Let $f: R \rightarrow R$ be a function given by $f(x+y)=f(x) f(y)$ for all $x, y \in R$ If $f(x)=1+x g(x) \cdot \log _e 2$, where $\lim\limits _{x \rightarrow 0} g(x)=1$. Then, $f^{\prime}(x)=$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

Let

f: R \rightarrow R

be a function given by

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

for all

x, y \in R

f(x)=1+x g(x) \cdot \log _e 2

, where

\lim\limits _{x \rightarrow 0} g(x)=1

. Then,

f^{\prime}(x)=

View answer and explanation

Correct option: A

\log _e 2^{f(x)}

Option A is correct because using the given condition

f(x)=1+x g(x) \cdot \log _e 2

, where

\lim\limits _{x \rightarrow 0} g(x)=1

, and the functional equation, we find

f^{\prime}(x)=\log _e 2^{f(x)}

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