QJEEAdv20Ma61c7: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Ma61c7 Solution: Let f : R R and g : R R be functions satisfying f(x + y) = f(x) …

Question

Verified Solution • 2025–26 Session moderate

Let f : R

\to

R and g : R

\to

R be functions
satisfying f(x + y) = f(x) + f(y) + f(x)f(y)
and f(x) = xg(x) for all x, y

\in

R.
If

\mathop {\lim }\limits_{x \to 0} g(x) = 1

, then which of the following statements is/are TRUE?

Accepted Answer

The given function f : R

\to

R is satisfying as

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

So,

f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}

= \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) + f(x)f(h) - f(x)} \over h}

= \mathop {\lim }\limits_{h \to 0} {{f(h)} \over h}(1 + f(x))

\because

f(x) = xg(x) \Rightarrow g(x) = {{f(x)} \over x}

\therefore

\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} g(x) = 1

(given)

\therefore

f'(x) = 1 + f(x)) \Rightarrow {{f'(x)} \over {1 + f(x)}} = 1

\Rightarrow {\log _e}(1 + f(x)) = x + c

\Rightarrow 1 + f(x) = {e^{x + c}}

\Rightarrow f(x) = {e^{x + c}} - 1

\because

f(0) = 0 \Rightarrow c = 0

Therefore,

f(x) = {e^x} - 1

is differentiable at every

x \in R

.

And

f'(x) = {e^x} \Rightarrow f'(0) = 1

Now,

g(x) = f{{(x)} \over x} = {{{e^x} - 1} \over x}

and if g(0) = 1

LHD (at x = 0) of

g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 - h) - g(0)} \over { - h}}

= \mathop {\lim }\limits_{h \to 0} {{{{{e^{ - h}} - 1} \over { - h}} - 1} \over { - h}}

= \mathop {\lim }\limits_{h \to 0} {{{e^{ - h}} - 1 + h} \over {{h^2}}} = {1 \over 2}

and, RHD (at x = 0) of

g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 + h) - g(0)} \over h}

= \mathop {\lim }\limits_{h \to 0} {{{e^h} - 1 - h} \over {{h^2}}} = {1 \over 2}

So, if g(0) = 1, then g is differentiable at every x

\in

R.

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