Question

# If $f\left( x \right) = {\cos ^2}x + {\sec ^2}x$, then A. $f\left( x \right) < 1$B. $f\left( x \right) = 1$C. $1 < f\left( x \right) < 2$D. $f\left( x \right) \geqslant 2$

Hint: In this question, first of fid the maximum and maximum of ${\cos ^2}\theta$ and ${\sec ^2}\theta$ by using the concept that the values of $\cos \theta$ will always lies between $\left[ { - 1,1} \right]$ and the values of $\sec \theta$ will always between $\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$. Then find out the maximum value of the given function to reach the solution of the given problem.

Given that $f\left( x \right) = {\cos ^2}x + {\sec ^2}x$
We know that the values of $\cos \theta$ will always lies between $\left[ { - 1,1} \right]$
So, the values of ${\cos ^2}\theta$ lies between $\left[ {0,1} \right]$
Hence, the maximum value of ${\cos ^2}\theta$ is 1 and the minimum value of ${\cos ^2}\theta$ is 0.
Also, we know that the values of $\sec \theta$ will always between $\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$
So, the values of ${\sec ^2}\theta$ lies between $\left( {1,\infty } \right)$
Hence, the maximum value of ${\sec ^2}\theta$ is $\infty$ and the minimum value of ${\sec ^2}\theta$ is 1.
Since ${\cos ^2}\theta = \dfrac{1}{{{{\sec }^2}\theta }}$, when ${\cos ^2}\theta$ has its minima ${\sec ^2}\theta$ has its maxima and vice-versa.
Let us consider the maximum value of the given function $f\left( x \right)$
$\Rightarrow f{\left( x \right)_{\max }} = \max {\cos ^2}\theta + \min {\sec ^2}\theta \\ \Rightarrow f{\left( x \right)_{\max }} = 1 + 1 \\ \Rightarrow f{\left( x \right)_{\max }} = 2 \\$
Therefore, $f\left( x \right) \geqslant 2$.
Thus, the correct option is D. $f\left( x \right) \geqslant 2$

Note: As ${\cos ^2}\theta = \dfrac{1}{{{{\sec }^2}\theta }}$, when ${\cos ^2}\theta$ has its minima ${\sec ^2}\theta$ has its maxima and when ${\cos ^2}\theta$ has its maxima ${\sec ^2}\theta$ has its minima. Always remember the range and domain of various trigonometric functions to solve these kinds of questions.