Question

If $f\left(x\right)={\cos }^{2}x+{\sec }^{2}x$, then

(a) f(x) < 1 (b) f(x) = 1 (c) 1 < f(x) < 2 (d) f(x) ≥ 2

Answer 1

$$\begin{gathered} f\left(x\right)={\cos }^{2}x+{\sec }^{2}x \\ ={\cos }^{2}x+{\sec }^{2}x-2\cos x\sec x+2\cos x\sec x \\ ={\left(\sec x-\cos x\right)}^2+2 \\ \therefore f\left(x\right)\ge 2\forall x\left[{\left(\sec x-\cos x\right)}^2\ge 0\forall x\right] \end{gathered}$$

Hence, the correct option is answer D.