Question:
If f(x) = cos2x + sec2x, then
A. f(x) < 1
B. f(x) = 1
C. 2 < f(x) < 1
D. f(x) ≥ 2
[Hint: A.M ≥ G.M.]
Solution:
D. f(x) ≥ 2
Explanation:
According to the question,
We have, f(x) = cos2x + sec2x
We know that, A.M ≥ G.M.\
$\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq \sqrt{\cos ^{2} x \sec ^{2} x}$
$\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq \sqrt{\cos ^{2} x \frac{1}{\cos ^{2} x}}$
$\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq 1$
⇒ cos02x + sec2x ≥ 2
⇒ f(x) ≥ 2
Thus, option (D) f(x) ≥ 2 is the correct answer.
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