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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$26$

Vedclass · Answer

(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:

Similar Questions

If $\frac{(\cot A \cot B - 1)}{(\cot B + \cot A)} = x,$ then the value of $x$ is

If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$ where $0 < \alpha, \beta < \frac{\pi}{2}$,then $2\beta$ is equal to:

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

In a $\Delta ABC$,$\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $D$ divides $BC$ internally in the ratio $1:3$,then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module