Answer
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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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
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