Question

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

 

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Answer 1

(a) 1

Here,

\[f\left( x \right) = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}}\]

\[ \Rightarrow \sqrt{1 - x^2} f\left( x \right) = \sin^{- 1} x\]

\[\text { Diffferentiating w . r . t . x, we get }\]

\[\sqrt{1 - x^2} f^{'} \left( x \right) - \frac{x f\left( x \right)}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{1 - x^2}}\]

\[ \Rightarrow \left( 1 - x^2 \right) f^{'} \left( x \right) - xf\left( x \right) = 1\] 

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