2.Sol:
It has only one common root
i.e., it has only one solution.CONTINUITY & DIFFERENTIABILITY
[ONLINE QUESTIONS]1.Sol: (^2)
( )^11 ; 0
x 1
f x k x
x e
f x( ) is continuous at x 0
i.e., 0 2
(0) lim^11
x x 1
f k
x e
2
0 2 2
(1 2 ) (2 ) .....( 1 ( 1)^1
lim 2!
2 1
2
x x
x x x k
x e
x
Clearly k 3 and f(0) 1
S( , ) R R f t: ( ) (1 e| |t)sin2| |,t t R
f t t e( ) (| | | |t)sin 2 | |t
| | sin 2 0
| | sin 2 0
t
t
t e t t
e t t
'( ) | | sin 2 (| | )(2cos2 )^0
| | sin 2 (| | )( cos2 ) 0
t t
t t
f t e t e t t
e t e t t
Given f t( ) is differentiable
i.e. LHD = RHD at t^0
| | sin(20) (| | e)cos( )
| | sin 2( ) 2cos(0)(| | e e)
0 (| | )2 0 2(| | e)
| | 0
i.e. | | .
S( , ) R and (0, )
Set S is subset of R [0, )
3.Sol:
2 2
lim lim
x x^2
f x f x f
4 2
5 5
k
i.e.,
(^21)
5
k
3
5
k
4.Sol: Given
1 1
2 x y y^55
differentiating with respect to x,
we get
4 6
2 1 5 1 5
5 5
dy
y y
dx
i.e.,
1 1
5 5 2
2
(^10) 4 4
2 1
dy y y y x
dx x
dy 2 x (^2) 1 10y
dx
dy x (^2) 1 5y
dx
again differentiating with respect to x,
we get
2
2
(^22)
2 1 4 10
2 1
d y x dy x
dx dx x
(^222)
2 1 5 1
d y dy
x x x
dx dx
2 2
1 2 25
x d y dyx y
dxdx
i.e.,
CONTINUITY & DIFFERENTIABILITY
[ONLINE QUESTIONS]
1.Sol:
2.Sol:
3.Sol:
4.Sol:
5.Sol:
2 2
1 2 25 0
d y dy
x x y
dxdx
i.e.,1, k 25
k 1 25 24
5.Sol: Given that f x f x f x 3 ' ''
The trick to such questions is to assume a function
that will help us to solve the answer fast.
Let f x ax bx cx d ^3 ^2
f x ax bx c' 3 2 2
and f x ax b'' 6 2
f x f x f x 3 ' ''
(^3) , 0
2
a b c d