The minimum value of 8 cos^2 x + 18 sec^2 x f | vedclass

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(C) Let $y = 8 \cos^2 x + 18 \sec^2 x$. Since $8 \cos^2 x$ and $18 \sec^2 x$ are both positive,we can apply the Arithmetic Mean-Geometric Mean ($AM$-$

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(C) Let $y = 8 \cos^2 x + 18 \sec^2 x$. Since $8 \cos^2 x$ and $18 \sec^2 x$ are both positive,we can apply the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality. For positive real numbers $a$ and $b$,$\frac{a+b}{2} \ge \sqrt{ab}$. Here,$a = 8 \cos^2 x$ and $b = 18 \sec^2 x$. $y = 8 \cos^2 x + 18 \sec^2 x \ge 2 \sqrt{(8 \cos^2 x)(18 \sec^2 x)}$. $y \ge 2 \sqrt{8 	imes 18 	imes \cos^2 x 	imes \frac{1}{\cos^2 x}}$. $y \ge 2 \sqrt{144}$. $y \ge 2 	imes 12 = 24$. Wait,checking the equality condition: $8 \cos^2 x = 18 \sec^2 x \implies \cos^4 x = \frac{18}{8} = \frac{9}{4} = 2.25$. Since $\cos^4 x \le 1$,this equality is impossible. Let $t = \cos^2 x$,where $t \in (0, 1]$. Then $y = 8t + \frac{18}{t}$. Let $f(t) = 8t + \frac{18}{t}$. $f'(t) = 8 - \frac{18}{t^2} = 0 \implies t^2 = \frac{18}{8} = \frac{9}{4} \implies t = 1.5$ (not in domain). Since $f'(t) The minimum occurs at the boundary $t = 1$. $y_{min} = 8(1) + \frac{18}{1} = 26$.

The minimum value of $8 \cos^2 x + 18 \sec^2 x$ for all $x \in R$ wherever it is defined,is:

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