Let y=4 sin^2 theta - cos 2 theta. If l and m | vedclass

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

Question

(A) Given $y = 4 \sin^2 	heta - \cos 2 	heta$. Using the identity $\cos 2 	heta = 1 - 2 \sin^2 	heta$,we get: $y = 4 \sin^2 	heta - (1 - 2 \sin^2

Vedclass · Accepted Answer

$lm = \frac{m}{l}$

Vedclass · Answer

(A) Given $y = 4 \sin^2 	heta - \cos 2 	heta$. Using the identity $\cos 2 	heta = 1 - 2 \sin^2 	heta$,we get: $y = 4 \sin^2 	heta - (1 - 2 \sin^2 	heta) = 6 \sin^2 	heta - 1$. Since $0 \leq \sin^2 	heta \leq 1$,we have: $0 \leq 6 \sin^2 	heta \leq 6$. Subtracting $1$ from all parts: $-1 \leq 6 \sin^2 	heta - 1 \leq 5$. Thus,the minimum value $l = -1$ and the maximum value $m = 5$. Now,$lm = (-1)(5) = -5$. Also,$\frac{m}{l} = \frac{5}{-1} = -5$. Therefore,$lm = \frac{m}{l}$.

Let $y=4 \sin^2 \theta - \cos 2 \theta$. If $l$ and $m$ are the minimum and maximum values of $y$ respectively,then

Similar Questions

The maximum value of $e^{(2 + \sqrt{3} \cos x + \sin x)}$ is

The maximum and minimum values of the function $|\sin 4x + 3|$ are

If $A + B + C = \pi ,$ then $\tan ^2 \frac{A}{2} + \tan ^2 \frac{B}{2} + \tan ^2 \frac{C}{2}$ is always

The maximum value of $\sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$ in the interval $\left( 0, \frac{\pi}{2} \right)$ is attained at

The maximum value of the function $f(x) = 3 \sin^{12} x + 4 \cos^{16} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module