The range of the function f(x) = sqrt{|sin^{- | vedclass

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(A) Let $g(x) = |\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||$. We know that for any $x \in R$,$|\sin x| \in [0, 1]$ and $|\cos x| \in [0, 1]$. Thus,$\sin

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$\{0\}$

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(A) Let $g(x) = |\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||$. We know that for any $x \in R$,$|\sin x| \in [0, 1]$ and $|\cos x| \in [0, 1]$. Thus,$\sin^{-1}|\sin x| \in [0, \frac{\pi}{2}]$ and $\cos^{-1}|\cos x| \in [0, \frac{\pi}{2}]$. Since both terms are non-negative,we can write $g(x) = \sin^{-1}|\sin x| - \cos^{-1}|\cos x|$. Using the identity $\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$,we have $\sin^{-1}|\sin x| = \frac{\pi}{2} - \cos^{-1}|\sin x|$. Also,$\cos^{-1}|\cos x| = \sin^{-1} \sqrt{1 - \cos^2 x} = \sin^{-1}|\sin x|$. Therefore,$g(x) = \sin^{-1}|\sin x| - \sin^{-1}|\sin x| = 0$. Since $g(x) = 0$ for all $x \in R$,the function $f(x) = \sqrt{0} = 0$. Thus,the range of the function is $\{0\}$.

The range of the function $f(x) = \sqrt{|\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||}$ is

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