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

The value of $x$ for the maximum value of $\sqrt{3} \cos x + \sin x$ is .....$^o$

If $A + B = \frac{\pi}{4}$,then $(1 + \tan A)(1 + \tan B) = $

$A$ circular wire of radius $7\,cm$ is cut and bent again into an arc of a circle of radius $12\,cm$. The angle subtended by the arc at the centre is ......$^o$

In a right-angled triangle,the hypotenuse is $2 \sqrt{2}$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

The value of $\csc \frac{\pi}{18} - \sqrt{3} \sec \frac{\pi}{18}$ is a

Vedclass Test Series

Exam Paper Generator

Online Exam Module