JEE Advanced Model Paper 2 with solutions for online practice

I. Lf' (0) =

Lt

x→0

f(0−h)−f(0)

−h

=

Lt

h→0

−h(e 1 x −e 1 h )

−h(e 1 x +e 1 h )

=

Lt

h→0

e− 2 h −1

e− 2 h +1

= -1
Rf' (0) =

Lt

h→0

f(0−h)−f(0)

−h

=

Lt

h→0

h(e 1 h −e− 1 h )

h(e 1 h +e− 1 h )

=

Lt

h→0

(1−e− 2 h )

(1+e− 2 h )

= 1
Since Lf'(0) ≠ Rf'(0)
∴ f is not differentiable at x = 0
But Lf'(0) and Rf'(0) one finite ,
⇒ f is continuous at x = 0
∴ f is continuous every where but not differentiable at x = 0
II . f(x) = exloga = elogeax = ah
Lf'(0) =

Lt

h→0

f(0−h)−f(0)

−h

=

Lt

h→0

ah−1

−h

= logea
Rf'(0) =

Lt

h→0

F(0+h)−f(0)

h

=

Lt

h→0

ah−1

h

= logea
Since Lf'(0) = Rf'(0)
∴ f(x) is differentiable at x = 0
∴ energy differentiable function is continuous
⇒ f(x) is continuous a x = 0
III. LHL =

Lt

h→0

f(2 - h) =

Lt

h→0

[(2−h)2] + [−2+h]2 =

Lt

h→0

{37(−2)2} = 7
RHL =

Lt

h→0

f(2 + h) =

Lt

h→0

[(2+h)2] + [−2−h]2 =

Lt

h→0

{4 + (−3)2} = 13
∴ LHL ≠ RHL
∴ f(x) is not continuous at x = 2
So , f(x) is not differentiable at x = 2
IV. f(x) =

1

1+x

Lt

h→0

f(- 1 - h) =

Lt

h→0

1

1−1−h

→ - ∞
⇒ f(x) is not continuous at x = -1
Also

Lt

h→0

f(−1−h)−f(−1)

(−1−h)−(−1)

=

Lt

h→0

− 1 h −1

−h

→ ∞
⇒ f(h) is not differentiable at x = -1