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:

For composite functions, always find the range from inside to outside step-by-step.

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 behavior of the argument of the secant function: 

1. The function inside the secant is \( \frac{\pi}{4} \cos^2 x \). We know that the range of \( \cos x \) is \([-1, 1]\), which implies that the range of \( \cos^2 x \) is \([0, 1]\) since it is just the square of \(\cos x\).
2. Multiplying \( \cos^2 x \) by \(\frac{\pi}{4}\), we get \( 0 \leq \frac{\pi}{4} \cos^2 x \leq \frac{\pi}{4} \).
3. The secant function, \( \sec y \), is defined as \( \frac{1}{\cos y} \) and its range is:
   • \([1, \infty)\) when \(-\frac{\pi}{2} < y < \frac{\pi}{2}\).
   • \((-\infty, -1] \cup [1, \infty)\) in general.
4. For values of \(\cos y\) between 0 and 1 inclusive, \(\sec y\) will range between \([1, \infty)\).
5. However, due to the multiplication by \(\frac{\pi}{4}\), the actual argument \( \frac{\pi}{4} \cos^2 x \) varies between \(0\) to \( \frac{\pi}{4} \).
6. Within this interval, the secant will not approach infinity since \(\cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}\). This corresponds to a maximum value of \(\sec\left( \frac{\pi}{4} \right) = \sqrt{2}\).
7. Therefore, the range of the secant function over this interval is \([1, \sqrt{2}]\).

Thus, the range of the function \( f(x) = \sec\left( \frac{\pi}{4} \cos^2 x \right) \) is \([1, \sqrt{2}]\).