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

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

Question

(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

$10$

Vedclass · Answer

(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:

Similar Questions

If $A + B + C = \pi$ and $\cos A = \cos B \cos C,$ then $\tan B \tan C$ is equal to

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

If $\frac{\sin^4 A}{a} + \frac{\cos^4 A}{b} = \frac{1}{a + b},$ then the value of $\frac{\sin^8 A}{a^3} + \frac{\cos^8 A}{b^3}$ is equal to

If $A+B+C=2S$,then $\sin(2S-A)+\sin(2S-B)+\sin(2S-C)-\sin(2S) = $

If $A+B+C=\frac{\pi}{3}$,then $\sin \left(\frac{\pi-6A}{6}\right)+\sin \left(\frac{\pi-6B}{6}\right)+\sin C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module