The domain of the function cos^{-1}left(frac{ | vedclass

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(D) For the function $\cos^{-1}(u)$ to be defined,we must have $-1 \leq u \leq 1$. Thus,$-1 \leq \frac{2 \sin^{-1}(\frac{1}{4x^2-1})}{\pi} \leq 1$. Mu

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$(-\infty, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup \{0\}$

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(D) For the function $\cos^{-1}(u)$ to be defined,we must have $-1 \leq u \leq 1$. Thus,$-1 \leq \frac{2 \sin^{-1}(\frac{1}{4x^2-1})}{\pi} \leq 1$. Multiplying by $\frac{\pi}{2}$,we get $-\frac{\pi}{2} \leq \sin^{-1}(\frac{1}{4x^2-1}) \leq \frac{\pi}{2}$. Since the range of $\sin^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,this inequality is always satisfied for all $y$ in the domain of $\sin^{-1}$,which is $[-1, 1]$. Therefore,we need $-1 \leq \frac{1}{4x^2-1} \leq 1$. Case $1$: $\frac{1}{4x^2-1} \leq 1 \implies \frac{1 - (4x^2-1)}{4x^2-1} \leq 0 \implies \frac{2-4x^2}{4x^2-1} \leq 0 \implies \frac{2(1-2x^2)}{(2x-1)(2x+1)} \geq 0$. This holds for $x \in (-\infty, -\frac{1}{\sqrt{2}}] \cup (-\frac{1}{2}, \frac{1}{2}) \cup [\frac{1}{\sqrt{2}}, \infty)$. Case $2$: $\frac{1}{4x^2-1} \geq -1 \implies \frac{1 + 4x^2 - 1}{4x^2-1} \geq 0 \implies \frac{4x^2}{4x^2-1} \geq 0$. This holds for $x \in (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty) \cup \{0\}$. Taking the intersection of both cases,we get $x \in (-\infty, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup \{0\}$.

The domain of the function $\cos^{-1}\left(\frac{2 \sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\right)$ is

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