Please solve RD Sharma Class 12 Chapter Functions Exercise 2.3 Question 1 Sub question (v) maths textbook solution.

\begin{aligned} &f \circ g(x)=\sin ^{-1}\left(x^{2}\right) \\ &g \circ f(x)=\left(\sin ^{-1} x\right)^{2} \end{aligned}

Hint :

$\\\text {Domain} (f \circ g)=\{x: x \in \text {domain of g and} \; g(x) \in \text {domain of f}\}\\ \text {Domain of}\; (f \circ g)=\left\{x: x \in R\right. and \left.x^{2} \in[-1,1]\right\}\\ \text {Domain of}\; f \circ g=[-1,1]$

Given : Here given that

$f(x)=\sin ^{-1} x \; \text {and}\; g(x)=x^{2}$

Here, we have to compute $f\; o\; g(x) \; \text {and}\; g\; o\; f(x)$

Solution :

Here we have to compute  $f\; o\; g$ :

\begin{aligned} &\because f \circ g:[-1,1] \in R \\ &f \circ g=f\{g(x)\} \\ &\quad=f\left\{x^{2}\right\} \\ &\quad=\sin ^{-1}\left(x^{2}\right) \end{aligned}

Now, we have to find out $g\; o\; f$:

Clearly, the range of f  is subset of the domain of g.

\begin{aligned} &f \circ f:[-1,1] \rightarrow R \\ &\qquad \begin{aligned} g \circ f(x) &=g\{f(x)\} \\ =g\left \{ \sin^{-1}x \right \} &=\left(\sin ^{-1} x\right)^{2} \end{aligned} \end{aligned}
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