If $y = \cos^{2} x + s e c^{2} x$ , then

$y \leq 2$

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$y \leq 1$

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$y \geq 2$

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$1 < y < 2$

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Solution

The correct option is C
$y \geq 2$

Explanation for the correct option:
Find the value of $y$ :
Given that,
$y = \cos^{2} x + s e c^{2} x$
Consider $\cos^{2} (x)$ and $s e c^{2} x$ as two separate terms
Also, it is known that Arithmetic mean (AM) is greater than or equal to the Geometric mean (GM)
$\begin{array}{rcl} A M & \geq & G M \\ \Rightarrow & \frac{\cos^{2} x + s e c^{2} x}{2} \geq \sqrt{(\cos^{2} x) (s e c^{2} x)} \\ \Rightarrow & \frac{\cos^{2} x + s e c^{2} x}{2} \geq \sqrt{1} \\ \Rightarrow & \cos^{2} x + s e c^{2} x \geq 2 \\ \Rightarrow & y \geq 2 \end{array}$
Hence, the correct option is C.

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