Q60 / 90 · JEE Main 2012 (07 May Online) · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

Relevant Questions