If $\int \frac{1}{\sqrt{2 a x-x^2}} d x=fog(x)+C$, then
Answer & explanation
Correct answer: option 2
We have,
$\int \frac{1}{\sqrt{2 a x-x^2}} d x=\int \frac{1}{\sqrt{a^2-(x-a)^2}} d(x-a)=\sin ^{-1}\left(\frac{x-a}{a}\right)+C$
$\Rightarrow fog(x)+C=\sin ^{-1}\left(\frac{x-a}{a}\right)+C$
$\Rightarrow f(g(x))=\sin ^{-1}\left(\frac{x-a}{a}\right)$
$\Rightarrow f(x)=\sin ^{-1} x \text { and } g(x)=\frac{x-a}{a}$