Question

if $y=({\sin }^{-1}x{)}^2+k{\sin }^{-1}x$, then $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$ =2

• A. $0$
• B. $1$
• C. $2$
• D. $y$

Answer 1

The correct option is C $2$

We have,

$y={\left({\sin }^{-1}x\right)}^2+k{\sin }^{-1}x$

On differentiation this equation with respect to x and we get,

$\frac{dy}{dx}=\frac{d}{dx}[{\left({\sin }^{-1}x\right)}^2+k{\sin }^{-1}x]$

$\frac{dy}{dx}=2{\sin }^{-1}x\frac{d}{dx}\left({\sin }^{-1}x\right)+k\frac{d}{dx}\left({\sin }^{-1}x\right)$

$\frac{dy}{dx}=2{\sin }^{-1}x\times \frac{1}{\sqrt{1-x^2}}+k\times \frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{2{\sin }^{-1}x+k}{\sqrt{1-x^2}}$

Again differentiation and we get,

$\frac{d^{2}y}{d{x}^2}=\frac{\sqrt{1-x^2}\frac{d}{dx}(2{\sin }^{-1}x+k)-(2{\sin }^{-1}x+k)\frac{d}{dx}\left(\sqrt{1-x^2}\right)}{{\left(\sqrt{1-x^2}\right)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{\sqrt{1-x^2}(2\times \frac{1}{\sqrt{1-x^2}}+0)-(2{\sin }^{-1}x+k)\frac{1}{2\sqrt{1-x^2}}\times (0-2x)}{1-x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{2+\frac{x(2{\sin }^{-1}x+k)}{\sqrt{1-x^2}}}{1-x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{2\sqrt{1-x^2}+x(2{\sin }^{-1}x+k)}{(1-x^2)\sqrt{1-x^2}}$

Then,

$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$

$=(1-x^2)\frac{2\sqrt{1-x^2}+x(2{\sin }^{-1}x+k)}{(1-x^2)\sqrt{1-x^2}}-x\left(\frac{2{\sin }^{-1}x+k}{\sqrt{1-x^2}}\right)$

$=\frac{2\sqrt{1-x^2}+x(2{\sin }^{-1}x+k)}{\sqrt{1-x^2}}-\frac{x(2{\sin }^{-1}x+k)}{\sqrt{1-x^2}}$

$=\frac{2\sqrt{1-x^2}+x(2{\sin }^{-1}x+k)-x(2{\sin }^{-1}x+k)}{\sqrt{1-x^2}}$

$=\frac{2\sqrt{1-x^2}}{\sqrt{1-x^2}}$

$=2$

Hence, this is the answer.