If f(x) = cos2x + sec2x,
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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