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The range of $f(x)=\sec \left(\frac{\pi}{4} \cos ^2 x\right),-\infty \lt x \lt \infty$ is
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The correct answer is:
$[1, \sqrt{2}]$
$f(x)=\sec \left(\frac{\pi}{4} \cos ^2 x\right)$
We know that, $0 \leq \cos ^2 x \leq 1$ at $\cos x=0, f(x)=1$ and at $\cos x=1, f(x)=\sqrt{2} ; \therefore 1 \leq x \leq \sqrt{2} \Rightarrow$ $x \in[1, \sqrt{2}]$.
We know that, $0 \leq \cos ^2 x \leq 1$ at $\cos x=0, f(x)=1$ and at $\cos x=1, f(x)=\sqrt{2} ; \therefore 1 \leq x \leq \sqrt{2} \Rightarrow$ $x \in[1, \sqrt{2}]$.
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