If g(f(x)) = |sin x| and f(g(x)) = (sin sqrt{ | vedclass

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(A) Given $g(f(x)) = |\sin x| = \sqrt{\sin^2 x}$.Also,$f(g(x)) = \sin^2 \sqrt{x}$.Comparing $g(f(x)) = \sqrt{\sin^2 x}$ with $g(f(x))$,we can infer $f

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

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(A) Given $g(f(x)) = |\sin x| = \sqrt{\sin^2 x}$.Also,$f(g(x)) = \sin^2 \sqrt{x}$.Comparing $g(f(x)) = \sqrt{\sin^2 x}$ with $g(f(x))$,we can infer $f(x) = \sin^2 x$ and $g(x) = \sqrt{x}$.Let us verify this:If $f(x) = \sin^2 x$ and $g(x) = \sqrt{x}$,then $g(f(x)) = g(\sin^2 x) = \sqrt{\sin^2 x} = |\sin x|$.And $f(g(x)) = f(\sqrt{x}) = \sin^2(\sqrt{x}) = (\sin \sqrt{x})^2$.Both conditions are satisfied.Therefore,the correct option is $A$.

If $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$,then

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