If alpha leq 2 sin^{-1} x + cos^{-1} x leq be | vedclass

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(C) Given the expression $f(x) = 2 \sin^{-1} x + \cos^{-1} x$. We can rewrite this as $f(x) = \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x)$. Since we kno

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$\alpha = 0, \beta = \pi$

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(C) Given the expression $f(x) = 2 \sin^{-1} x + \cos^{-1} x$. We can rewrite this as $f(x) = \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x)$. Since we know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression becomes $f(x) = \sin^{-1} x + \frac{\pi}{2}$. The range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Adding $\frac{\pi}{2}$ to all parts of the inequality,we get $-\frac{\pi}{2} + \frac{\pi}{2} \leq \sin^{-1} x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$. This simplifies to $0 \leq 2 \sin^{-1} x + \cos^{-1} x \leq \pi$. Comparing this with $\alpha \leq 2 \sin^{-1} x + \cos^{-1} x \leq \beta$,we find $\alpha = 0$ and $\beta = \pi$.

If $\alpha \leq 2 \sin^{-1} x + \cos^{-1} x \leq \beta$,then

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