The range of the real valued function f(x) = | vedclass

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(D) We know that for $x \in [-1, 1]$,$\operatorname{Cos}^{-1}(-x) + \operatorname{Sin}^{-1}(-x) = \frac{\pi}{2}$.Given the function $f(x) = \operatorn

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$\{0, \pi\}$

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(D) We know that for $x \in [-1, 1]$,$\operatorname{Cos}^{-1}(-x) + \operatorname{Sin}^{-1}(-x) = \frac{\pi}{2}$.Given the function $f(x) = \operatorname{Cos}^{-1}(-x) + \operatorname{Sin}^{-1}(-x) + \operatorname{Cosec}^{-1}(x)$,we substitute the identity:$f(x) = \frac{\pi}{2} + \operatorname{Cosec}^{-1}(x)$.The domain of $\operatorname{Cosec}^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$.Case $1$: If $x \in [1, \infty)$,then $x = 1$ is the only value in the domain of $\operatorname{Cos}^{-1}(-x)$ and $\operatorname{Sin}^{-1}(-x)$ (which is $[-1, 1]$). Thus,$x = 1$.$f(1) = \frac{\pi}{2} + \operatorname{Cosec}^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.Case $2$: If $x \in (-\infty, -1]$,then $x = -1$ is the only value in the domain of $\operatorname{Cos}^{-1}(-x)$ and $\operatorname{Sin}^{-1}(-x)$. Thus,$x = -1$.$f(-1) = \frac{\pi}{2} + \operatorname{Cosec}^{-1}(-1) = \frac{\pi}{2} - \frac{\pi}{2} = 0$.Therefore,the range of the function is $\{0, \pi\}$.

The range of the real valued function $f(x) = \operatorname{Cos}^{-1}(-x) + \operatorname{Sin}^{-1}(-x) + \operatorname{Cosec}^{-1}(x)$ is

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