NIMCET 2018 — Mathematics PYQ

Calculation:

By using the definition of modulus function, the given function can be written as: $f(x)=\{\begin{array}{llc}\frac{x}{1+x},& amp;x& gt;0\\ 0,& amp;x=0& \\ \frac{x}{1-x},& amp;x& lt;0\end{array}$

Since the expressions for $f(x)$ change for x>0 and x<0, let us compare the limits of the derivatives as $x\to 0$.

For x>0, $f(x)=\frac{x}{1+x}$.

$\Rightarrow f^{\prime }(x)=x\left[\frac{d}{dx}\left(\frac{1}{1+x}\right)\right]+\left(\frac{d}{dx}x\right)\frac{1}{1+x}$

$\Rightarrow f^{\prime }(x)=x\frac{(-1)}{(1+x{)}^2}+\frac{1}{1+x}$

$\Rightarrow f^{\prime }(x)=\frac{1}{(1+x{)}^2}$

$\Rightarrow {\lim }_{x\to 0^{+}}{f}^{\prime }(x)=1$.

Similarly, for x<0, $f(x)=\frac{x}{1-x}$.

$\Rightarrow {\lim }_{x\to 0^{-}}{f}^{\prime }(x)={\lim }_{x\to 0^{-}}\frac{1}{(1-x{)}^2}=1$.

Since ${\lim }_{x\to 0^{+}}{f}^{\prime }(x)={\lim }_{x\to 0^{-}}{f}^{\prime }(x)=1$, the function $f(x)$ is differentiable at $x=0$, and $f(0)=1$.

Also, ${\lim }_{x\to \infty }{f}^{\prime }(x)={\lim }_{x\to \infty }{f}^{\prime }(x)=0$.

$\therefore$ The function is differentiable in $(-\infty ,\infty )$, i.e. it is differentiable everywhere.