Question

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

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

Hint:

\(\sec\theta \ge 1\) for \(\theta \in [0, \pi/2)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
\(\cos^2 x \in [0, 1] \rightarrow \frac{\pi}{4}\cos^2 x \in [0, \pi/4]\).
Step 2: Detailed Explanation:
Let \(t = (\pi/4)\cos^2 x\), \(t \in [0, \pi/4]\)
\(\sec t\) is increasing on \([0, \pi/4]\)? Actually \(\sec\) is increasing on \([0, \pi/2)\)
\(\sec(0) = 1\), \(\sec(\pi/4) = \sqrt{2}\)
So range is \([1, \sqrt{2}]\)
Step 3: Final Answer:
\([1, \sqrt{2}]\).