Question

If $y={\sin }^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2}),0\le x\le \frac{-1+\sqrt{5}}{2}$ and $\frac{dy}{dx}=\frac{1}{2\sqrt{x(1-x)}}+p$, then $p=$

• A. $0$
• B. $\frac{1}{\sqrt{1-x}}$
• C. ${\sin }^{-1}\sqrt{x}$
• D. $\frac{1}{\sqrt{1-x^2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{1-x^2}}$

$\begin{array}{c}y={\sin }^{-1}\left(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2}\right)\\ y={\sin }^{-1}\left(x\right)+{\sin }^{-1}\left(\sqrt{x}\right)\\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x}}\frac{1}{2\sqrt{x}}\\ \frac{dy}{dx}=\frac{1}{2\sqrt{x\left(1-x\right)}}+\frac{1}{\sqrt{1-x^2}}\\ P=\frac{1}{\sqrt{1-x^2}}\end{array}$