Question

The domain of derivative of the function $f\left(x\right)=|{\sin }^{-1}(2x^2-1)|$, is

• A. $(-1,1)$
• B. $(-1,1)\sim \{0,\pm \frac{1}{\sqrt{2}}\}$
• C. $(-1,1)\sim \left\{0\right\}$
• D. $(-1,1)\sim \{\pm \frac{1}{\sqrt{2}}\}$

Answer 1

The correct option is B $(-1,1)\sim \{0,\pm \frac{1}{\sqrt{2}}\}$

$y=|{\sin }^{-1}(2x^2-1)|$

$\therefore \frac{dy}{dx}=\frac{|{\sin }^{-1}(2x^2-1)|}{{\sin }^{-1}(2x^2-1)}\cdot \frac{4x}{|2x|\sqrt{1-x^2}}$
which would exists, if

$|2x|\ne 0,{\sin }^{-1}(2x^2-1)\ne 0$ and $1-x^2>0$

$\Rightarrow x\ne 0,2x^2-1\ne 0$ and $|x|<1$

$\Rightarrow x\ne 0,\pm \frac{1}{\sqrt{2}}$ and $|x|<1$

$\Rightarrow xϵ(-1,1)\sim \{0,\pm \frac{1}{\sqrt{2}}\}$

Hence, (b) is the correct answer.