If f(x) = max(sin x, sin^{-1}(cos x)),then | vedclass

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(A) Given,$f(x) = \max(\sin x, \sin^{-1}(\cos x))$.We know that $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x))$.For $x \in [-\pi, \pi]$,this

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$f$ is continuous everywhere

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(A) Given,$f(x) = \max(\sin x, \sin^{-1}(\cos x))$.We know that $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x))$.For $x \in [-\pi, \pi]$,this simplifies to:$\sin^{-1}(\cos x) = \begin{cases} \frac{\pi}{2} - x, & 0 \le x \le \pi \ \frac{\pi}{2} + x, & -\pi \le x Both $\sin x$ and $\sin^{-1}(\cos x)$ are continuous functions on their respective domains. The maximum of two continuous functions is also a continuous function.By plotting the graphs of $y = \sin x$ and $y = \sin^{-1}(\cos x)$,we observe that the two curves intersect at certain points,but there are no breaks or jumps in the resulting function $f(x) = \max(\sin x, \sin^{-1}(\cos x))$.Therefore,$f(x)$ is continuous everywhere.

If $f(x) = \max(\sin x, \sin^{-1}(\cos x))$,then

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