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Page 1
PAGE # 1
PART : MATHEMATICS
1. Let y = f(x) is a solution of differential equation
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? and f(0) = 0 then f(1) is equal to :
;fn y = f(x) vody lehdj.k
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? dk gy gS rFkk f(0) = 0 rc f(1) dk eku gS&
(1) ?n2 (2) 2 + ?n2 (3) 1 + ?n2 (4) 3 + ?n2
Ans. (3)
Sol. e
y
= t
e
y
dx
dy
=
dx
dt
dx
dt
– t = e
x
IF =
?
? dx . 1
e = e
–x
t(e
–x
) = dx e . e
x x ?
?
e
y–x
= x + c
Put x = 0, y = 0 j[kus ij then c = 1
e
y–x
= x + 1
y = x + ?n(x + 1)
at x = 1 j[kus ij, y = 1 + ?n(2)
2. If ?? ?is a roots of equation x
2
+ x + 1 = 0 and A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
then A
31
equal to :
;fn ?? ?lehdj.k x
2
+ x + 1 = 0 dk ,d ewy gS rFkk A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
rc A
31
cjkcj gS&
(1) A (2) A
2
(3) A
3
(4) A
4
Ans. (3)
Sol. A
2
=
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
2
2
1
1
1 1 1
1
1
1 1 1
3
1
=
?
?
?
?
?
?
?
?
?
?
0 1 0
1 0 0
0 0 1
? A
4
= I
? A
30
= A
28
× A
3
= A
3
Page 2
PAGE # 1
PART : MATHEMATICS
1. Let y = f(x) is a solution of differential equation
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? and f(0) = 0 then f(1) is equal to :
;fn y = f(x) vody lehdj.k
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? dk gy gS rFkk f(0) = 0 rc f(1) dk eku gS&
(1) ?n2 (2) 2 + ?n2 (3) 1 + ?n2 (4) 3 + ?n2
Ans. (3)
Sol. e
y
= t
e
y
dx
dy
=
dx
dt
dx
dt
– t = e
x
IF =
?
? dx . 1
e = e
–x
t(e
–x
) = dx e . e
x x ?
?
e
y–x
= x + c
Put x = 0, y = 0 j[kus ij then c = 1
e
y–x
= x + 1
y = x + ?n(x + 1)
at x = 1 j[kus ij, y = 1 + ?n(2)
2. If ?? ?is a roots of equation x
2
+ x + 1 = 0 and A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
then A
31
equal to :
;fn ?? ?lehdj.k x
2
+ x + 1 = 0 dk ,d ewy gS rFkk A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
rc A
31
cjkcj gS&
(1) A (2) A
2
(3) A
3
(4) A
4
Ans. (3)
Sol. A
2
=
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
2
2
1
1
1 1 1
1
1
1 1 1
3
1
=
?
?
?
?
?
?
?
?
?
?
0 1 0
1 0 0
0 0 1
? A
4
= I
? A
30
= A
28
× A
3
= A
3
PAGE # 2
3. The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least
once.
vadks 1, 3, 5, 7, 9 dk iz;ksx dj cukbZ tk ldus okyh N% vadks dh la[;kvksa dh la[;k fd izR;sd vad de ls de
,d ckj iz;qDr gks] gksxh&
Ans. 1800
Sol. 1, 3, 5, 7, 9
For digit to repeat we have
5
C1 choice
And six digits can be arrange in
2
6
ways.
Hence total such numbers =
2
6 5
Sol. 1, 3, 5, 7, 9
Ikqujko`fÙk okys vad ds fy,
5
C1 fodYi gSA
vkSj N% vadksa dh O;oLFkk ds rjhds
2
6
vr% dqy la[;k,sa =
2
6 5
4. The area that is enclosed in the circle x
2
+ y
2
= 2 which is not common area enclosed by y = x & y
2
= x
is
o`Ùk x
2
+ y
2
= 2 ds vUrxZr ifjc) og {ks=kQy tks y = x & y
2
= x }kjk ifjc) {ks=kQy ds lkFk mHk;fu”"B u gks&
(1)
12
1
(24 ? – 1) (2)
6
1
(12 ? – 1) (3)
12
1
(6 ? – 1) (4)
12
1
(12 ? – 1)
Ans. (2)
Sol. Total area – enclosed area
dqqy {ks=kQy – mHk;fu”"B {ks=kQy
2 ? –
?
?
1
0
dx x x
2 ? –
1
0
2 2 / 3
2
x
3
x 2
?
?
?
?
?
?
?
?
?
2 ? – ?
?
?
?
?
?
?
2
1
3
2
? 2 ? – ?
?
?
?
?
?
6
1
?
6
1 12 ? ?
Page 3
PAGE # 1
PART : MATHEMATICS
1. Let y = f(x) is a solution of differential equation
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? and f(0) = 0 then f(1) is equal to :
;fn y = f(x) vody lehdj.k
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? dk gy gS rFkk f(0) = 0 rc f(1) dk eku gS&
(1) ?n2 (2) 2 + ?n2 (3) 1 + ?n2 (4) 3 + ?n2
Ans. (3)
Sol. e
y
= t
e
y
dx
dy
=
dx
dt
dx
dt
– t = e
x
IF =
?
? dx . 1
e = e
–x
t(e
–x
) = dx e . e
x x ?
?
e
y–x
= x + c
Put x = 0, y = 0 j[kus ij then c = 1
e
y–x
= x + 1
y = x + ?n(x + 1)
at x = 1 j[kus ij, y = 1 + ?n(2)
2. If ?? ?is a roots of equation x
2
+ x + 1 = 0 and A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
then A
31
equal to :
;fn ?? ?lehdj.k x
2
+ x + 1 = 0 dk ,d ewy gS rFkk A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
rc A
31
cjkcj gS&
(1) A (2) A
2
(3) A
3
(4) A
4
Ans. (3)
Sol. A
2
=
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
2
2
1
1
1 1 1
1
1
1 1 1
3
1
=
?
?
?
?
?
?
?
?
?
?
0 1 0
1 0 0
0 0 1
? A
4
= I
? A
30
= A
28
× A
3
= A
3
PAGE # 2
3. The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least
once.
vadks 1, 3, 5, 7, 9 dk iz;ksx dj cukbZ tk ldus okyh N% vadks dh la[;kvksa dh la[;k fd izR;sd vad de ls de
,d ckj iz;qDr gks] gksxh&
Ans. 1800
Sol. 1, 3, 5, 7, 9
For digit to repeat we have
5
C1 choice
And six digits can be arrange in
2
6
ways.
Hence total such numbers =
2
6 5
Sol. 1, 3, 5, 7, 9
Ikqujko`fÙk okys vad ds fy,
5
C1 fodYi gSA
vkSj N% vadksa dh O;oLFkk ds rjhds
2
6
vr% dqy la[;k,sa =
2
6 5
4. The area that is enclosed in the circle x
2
+ y
2
= 2 which is not common area enclosed by y = x & y
2
= x
is
o`Ùk x
2
+ y
2
= 2 ds vUrxZr ifjc) og {ks=kQy tks y = x & y
2
= x }kjk ifjc) {ks=kQy ds lkFk mHk;fu”"B u gks&
(1)
12
1
(24 ? – 1) (2)
6
1
(12 ? – 1) (3)
12
1
(6 ? – 1) (4)
12
1
(12 ? – 1)
Ans. (2)
Sol. Total area – enclosed area
dqqy {ks=kQy – mHk;fu”"B {ks=kQy
2 ? –
?
?
1
0
dx x x
2 ? –
1
0
2 2 / 3
2
x
3
x 2
?
?
?
?
?
?
?
?
?
2 ? – ?
?
?
?
?
?
?
2
1
3
2
? 2 ? – ?
?
?
?
?
?
6
1
?
6
1 12 ? ?
PAGE # 3
5. If sum of all the coefficient of even powers in (1– x + x
2
– x
3
……..x
2n
) (1 + x + x
2
+x
3
……….+ x
2n
) is 61
then n is equal to
(1– x + x
2
– x
3
……..x
2n
) (1 + x + x
2
+x
3
……….+ x
2n
) esa le?kkrksa ds lHkh xq.kkadks dk ;ksxQy 61 gSA rc n dk
eku cjkcj gS&
(1) 30 (2) 32 (3) 28 (4) 36
Ans. (1)
Sol. Let ekuk (1– x + x
2
…..) (1 + x + x
2
……) = a0 + a1 x + a2x
2
+……
put x = 1 j[kus ij
1(2n+1) = a0 + a1 + a2 +……a2n ……..(i)
put x = –1 j[kus ij
(2n+1??1 = a0 – a1 + a2 +……a2n ……..(ii)
Form (i) + (ii) ls
4n + 2= 2(a0 + a2 +….)
= 2 ? ?61
? 2n+1 = 61 ? n = 30
6. If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the
value of m + n is
izFke n izkd`r la[;kvksa dk izlj.k 10 gS rFkk izFke m le izkd`r la[;kvksa dk izlj.k 16 gS rks m + n dk eku gS&
Ans. 18
Sol. Var (1, 2, ……..,n) = 10 ?
n
n ......... 2 1
2 2 2
? ? ?
– 10
n
n ......... 2 1
2
? ?
?
?
?
?
? ? ? ?
? 10
2
1 n
6
) 1 n 2 )( 1 n (
2
? ?
?
?
?
?
? ?
?
? ?
? n
2
– 1 = 120 ? n = 11
Var (2, 4, 6,…………,2m) = 16 ? var (1, 2,……..,m) = 4
? m
2
– 1 = 48 ? m = 7 ? m + n = 18
7. Evaluate
2 x
lim
?
x 1
2
x
1 x x
3 3
12 3 3
?
?
?
?
? ?
Kkr dhft,A
Ans. 72
Sol. Put
2
x
3 = t j[kus ij
?
3 t
lim
?
t
1
t
3
12
3
t 4
2
2
? ?
?
=
3 t
lim
? ) t 3 ( 3
t ) 9 t ( 4
2 2
? ?
?
=
3 t
lim
? 3
) t 3 ( t 4
2
?
=
3
6 9 4 ? ?
= 72
Page 4
PAGE # 1
PART : MATHEMATICS
1. Let y = f(x) is a solution of differential equation
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? and f(0) = 0 then f(1) is equal to :
;fn y = f(x) vody lehdj.k
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? dk gy gS rFkk f(0) = 0 rc f(1) dk eku gS&
(1) ?n2 (2) 2 + ?n2 (3) 1 + ?n2 (4) 3 + ?n2
Ans. (3)
Sol. e
y
= t
e
y
dx
dy
=
dx
dt
dx
dt
– t = e
x
IF =
?
? dx . 1
e = e
–x
t(e
–x
) = dx e . e
x x ?
?
e
y–x
= x + c
Put x = 0, y = 0 j[kus ij then c = 1
e
y–x
= x + 1
y = x + ?n(x + 1)
at x = 1 j[kus ij, y = 1 + ?n(2)
2. If ?? ?is a roots of equation x
2
+ x + 1 = 0 and A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
then A
31
equal to :
;fn ?? ?lehdj.k x
2
+ x + 1 = 0 dk ,d ewy gS rFkk A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
rc A
31
cjkcj gS&
(1) A (2) A
2
(3) A
3
(4) A
4
Ans. (3)
Sol. A
2
=
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
2
2
1
1
1 1 1
1
1
1 1 1
3
1
=
?
?
?
?
?
?
?
?
?
?
0 1 0
1 0 0
0 0 1
? A
4
= I
? A
30
= A
28
× A
3
= A
3
PAGE # 2
3. The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least
once.
vadks 1, 3, 5, 7, 9 dk iz;ksx dj cukbZ tk ldus okyh N% vadks dh la[;kvksa dh la[;k fd izR;sd vad de ls de
,d ckj iz;qDr gks] gksxh&
Ans. 1800
Sol. 1, 3, 5, 7, 9
For digit to repeat we have
5
C1 choice
And six digits can be arrange in
2
6
ways.
Hence total such numbers =
2
6 5
Sol. 1, 3, 5, 7, 9
Ikqujko`fÙk okys vad ds fy,
5
C1 fodYi gSA
vkSj N% vadksa dh O;oLFkk ds rjhds
2
6
vr% dqy la[;k,sa =
2
6 5
4. The area that is enclosed in the circle x
2
+ y
2
= 2 which is not common area enclosed by y = x & y
2
= x
is
o`Ùk x
2
+ y
2
= 2 ds vUrxZr ifjc) og {ks=kQy tks y = x & y
2
= x }kjk ifjc) {ks=kQy ds lkFk mHk;fu”"B u gks&
(1)
12
1
(24 ? – 1) (2)
6
1
(12 ? – 1) (3)
12
1
(6 ? – 1) (4)
12
1
(12 ? – 1)
Ans. (2)
Sol. Total area – enclosed area
dqqy {ks=kQy – mHk;fu”"B {ks=kQy
2 ? –
?
?
1
0
dx x x
2 ? –
1
0
2 2 / 3
2
x
3
x 2
?
?
?
?
?
?
?
?
?
2 ? – ?
?
?
?
?
?
?
2
1
3
2
? 2 ? – ?
?
?
?
?
?
6
1
?
6
1 12 ? ?
PAGE # 3
5. If sum of all the coefficient of even powers in (1– x + x
2
– x
3
……..x
2n
) (1 + x + x
2
+x
3
……….+ x
2n
) is 61
then n is equal to
(1– x + x
2
– x
3
……..x
2n
) (1 + x + x
2
+x
3
……….+ x
2n
) esa le?kkrksa ds lHkh xq.kkadks dk ;ksxQy 61 gSA rc n dk
eku cjkcj gS&
(1) 30 (2) 32 (3) 28 (4) 36
Ans. (1)
Sol. Let ekuk (1– x + x
2
…..) (1 + x + x
2
……) = a0 + a1 x + a2x
2
+……
put x = 1 j[kus ij
1(2n+1) = a0 + a1 + a2 +……a2n ……..(i)
put x = –1 j[kus ij
(2n+1??1 = a0 – a1 + a2 +……a2n ……..(ii)
Form (i) + (ii) ls
4n + 2= 2(a0 + a2 +….)
= 2 ? ?61
? 2n+1 = 61 ? n = 30
6. If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the
value of m + n is
izFke n izkd`r la[;kvksa dk izlj.k 10 gS rFkk izFke m le izkd`r la[;kvksa dk izlj.k 16 gS rks m + n dk eku gS&
Ans. 18
Sol. Var (1, 2, ……..,n) = 10 ?
n
n ......... 2 1
2 2 2
? ? ?
– 10
n
n ......... 2 1
2
? ?
?
?
?
?
? ? ? ?
? 10
2
1 n
6
) 1 n 2 )( 1 n (
2
? ?
?
?
?
?
? ?
?
? ?
? n
2
– 1 = 120 ? n = 11
Var (2, 4, 6,…………,2m) = 16 ? var (1, 2,……..,m) = 4
? m
2
– 1 = 48 ? m = 7 ? m + n = 18
7. Evaluate
2 x
lim
?
x 1
2
x
1 x x
3 3
12 3 3
?
?
?
?
? ?
Kkr dhft,A
Ans. 72
Sol. Put
2
x
3 = t j[kus ij
?
3 t
lim
?
t
1
t
3
12
3
t 4
2
2
? ?
?
=
3 t
lim
? ) t 3 ( 3
t ) 9 t ( 4
2 2
? ?
?
=
3 t
lim
? 3
) t 3 ( t 4
2
?
=
3
6 9 4 ? ?
= 72
PAGE # 4
8. If f(x) is continuous and differentiable in x ? [–7, 0] and f ?(x) ?? ?? ?x ? [–7, 0], also f(–7) = –3 then range
of f(–1) + f(0)
;fn f(x), x ? [–7, 0] esa lrr~ rFkk vodyuh; gS rFkk f ?(x) ???? ?x ? [–7, 0] vkSj f(–7) = –3 rks f(–1) + f(0) dk
ifjlj gS&
(1) [–5, –7] (2) (– ?, 6] (3) (– ?, 20] (4) [-5, 3]
Ans. (3)
Sol. Lets use LMVT for x ? [–7, –1] esa ykaxzkt e/;eku izes; ls
2
) 7 1 (
) 7 ( f ) 1 ( f
?
? ?
? ? ?
2
6
3 ) 1 ( f
?
? ?
? f(–1) ? 9
Also use LMVT for x ? [–7, 0] esa ykaxzkt e/;eku izes; ls
2
) 7 0 (
) 7 ( f ) 0 ( f
?
?
? ?
2
7
3 ) 0 ( f
?
?
? f(0) ? 11 ? f(0) + f(–1) ? 20
9. If y = mx + 4 is common tangent to parabolas y
2
= 4x and x
2
= 2by. Then value of b is
;fn y = mx + 4 ijoy; y
2
= 4x rFkk x
2
= 2by dh mHk;fu”"B Li’'kZ js[kk gS rks b dk eku gS&
(1) –64 (2) –32 (3) –128 (4) 16
Ans. (3)
Sol. y = mx + 4 ……(i)
y
2
= 4x tangent dh Li'kZ js[kk y = mx +
m
a
? y = mx +
m
1
………(ii)
from (i) and (ii) ls
4 =
m
1
? m =
4
1
So line y =
4
1
x + 4 is also tangent to parabola x
2
= 2by, so solve
vr% js[kk y =
4
1
x + 4 ijoy; x
2
= 2by dh Li'kZ js[kk gS vr%
x
2
= 2b ?
?
?
?
?
? ?
4
16 x
? 2x
2
– bx – 16b = 0 ? D = 0 ? b
2
– 4 × 2 × (–16b) = 0
? b
2
+ 32 × 4b = 0
b = –128, b = 0 (not possible vlaHko)
Page 5
PAGE # 1
PART : MATHEMATICS
1. Let y = f(x) is a solution of differential equation
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? and f(0) = 0 then f(1) is equal to :
;fn y = f(x) vody lehdj.k
x y
e 1
dx
dy
e ? ?
?
?
?
?
?
? dk gy gS rFkk f(0) = 0 rc f(1) dk eku gS&
(1) ?n2 (2) 2 + ?n2 (3) 1 + ?n2 (4) 3 + ?n2
Ans. (3)
Sol. e
y
= t
e
y
dx
dy
=
dx
dt
dx
dt
– t = e
x
IF =
?
? dx . 1
e = e
–x
t(e
–x
) = dx e . e
x x ?
?
e
y–x
= x + c
Put x = 0, y = 0 j[kus ij then c = 1
e
y–x
= x + 1
y = x + ?n(x + 1)
at x = 1 j[kus ij, y = 1 + ?n(2)
2. If ?? ?is a roots of equation x
2
+ x + 1 = 0 and A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
then A
31
equal to :
;fn ?? ?lehdj.k x
2
+ x + 1 = 0 dk ,d ewy gS rFkk A =
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
1
1
1 1 1
3
1
rc A
31
cjkcj gS&
(1) A (2) A
2
(3) A
3
(4) A
4
Ans. (3)
Sol. A
2
=
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
2
2
2
2
1
1
1 1 1
1
1
1 1 1
3
1
=
?
?
?
?
?
?
?
?
?
?
0 1 0
1 0 0
0 0 1
? A
4
= I
? A
30
= A
28
× A
3
= A
3
PAGE # 2
3. The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least
once.
vadks 1, 3, 5, 7, 9 dk iz;ksx dj cukbZ tk ldus okyh N% vadks dh la[;kvksa dh la[;k fd izR;sd vad de ls de
,d ckj iz;qDr gks] gksxh&
Ans. 1800
Sol. 1, 3, 5, 7, 9
For digit to repeat we have
5
C1 choice
And six digits can be arrange in
2
6
ways.
Hence total such numbers =
2
6 5
Sol. 1, 3, 5, 7, 9
Ikqujko`fÙk okys vad ds fy,
5
C1 fodYi gSA
vkSj N% vadksa dh O;oLFkk ds rjhds
2
6
vr% dqy la[;k,sa =
2
6 5
4. The area that is enclosed in the circle x
2
+ y
2
= 2 which is not common area enclosed by y = x & y
2
= x
is
o`Ùk x
2
+ y
2
= 2 ds vUrxZr ifjc) og {ks=kQy tks y = x & y
2
= x }kjk ifjc) {ks=kQy ds lkFk mHk;fu”"B u gks&
(1)
12
1
(24 ? – 1) (2)
6
1
(12 ? – 1) (3)
12
1
(6 ? – 1) (4)
12
1
(12 ? – 1)
Ans. (2)
Sol. Total area – enclosed area
dqqy {ks=kQy – mHk;fu”"B {ks=kQy
2 ? –
?
?
1
0
dx x x
2 ? –
1
0
2 2 / 3
2
x
3
x 2
?
?
?
?
?
?
?
?
?
2 ? – ?
?
?
?
?
?
?
2
1
3
2
? 2 ? – ?
?
?
?
?
?
6
1
?
6
1 12 ? ?
PAGE # 3
5. If sum of all the coefficient of even powers in (1– x + x
2
– x
3
……..x
2n
) (1 + x + x
2
+x
3
……….+ x
2n
) is 61
then n is equal to
(1– x + x
2
– x
3
……..x
2n
) (1 + x + x
2
+x
3
……….+ x
2n
) esa le?kkrksa ds lHkh xq.kkadks dk ;ksxQy 61 gSA rc n dk
eku cjkcj gS&
(1) 30 (2) 32 (3) 28 (4) 36
Ans. (1)
Sol. Let ekuk (1– x + x
2
…..) (1 + x + x
2
……) = a0 + a1 x + a2x
2
+……
put x = 1 j[kus ij
1(2n+1) = a0 + a1 + a2 +……a2n ……..(i)
put x = –1 j[kus ij
(2n+1??1 = a0 – a1 + a2 +……a2n ……..(ii)
Form (i) + (ii) ls
4n + 2= 2(a0 + a2 +….)
= 2 ? ?61
? 2n+1 = 61 ? n = 30
6. If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the
value of m + n is
izFke n izkd`r la[;kvksa dk izlj.k 10 gS rFkk izFke m le izkd`r la[;kvksa dk izlj.k 16 gS rks m + n dk eku gS&
Ans. 18
Sol. Var (1, 2, ……..,n) = 10 ?
n
n ......... 2 1
2 2 2
? ? ?
– 10
n
n ......... 2 1
2
? ?
?
?
?
?
? ? ? ?
? 10
2
1 n
6
) 1 n 2 )( 1 n (
2
? ?
?
?
?
?
? ?
?
? ?
? n
2
– 1 = 120 ? n = 11
Var (2, 4, 6,…………,2m) = 16 ? var (1, 2,……..,m) = 4
? m
2
– 1 = 48 ? m = 7 ? m + n = 18
7. Evaluate
2 x
lim
?
x 1
2
x
1 x x
3 3
12 3 3
?
?
?
?
? ?
Kkr dhft,A
Ans. 72
Sol. Put
2
x
3 = t j[kus ij
?
3 t
lim
?
t
1
t
3
12
3
t 4
2
2
? ?
?
=
3 t
lim
? ) t 3 ( 3
t ) 9 t ( 4
2 2
? ?
?
=
3 t
lim
? 3
) t 3 ( t 4
2
?
=
3
6 9 4 ? ?
= 72
PAGE # 4
8. If f(x) is continuous and differentiable in x ? [–7, 0] and f ?(x) ?? ?? ?x ? [–7, 0], also f(–7) = –3 then range
of f(–1) + f(0)
;fn f(x), x ? [–7, 0] esa lrr~ rFkk vodyuh; gS rFkk f ?(x) ???? ?x ? [–7, 0] vkSj f(–7) = –3 rks f(–1) + f(0) dk
ifjlj gS&
(1) [–5, –7] (2) (– ?, 6] (3) (– ?, 20] (4) [-5, 3]
Ans. (3)
Sol. Lets use LMVT for x ? [–7, –1] esa ykaxzkt e/;eku izes; ls
2
) 7 1 (
) 7 ( f ) 1 ( f
?
? ?
? ? ?
2
6
3 ) 1 ( f
?
? ?
? f(–1) ? 9
Also use LMVT for x ? [–7, 0] esa ykaxzkt e/;eku izes; ls
2
) 7 0 (
) 7 ( f ) 0 ( f
?
?
? ?
2
7
3 ) 0 ( f
?
?
? f(0) ? 11 ? f(0) + f(–1) ? 20
9. If y = mx + 4 is common tangent to parabolas y
2
= 4x and x
2
= 2by. Then value of b is
;fn y = mx + 4 ijoy; y
2
= 4x rFkk x
2
= 2by dh mHk;fu”"B Li’'kZ js[kk gS rks b dk eku gS&
(1) –64 (2) –32 (3) –128 (4) 16
Ans. (3)
Sol. y = mx + 4 ……(i)
y
2
= 4x tangent dh Li'kZ js[kk y = mx +
m
a
? y = mx +
m
1
………(ii)
from (i) and (ii) ls
4 =
m
1
? m =
4
1
So line y =
4
1
x + 4 is also tangent to parabola x
2
= 2by, so solve
vr% js[kk y =
4
1
x + 4 ijoy; x
2
= 2by dh Li'kZ js[kk gS vr%
x
2
= 2b ?
?
?
?
?
? ?
4
16 x
? 2x
2
– bx – 16b = 0 ? D = 0 ? b
2
– 4 × 2 × (–16b) = 0
? b
2
+ 32 × 4b = 0
b = –128, b = 0 (not possible vlaHko)
PAGE # 5
10. If ? and ? are the roots of equation (k+1) tan
2
x – ? 2 tanx = 1 ?– k and tan
2
( ?+ ?) = 50. Find value of
? .
;fn ? rFkk ? lehdj.k (k+1) tan
2
x – ? 2 tanx = 1 ?– k ds ewy gS rFkk tan
2
( ?+ ?) = 50 rks ? ?dk eku gS&
(1) 10 (2) 5 (3) 7 (4) 12
Ans. (1)
(k+1) tan
2
x – ? 2 tanx + (k–1)=0
tan ? ?+ tan ? ?=
1 k
2
?
?
tan ? ?tan ? =
1 k
1 k
?
?
tan ( ?+ ?)= (k–1)=0
2
2
2
1 k
1 k
1
1 k
2
?
?
?
?
?
?
?
?
?
tan
2
( ?+ ?) = 50
2
2
?
?
? ?= 10
11. Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2)
fcUnqvksa (2, 1, 0), (6, 3, 3) rFkk (5, 2, 2) dks lekfgr djus okys lery esa fcUnq (2, 1, 6) dk izfrfcEc gS&
(1) (6, 5, –2) (2) (6, –5, 2) (3) (2, –3, 4) (4) (2, –5, 6)
Ans. (1)
Sol. Plane is lery x + y – 2z = 3 ?
1
2 x ?
=
1
1 y ?
=
2
6 z
?
?
=
6
) 3 12 1 2 ( 2 ? ? ? ?
? (x, y, z) = (6, 5, –2)
12. Let ekuk (x)
k
+ (y)
k
= (a)
k
where tgka a, k > 0 and rFkk 0
x
y
dx
dy 3
1
? ?
?
?
?
?
?
? , then find rc k cjkcj gS&
(1)
3
1
(2)
3
2
(3)
3
4
(1) 2
Ans. (2)
Sol. k.x
k – 1
+ k.y
k – 1
dx
dy
= 0
1 k
y
x
dx
dy
?
?
?
?
?
?
?
?
?
? ?
1 k
y
x
dx
dy
?
?
?
?
?
?
?
?
?
? =0
k – 1 =
3
1
?
k = 1
3
1
? =
3
2
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