Minimum value of 8 cos^2 x + 18 sec^2 x for a | vedclass

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(C) Let $y = 8 \cos^2 x + 18 \sec^2 x$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for positive terms:$\frac{8 \cos^2 x + 18 \sec^

Vedclass · Accepted Answer

$26$

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(C) Let $y = 8 \cos^2 x + 18 \sec^2 x$.Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for positive terms:$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq \sqrt{8 \cos^2 x \cdot 18 \sec^2 x}$$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq \sqrt{144 \cdot \cos^2 x \cdot \frac{1}{\cos^2 x}}$$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq \sqrt{144}$$\frac{8 \cos^2 x + 18 \sec^2 x}{2} \geq 12$$8 \cos^2 x + 18 \sec^2 x \geq 24$.However,we must check if the equality holds. The equality in $AM$-$GM$ holds when $8 \cos^2 x = 18 \sec^2 x$.$\cos^4 x = \frac{18}{8} = \frac{9}{4} = 2.25$.Since $\cos^4 x$ cannot exceed $1$,this equality is impossible. Thus,we analyze the function $f(x) = 8 \cos^2 x + 18 \sec^2 x$. Let $t = \cos^2 x$. Then $y = 8t + \frac{18}{t}$,where $t \in (0, 1]$.Let $f(t) = 8t + \frac{18}{t}$. The derivative $f'(t) = 8 - \frac{18}{t^2}$. Setting $f'(t) = 0$ gives $t^2 = \frac{18}{8} = 2.25$,so $t = 1.5$,which is outside the domain $(0, 1]$.Since $f'(t) The minimum value occurs at the largest possible value of $t$,which is $t = 1$ (i.e.,$\cos^2 x = 1$).$y_{min} = 8(1) + \frac{18}{1} = 8 + 18 = 26$.

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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