Limit Continuity And Differentiability Ques 35

Answer:

Correct Answer: 35.(a, b, d)

Solution: (a, b, d) Since, $x+|y|=2 y \Rightarrow \begin{cases}x+y=2 y, & \text { when } y>0 \\ x-y=2 y, & \text { when } y<0\end{cases}$

$\Rightarrow \quad\left\{\begin{array}{cc}y=x, & \text { when } y>0 \Rightarrow x>0 \\ y=x / 3, & \text { when } y<0 \Rightarrow x<0\end{array}\right.$

which could be plotted as,

[Image]

Clearly, $y$ is continuous for all $x$ but not differentiable at $x=0$.

Also, $\quad \frac{d y}{d x}=\left\{\begin{array}{cc}1, & x>0 \\ 1 / 3, & x<0\end{array}\right.$

Thus, $f(x)$ is defined for all $x$, continuous at $x=0$, differentiable for all $x \in R-\{0\}, \frac{d y}{d x}=\frac{1}{3}$ for $x<0$.