If M_1 and M_2 are the maximum values of frac | vedclass

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(B) For the first expression,we know that $-\sqrt{11^2 + 60^2} \leq 11 \cos 2x + 60 \sin 2x \leq \sqrt{11^2 + 60^2}$.This simplifies to $-61 \leq 11 \

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(B) For the first expression,we know that $-\sqrt{11^2 + 60^2} \leq 11 \cos 2x + 60 \sin 2x \leq \sqrt{11^2 + 60^2}$.This simplifies to $-61 \leq 11 \cos 2x + 60 \sin 2x \leq 61$.To maximize $\frac{1}{11 \cos 2x + 60 \sin 2x + 69}$,we need to minimize the denominator.The minimum value of the denominator is $69 - 61 = 8$.Thus,$M_1 = \frac{1}{8}$.For the second expression,$3 \cos^2 5x + 4 \sin^2 5x = 3(\cos^2 5x + \sin^2 5x) + \sin^2 5x = 3 + \sin^2 5x$.The maximum value occurs when $\sin^2 5x = 1$,so $M_2 = 3 + 1 = 4$.Therefore,$\frac{M_1}{M_2} = \frac{1/8}{4} = \frac{1}{32}$.

If $M_1$ and $M_2$ are the maximum values of $\frac{1}{11 \cos 2x + 60 \sin 2x + 69}$ and $3 \cos^2 5x + 4 \sin^2 5x$ respectively,then $\frac{M_1}{M_2} = $

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