Question

If ${\sin }^{-1}x+|y|=2y$, then $y$ as a function of $x$ is

• A. defined for $-1\le x\le 1$
• B. continuous at $x=0$
• C. differentiable for all $x$ in $(-1,1)$
• D. such that $\frac{dy}{dx}=\frac{1}{3\sqrt{1-x^2}}$ for $-1<x<0$

Answer 1

Answer: (A), (B), (D)

such that $\frac{dy}{dx}=\frac{1}{3\sqrt{1-x^2}}$ for $-1<x<0$

${\sin }^{-1}x+|y|=2y$
$\Rightarrow 2y-|y|={\sin }^{-1}x$
Let $y\ge 0$
Then, $y={\sin }^{-1}x$
$y\ge 0\Rightarrow x\in [0,1]$

Let $y<0$
Then, ${\sin }^{-1}x=3y$
$\Rightarrow y=\frac{1}{3}{\sin }^{-1}x$
$y<0\Rightarrow x\in [-1,0)$
$y=\{\begin{array}{c}\frac{1}{3}{\sin }^{-1}x,-1\le x<0\\ {\sin }^{-1}x,0\le x\le 1\end{array}$

$\Rightarrow \frac{dy}{dx}=\{\begin{array}{c}\frac{1}{3}\frac{1}{\sqrt{1-x^2}},-1\le x<0\\ \frac{1}{\sqrt{1-x^2}},0\le x\le 1\end{array}$
Function is continuous but not differentiable at $x=0$