
sin –
8
cos 4
2 2
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
8
cos
8
sin
8
cos
8
sin
2 2 4 4
– 3 ?
?
?
?
?
?
?
?
?
?
?
? ? ?
8
sin –
8
cos
2 2
= cos
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
3 –
8
cos
8
sin – 1 4
4
2 2
=
2
1
?
?
?
?
?
?
2
1
– 1 =
2 2
1
7. Find the value of
?
?
?
2
0
8 8
8
dx
x cos x sin
x sin x
?
?
?
2
0
8 8
8
dx
x cos x sin
x sin x
dk eku gS&
(1) ?
2
(2) 2 ?
2
(3) 3 ?
2
(4) 4 ?
2
Ans. (1)
Sol.
?
?
?
?
?
?
0
8 8
8
8 8
8
dx
x cos x sin
x sin ) x – 2 (
x cos x sin
x sin x
=
?
?
?
?
0
8 8
8
dx
x cos x sin
x sin 2
= dx
x cos x sin
x cos
x cos x sin
x sin
2
2 /
0
8 8
8
8 8
8
?
?
?
?
?
?
= dx 1 2
2 /
0
?
?
? = 2 ? ?×
2
?
= ?
2
8. If f(x) = a + bx + cx
2
where a, b, c ?R then
?
1
0
dx ) x ( f is
;fn f(x) = a + bx + cx
2
tgkaa a, b, c ?R rc
?
1
0
dx ) x ( f cjkcj gS
(1)
3
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
f 2 0 f 1 f
(2)
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
f 4 0 f 1 f
(3)
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
f 4 – 0 f 1 f
(4)
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
f 4 – 0 f – 1 f
Ans. (2)
Sol. ? ?
?
? ?
1
0
2
dx cx bx a = ax +
2
bx
2
+
1
0
3
3
cx
= a +
2
b
+
3
c
f(1) = a + b + c
f(0) = a
f ?
?
?
?
?
?
2
1
= a +
2
b
+
4
C
Now
6
1
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1
f 4 0 f 1 f
=
6
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ? ? ?
4
c
2
b
a 4 a c b a
=
6
1
? ? c 2 b 3 a 6 ? ? = a +
2
b
+
3
c
9. If number of 5 digit numbers which can be formed without repeating any digit while tenth place of all of
the numbers must be 2 is 336 k find value of k
fcuk fdlh vad dh iqujko`fÙk ds cuus okys 5 vadks dh la[;kvksas dh la[;k tcfd ngkbZ ds LFkku ij lHkh la[;kvksa esa
2 vkrk gks 336 k gS rks k dk eku gS&
(1) 8 (2) 7 (3) 6 (4) 5
Ans. (1)
Sol.
2
Number of numbers la[;kvksa dh la[;k = 8 ? ?8 ? ?7 ? ?6 = 2688 = 336k ? ?k = 8
10. A (3,–1), B(1,3), C(2,4) are vertices of ?ABC if D is centroid of ?ABC and P is point of intersection of
lines x + 3y – 1 = 0 and 3x – y + 1 = 0 then which of the following points lies on line joining D and P
f=kHkqt ABC ds '’kh"kZ A (3,–1), B(1,3), C(2,4) gS ;fn D f=kHkqt ABC dk dsUnzd gS rFkk P js[kkvksa x + 3y – 1 = 0
rFkk 3x – y + 1 = 0 dk izfrPNsn fcUnq gS rks fuEu esa ls dkSulk fcUnq D rFkk P dks tksM+us okyh js[kk ij fLFkr gS&
(1) (–9,–7) (2*) (–9,–6) (3) (9,6) (4) ? ? 6 ,– 9
Ans. (2)
Sol. D (2,2)
Point of intersection P ?
?
?
?
?
?
5
2
,
5
1
–
equation of line DP
8x – 11y + 6 = 0
Sol. D (2,2)
izfrPNsn fcUnq P ?
?
?
?
?
?
5
2
,
5
1
–
js[kk DP dk lehdj.k
8x – 11y + 6 = 0
11. If f(x) is twice differentiable and continuous function in x ? [a,b] also f'(x) > 0 and f ''(x) < 0 and c ? (a,b)
then
) c ( f – ) b ( f
) a ( f – ) c ( f
is greater than
;fn f(x) nks ckj vodyuh; rFkk lrr~ Qyu gS rFkk x ? [a,b] rFkk f'(x) > 0, f ''(x) < 0 rFkk c ? (a,b) rks
) c ( f – ) b ( f
) a ( f – ) c ( f
ftlls cM+k gS] og gS&
(1)
a – c
c – b
(2) 1 (3)
c – b
b a ?
(4)
c – b
a – c
Ans. (4)
Sol. Lets use LMVT for x ??[a,c]
a – c
) a ( f – ) c ( f
= f'( ?) , ? ? (a,c)
also use LMVT for x ? [c,b]
c – b
) c ( f – ) b ( f
= f'( ?) , ? ? (c,b)
? f ''(x) < 0 ? f '(x) is decreasing
f '( ?) > f '( ?)
a – c
) a ( f – ) c ( f
>
c – b
) c ( f – ) b ( f
) c ( f – ) b ( f
) a ( f – ) c ( f
>
c – b
a – c
( ? f(x) is increasing)
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