Question

If $f\left(\theta \right)=2\left({\sec }^2\theta +{\cos }^2\theta \right)$, then its value always

• A. $f\left(\theta \right)<2$
• B. $f\left(\theta \right)=2$
• C. $4>f\left(\theta \right)>2$
• D. $f\left(\theta \right)\ge 4$

Answer 1

Answer: (D)

Given, $f(\theta )=2\left({\sec }^2\theta +{\cos }^2\theta \right)$ ...(i)
Let $a=2{\sec }^2\theta$ and $b=2{\cos }^2\theta$
We know that, $\frac{a+b}{2}\ge \sqrt{ab}$
$2{\sec }^2\theta +2{\cos }^2\theta \ge 2\sqrt{2{\sec }^2\theta \cdot 2{\cos }^2\theta }$
$\Rightarrow 2\left({\sec }^2\theta +{\cos }^2\theta \right)\ge 2\sqrt{4}$
$\Rightarrow 2\left({\sec }^2\theta +{\cos }^2\theta \right)\ge 4$
$f(\theta )\ge 4$ [from Eq (i)]