If y = log_2 sin x, then the minimum value of | vedclass

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(D) We know that $\cosh y = \frac{e^y + e^{-y}}{2}$.Let $u = \cosh y = \frac{e^y + e^{-y}}{2}$.To find the minimum value of $u$, we analyze the functi

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$1$

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(D) We know that $\cosh y = \frac{e^y + e^{-y}}{2}$.Let $u = \cosh y = \frac{e^y + e^{-y}}{2}$.To find the minimum value of $u$, we analyze the function $\cosh y$.The function $\cosh y$ is a standard hyperbolic function defined as $\cosh y = \frac{e^y + e^{-y}}{2}$.We know that for any real number $y$, $e^y > 0$ and $e^{-y} > 0$.By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality, $\frac{e^y + e^{-y}}{2} \ge \sqrt{e^y \cdot e^{-y}} = \sqrt{e^0} = 1$.The equality holds when $e^y = e^{-y}$, which implies $e^{2y} = 1$, so $y = 0$.Since $y = \log_2 \sin x$, we check if $y=0$ is possible. $y=0 \implies \log_2 \sin x = 0 \implies \sin x = 2^0 = 1$.Since $\sin x = 1$ is possible for $x = \frac{\pi}{2}$, the value $y=0$ is attainable.Therefore, the minimum value of $\cosh y$ is $1$.

If $y = \log_2 \sin x$, then the minimum value of $\cosh y$ is

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