Let (g circ f)(x) = sin x and (f circ g)(x) = | vedclass

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(D) Given $(g \circ f)(x) = g(f(x)) = \sin x$ and $(f \circ g)(x) = f(g(x)) = (\sin \sqrt{x})^2$. We test the options:$(a)$ If $f(x) = \sin^2 x$ and $

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$f(x) = \sin \sqrt{x}, g(x) = x^2$

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(D) Given $(g \circ f)(x) = g(f(x)) = \sin x$ and $(f \circ g)(x) = f(g(x)) = (\sin \sqrt{x})^2$. We test the options:$(a)$ If $f(x) = \sin^2 x$ and $g(x) = x$,then $f(g(x)) = \sin^2 x$ and $g(f(x)) = \sin^2 x$. This does not match the given conditions.$(b)$ If $f(x) = \sin \sqrt{x}$ and $g(x) = \sqrt{x}$,then $f(g(x)) = \sin \sqrt{\sqrt{x}} = \sin x^{1/4}$ and $g(f(x)) = \sqrt{\sin \sqrt{x}}$. This does not match.$(c)$ If $f(x) = \sin^2 x$ and $g(x) = \sqrt{x}$,then $f(g(x)) = f(\sqrt{x}) = \sin^2 \sqrt{x} = (\sin \sqrt{x})^2$ and $g(f(x)) = g(\sin^2 x) = \sqrt{\sin^2 x} = |\sin x|$. This does not match the condition $g(f(x)) = \sin x$ exactly for all $x$.$(d)$ If $f(x) = \sin \sqrt{x}$ and $g(x) = x^2$,then $f(g(x)) = f(x^2) = \sin \sqrt{x^2} = \sin |x|$ and $g(f(x)) = g(\sin \sqrt{x}) = (\sin \sqrt{x})^2$. Note: The question implies finding the correct pair. Re-evaluating the provided options,none perfectly satisfy both conditions for all $x \in \mathbb{R}$. However,based on standard problem sets,there may be a typo in the question or options. Given the structure,option $(d)$ is the closest intended answer.

Let $(g \circ f)(x) = \sin x$ and $(f \circ g)(x) = (\sin \sqrt{x})^2$. Then,

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