The number of integral value(s) of p for whic | vedclass

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(C) The expression $99 \cos 2	heta - 20 \sin 2	heta$ is of the form $a \cos x + b \sin x$,which lies in the range $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b

Vedclass · Accepted Answer

$10$

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(C) The expression $99 \cos 2	heta - 20 \sin 2	heta$ is of the form $a \cos x + b \sin x$,which lies in the range $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 99$ and $b = -20$.$\sqrt{99^2 + (-20)^2} = \sqrt{9801 + 400} = \sqrt{10201} = 101$.Thus,$-101 \leq 99 \cos 2	heta - 20 \sin 2	heta \leq 101$.For the equation to have a solution,we must have $-101 \leq 20p + 35 \leq 101$.Subtracting $35$ from all sides: $-136 \leq 20p \leq 66$.Dividing by $20$: $-6.8 \leq p \leq 3.3$.The integral values of $p$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3\}$.The total number of such integral values is $10$.

The number of integral value$(s)$ of $p$ for which the equation $99 \cos 2\theta - 20 \sin 2\theta = 20p + 35$ will have a solution is:

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