JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Continuity and Differentiability

Let

f : R \to R

satisfy the equation

f (x + y) = f (x) \cdot f (y)

for all

x, y \in R

and

f (x) \neq = 0

for any

x \in R

. If the function

f

is differentiable at

x = 0

and

f^{'} (0) = 3

, then

lim_{h \to 0} \frac{1}{h} (f (h) - 1)

is equal to

Answer: 3

Solution

Setting

x = y = 0

f (x + y) = f (x) f (y)

gives

f (0) = f (0)^{2}

, and since

f

vanishes nowhere,

f (0) = 1

. The required limit is then exactly the difference quotient of

f

at the origin:

h \to 0 lim \frac{f ( h ) - 1}{h} = h \to 0 lim \frac{f ( 0 + h ) - f ( 0 )}{h} = f^{'} (0)

and

f

is differentiable there with

f^{'} (0) = 3

, so the limit equals

3

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer