Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given $x+|y|=2y$ $$\begin{array}{rl}& \Rightarrow x+y=2y\text{ or }x-y=2y\\ & \Rightarrow x=y\text{ or }x=3y\end{array}$$ This represent a straight line which passes through origin. Hence, $x+|y|=2y$ is continuous at $x=0$. Now, we check differentiability at $x=0$ $$\begin{array}{c}x+|y|=2y\Rightarrow x+y=2y,y\ge 0\\ x-y=2y,y<0\end{array}$$ Thus, $f(x)=\left\{\begin{array}{ll}x,& y<0\\ x/3,& y\ge 0\end{array}\right\}$ $$\begin{array}{rl}& \text{ Now, L.H.D. }=\underset{h\to 0^{-}}{\lim }\frac{f(x+h-)fx}{-h}()\\ & =\underset{h\to 0^{-}}{\lim }\frac{x+h-x}{-h}=-1\\ & \text{ R.H.D }=\underset{h\to 0^{+}}{\lim }\frac{f(x+h-)fx}{h}()\\ & =\underset{h\to 0^{+}}{\lim }\frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\underset{h\to 0^{+}}{\lim }\frac{1}{3}=\frac{1}{3}\\ & \end{array}$$ Since, L.H.D $\ne$ R.H.D. at $x=0$ $\therefore$ given function is not differentiable at $x=0$