Question

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

• A. $[1,\sqrt{2}]$
• B. $(1,\infty )$
• C. $[-\sqrt{2},-1]\cup [1,\sqrt{2}]$
• D. $(-\infty ,-1)\cup (1,\infty )$

Answer 1

The correct option is A $[1,\sqrt{2}]$

Given $f\left(x\right)=\sec \left(\frac{\pi }{4}{\cos }^{2}x\right)$
We know that, $0\le {\cos }^{2}x\le 1$
$\Rightarrow 0\le \frac{\pi }{4}{\cos }^{2}x\le \frac{\pi }{4}$
By graph of $y=\sec x$

Clearly, for $y=\sec x$ when $x\in [0,\frac{\pi }{4}],\sec x\in [1,\sqrt{2}]$
$\therefore$ range $\in [1,\sqrt{2}]$