UPSC NDA Math Model Paper 2

By using the definition of modulus function, the given function can be written as: $f(x)=\begin{cases} \frac{x}{1+x}, & x>0 \\ 0, & x=0 \\ \frac{x}{1-x}, & x<0 \end{cases}$ Since the expressions for f(x) change for x > 0 and x < 0, let us compare the limits of the derivatives as x → 0. For $x>0, f(x)=\frac{x}{1+x}$ $\Rightarrow f'(x)=x\left[\frac{d}{dx}\left(\frac{1}{1+x}\right)\right]$ $+\left(\frac{d}{dx}x\right)\frac{1}{1+x}$ $\Rightarrow f'(x)=x\frac{-1}{(1+x)^2}+\frac{1}{1+x}$ $\Rightarrow f'(x)=\frac{1}{(1+x)^2}$ $\Rightarrow \lim\limits_{x\to 0^+} f'(x)=1$ Similarly, for $x<0, f(x)=\frac{x}{1-x}$ $\Rightarrow \lim\limits_{x\to 0^-} f'(x)=\lim\limits_{x\to 0^-}\frac{1}{(1-x)^2}=1$ Since $\lim\limits_{x\to 0^+} f'(x)=\lim\limits_{x\to 0^-} f'(x)=1,$ the function $f(x)$ is differentiable at $x=0,$ and $f(0)=1$ Also, $\lim\limits_{x\to\infty^+} f'(x)$ $=\lim\limits_{x\to\infty^-} f'(x)=0$ ∴ The function is differentiable in (-∞, ∞), i.e. it is differentiable everywhere.