Practicing Success
If $4 x+3|y|=5 y$, then y as a function of x is |
differentiable at x = 0 continuous at x = 0 $\frac{d y}{d x}=2$ for all x none of these |
continuous at x = 0 |
We have, $4 x+3|y|=5 y$ $\Rightarrow 4 x+3 y=5 y$, if $y \geq 0$ and, $4 x-3 y=5 y$, if y < 0 $\Rightarrow y= \begin{cases}2 x, & x \geq 0 \\ \frac{x}{2}, & x<0\end{cases}$ Clearly, y is continuous at x = 0. We have, (LHD of y at x = 0) = $\frac{1}{2}$ and, (RHD of y at x = 0) = 2 So, y is not differentiable at x = 0. Also, $\frac{d y}{d x}= \begin{cases}2, & x>0 \\ \frac{1}{2}, & x<0\end{cases}$ |