The range of f(x) = sec left( frac{pi }{4} co | vedclass

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Question

(A) Given the function $f(x) = \sec \left( \frac{\pi}{4} \cos^2 x \right)$.We know that for any real $x$,the range of $\cos^2 x$ is $[0, 1]$.Therefore

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$[1, \sqrt{2}]$

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(A) Given the function $f(x) = \sec \left( \frac{\pi}{4} \cos^2 x ight)$.We know that for any real $x$,the range of $\cos^2 x$ is $[0, 1]$.Therefore,the argument of the secant function,$	heta = \frac{\pi}{4} \cos^2 x$,lies in the interval $[0, \frac{\pi}{4}]$.Since the secant function is strictly increasing on the interval $[0, \frac{\pi}{4}]$,we evaluate the function at the boundaries:At $\cos^2 x = 0$,$f(x) = \sec(0) = 1$.At $\cos^2 x = 1$,$f(x) = \sec \left( \frac{\pi}{4} ight) = \sqrt{2}$.Thus,the range of $f(x)$ is $[1, \sqrt{2}]$.

The range of $f(x) = \sec \left( \frac{\pi }{4} \cos^2 x \right)$ for $-\infty < x < \infty$ is

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