Question
If $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x},$ then:
-
$\text{f(x)}<1$
-
$\text{f(x)}=1$
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$2<\text{f(x)}<1$
-
$\text{f(x)}\geq2$
[Hint: $\text{A.M}\geq\text{G.M.}$]
If $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x},$ then:
$\text{f(x)}<1$
$\text{f(x)}=1$
$2<\text{f(x)}<1$
$\text{f(x)}\geq2$
[Hint: $\text{A.M}\geq\text{G.M.}$]
Solution:
Given that; $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x}$
We know that $\text{AM}\geq\text{GM}$
$\Rightarrow\frac{\cos^2\text{x}+\sec^2\text{x}}{2}\geq\sqrt{\cos^2\text{x}.\sec^2\text{x}}$
$\Rightarrow\frac{\cos^2\text{x}+\sec^2\text{x}}{2}\geq1$ $\big[\text{since}\sec\theta=\frac{1}{\cos\theta}\big]$
$\Rightarrow\cos^2\text{x}+\sec^2\text{x}\geq2$
$\Rightarrow\text{f(x)}\geq2$
Hence, the correct option is (d)
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