Question

If $f\left(x\right)=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$, then (1 − x)² f '' (x) − xf(x) =

(a) 1
(b) −1
(c) 0
(d) none of these

Answer 1

(a) 1

Here,
$$\begin{gathered} f\left(x\right)=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}} \\ \Rightarrow \sqrt{1-x^2}f\left(x\right)={\sin }^{-1}x \\ \mathrm{Diffferentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathrm{we}\mathrm{get} \\ \sqrt{1-x^2}{f}^{\text{'}}\left(x\right)-\frac{x\mathit{}f\left(x\right)}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}} \\ \Rightarrow \left(1-x^2\right)f^{\text{'}}\left(x\right)-xf\left(x\right)=1 \end{gathered}$$

DISCLAIMER : In the question instead of (1 − x)² f '' (x) − xf(x)
it should be (1 − x)² f ' (x) − xf(x)