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

Given $x+|y|=2 y$x+∣y∣=2y $$\begin{aligned} & \Rightarrow \quad x+y=2 y \text { or } x-y=2 y \\ & \Rightarrow \quad x=y \text { or } x=3 y \end{aligned}$$​⇒x+y=2y or x−y=2y⇒x=y or x=3y​ This represent a straight line which passes through origin. Hence, $x+|y|=2 y$x+∣y∣=2y is continuous at $x=0$x=0. Now, we check differentiability at $x=0$x=0 $$\begin{gathered} x+|y|=2 y \Rightarrow x+y=2 y, y \geq 0 \\ x-y=2 y, y < 0 \end{gathered}$$x+∣y∣=2y⇒x+y=2y,y≥0x−y=2y,y<0​ Thus, $f(x)=\left\{\begin{array}{ll}x, & y < 0 \\ x / 3, & y \geq 0\end{array}\right\}$f(x)={x,x/3,​y<0y≥0​} $$\begin{aligned} & \text { Now, L.H.D. }=\lim _{h \rightarrow 0^{-}} \frac{f(x+h-) f x}{-h}(\quad) \\ & =\lim _{h \rightarrow 0^{-}} \frac{x+h-x}{-h}=-1 \\ & \text { R.H.D }=\lim _{h \rightarrow 0^{+}} \frac{f(x+h-) f x}{h}(\quad) \\ & =\lim _{h \rightarrow 0^{+}} \frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{3}=\frac{1}{3} \\ & \end{aligned}$$​ Now, L.H.D. =h→0−lim​−hf(x+h−)fx​()=h→0−lim​−hx+h−x​=−1 R.H.D =h→0+lim​hf(x+h−)fx​()=h→0+lim​h3x+h​−3x​​=h→0+lim​31​=31​​ Since, L.H.D $\neq$= R.H.D. at $x=0$x=0 $\therefore$∴ given function is not differentiable at $x=0$x=0