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, $f \left(x\right)=\begin{Bmatrix}x,&y<0\\ \frac{x}{3},&y\ge0\end{Bmatrix}$
Now, $L.H.D.=$ $\displaystyle \lim_{h \to 0^-}$$\frac{f \left(x+h-\right)f \,x}{-h} \left(\quad\right)$
$=\displaystyle \lim_{h \to 0^-}$$\frac{x+h-x}{-h} =-1$
$\text{R.H.D}=\displaystyle \lim_{h \to 0^+}$$\frac{f \left(x+h-\right)f \,x}{-h} \left(\quad\right)$
$\displaystyle \lim_{h \to 0^+}$$\frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\displaystyle \lim_{h \to 0^+}$$\frac{1}{3}=\frac{1}{3}$
Since, $L.H.D \ne R.H.D.$ at $x = 0$
$\therefore $ given function is not differentiable at $x=0$