Q84 / 125 · BITSAT 2015 · Mathematics
Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in \mathbb{R}$
If $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$, then which one of the following is incorrect?
- A$\mathrm{f}(\mathrm{x})$ is continuous, $\forall \mathrm{x} \in \mathrm{R}$
- B$\mathrm{f}^{\prime}(\mathrm{x})$ is constant, $\forall \mathrm{x} \in \mathrm{R}$
- C$\mathrm{f}(\mathrm{x})$ is differentiable, $\forall \mathrm{x} \in \mathrm{R}$
- Correct answerD$\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.
Solution
Let $f(x+y)=f(x)+f(y), \forall x, y \in R$
Put $x=0=y$
$\Rightarrow f(0)=f(0)+f(0)$
$\Rightarrow f(0)=0$
Now, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$f^{\prime \prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)}{h}$
Now, $f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h}$
$$
\begin{array}{l}
\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})}{\mathrm{h}}=\mathrm{f}^{\prime}(0) \\
\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{f}(0)+\mathrm{C}
\end{array}
$$
But $f(0)=0$
$$
\therefore \mathrm{C}=0
$$
Hence, $f(x)=x f(0), \forall x \in R$
Clearly, $\mathrm{f}(\mathrm{x})$ is everywhere continuous and differentiable and $\mathrm{f}^{\prime}(\mathrm{x})$ is constant.
$\forall \mathrm{x} \in \mathrm{R}$
Hence, the correct answer is option: 'D'.