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

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(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

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

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(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

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