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

To determine the range of the function \(f(x) = \sec\left(\frac{\pi}{4}\cos^2 x\right)\), we need to analyze the expression within the secant function and understand how it affects the range.

1. First, let's consider the inner function \(\frac{\pi}{4}\cos^2 x\):
   • The range of \(\cos^2 x\) is \([0, 1]\), since cosine squared will yield values from 0 to 1.
   • Therefore, \(\frac{\pi}{4}\cos^2 x\) will take values in the range \([0, \frac{\pi}{4}]\).
2. The secant function, \(\sec\theta\), is defined as \(\frac{1}{\cos\theta}\) and tends to have values in two intervals when \(\cos\theta\) is in certain ranges:
   • \(\sec\theta\) is undefined for \(\theta = \frac{\pi}{2} + k\pi\), where \(k\) is an integer.
   • For \(\theta \in [0, \frac{\pi}{4}]\), \(\cos\theta\) is positive and decreases from 1 to \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\).
   • Thus, \(\sec\theta\) will increase from 1 to \(\sqrt{2}\).
3. Hence, the range of \(f(x)\) is \([1, \sqrt{2}]\).

Therefore, the correct option is \([1, \sqrt{2}]\).