Question

lf $x+|y|=2y,$ then $y$ as a function of $x,$ at $x=0$ is

• A. differentiable but not continuous
• B. continuous but not differentiable
• C. continuous as well as differentiable
• D. neither continuous nor differentiable

Answer 1

Answer: (B)

Given $x+|y|=2y$
$\Rightarrow x+y=2y$ or $x-y=2y$
$\Rightarrow x=y$ or $x=3y$
This represent a straight line which passes through origin.
Hence, $x+|y|=2y$ is continuous at $x=0$.
Now, we check differentiability at $x=0$
$x+|y|=2y\Rightarrow x+y=2y,y\ge 0$
$x-y=2y9y<0$
Thus, $f\left(x\right)=\left\{\begin{array}{cc}x,& y<0\\ \frac{x}{3},& y\ge 0\end{array}\right\}$
Now, $L.H.D.=$ $\underset{h\to 0^{-}}{\lim }$$\frac{f\left(x+h-\right)fx}{-h}\left(\right)$
$=\underset{h\to 0^{-}}{\lim }$$\frac{x+h-x}{-h}=-1$
$\text{R.H.D}=\underset{h\to 0^{+}}{\lim }$$\frac{f\left(x+h-\right)fx}{-h}\left(\right)$
$\underset{h\to 0^{+}}{\lim }$$\frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\underset{h\to 0^{+}}{\lim }$$\frac{1}{3}=\frac{1}{3}$
Since, $L.H.D\ne R.H.D.$ at $x=0$
$\therefore$ given function is not differentiable at $x=0$