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The larger of $\cos (\log \theta)$ and $\log (\cos \theta)$, if $e^{-\pi / 2} < \theta < \pi / 2$ is
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Verified Answer
The correct answer is:
$\cos (\log \theta)$
$$
\begin{aligned}
& \text { } \cos (\log \theta), \theta \in\left(e^{\frac{-\pi}{2}}, \frac{\pi}{2}\right) \\
& e^{-\frac{\pi}{2}} < \theta < \frac{\pi}{2} \\
& \Rightarrow \quad-\frac{\pi}{2} < \log \theta < \log \frac{\pi}{2} \leq \frac{\pi}{2}
\end{aligned}
$$
So, $\cos (\log \theta) \geq 0$
$$
\log (\cos \theta) \leq 0
$$
$$
0 \leq \cos \theta \leq 1 \text { in } \theta \in\left[e^{-\frac{\pi}{2}}, \frac{\pi}{2}\right]
$$
$$
\therefore \quad \log (\cos \theta) \leq 0
$$
Hence, $\cos (\log \theta) \geq \log (\cos \theta)$.
\begin{aligned}
& \text { } \cos (\log \theta), \theta \in\left(e^{\frac{-\pi}{2}}, \frac{\pi}{2}\right) \\
& e^{-\frac{\pi}{2}} < \theta < \frac{\pi}{2} \\
& \Rightarrow \quad-\frac{\pi}{2} < \log \theta < \log \frac{\pi}{2} \leq \frac{\pi}{2}
\end{aligned}
$$
So, $\cos (\log \theta) \geq 0$
$$
\log (\cos \theta) \leq 0
$$
$$
0 \leq \cos \theta \leq 1 \text { in } \theta \in\left[e^{-\frac{\pi}{2}}, \frac{\pi}{2}\right]
$$
$$
\therefore \quad \log (\cos \theta) \leq 0
$$
Hence, $\cos (\log \theta) \geq \log (\cos \theta)$.
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