$I f f (x) = c o s^{2} x + s e c^{2} x, t h e n$

f(x) <1

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f(x) = 1

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1 < f(x) < 2

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f(x) $\geq$ 2

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Solution

The correct option is D
f(x) $\geq$ 2

$S i n c e, {(x - \frac{1}{x})}^{2} \geq 0, \forall x ϵ R, w e h a v e x^{2} + \frac{1}{x^{2}} \geq 2 a n d H e n c e, f (x) = c o s^{2} x + \frac{1}{c o s^{2} x} \geq 2$

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