JEE Main 2024

Question

Let the set of all $a\in R$ such that the equation $\cos 2x+a\sin x=2a-7$ has a solution be $[p, q]$ and $r=	an 9^\circ-	an 27^\circ-\frac{1}{\cot 63^\circ}+	an 81^\circ$, then $pqr$ is equal to ________. A). 48 B). 24 C). 6 D). 4

Edvaya · Accepted Answer

The correct answer is 48. **Solution:**
Part 1:
$\cos 2x+a\sin x=2a-7$
$1-2\sin^2 x+a\sin x=2a-7$
$2\sin^2 x-a\sin x+2a-8=0$
Let $y=\sin x$, so $y\in[-1, 1]$.
$2y^2-ay+2a-8=0$
$2(y^2-4)-a(y-2)=0$
$(y-2)(2y+4-a)=0$
Since $y=\sin x\in[-1, 1]$, $y-2
e 0$. Therefore, $2y+4-a=0 \Rightarrow y=\frac{a-4}{2}$.
For a valid solution, $-1\le\frac{a-4}{2}\le 1 \Rightarrow -2\le a-4\le 2 \Rightarrow 2\le a\le 6$.
Thus, the interval is $[2, 6]$, which means $p=2, q=6$.

Part 2:
$r=	an 9^\circ-	an 27^\circ-	an 63^\circ+	an 81^\circ$
$r=(	an 9^\circ+\cot 9^\circ)-(	an 27^\circ+\cot 27^\circ)$
Using $	an	heta+\cot	heta=\frac{2}{\sin 2	heta}$:
$r=\frac{2}{\sin 18^\circ}-\frac{2}{\sin 54^\circ}$
We know $\sin 18^\circ=\frac{\sqrt{5}-1}{4}$ and $\sin 54^\circ=\cos 36^\circ=\frac{\sqrt{5}+1}{4}$.
$r=\frac{8}{\sqrt{5}-1}-\frac{8}{\sqrt{5}+1}=8\left[\frac{\sqrt{5}+1-(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}ight]=8\left[\frac{2}{4}ight]=4$.

Finally, $pqr=2 	imes 6 	imes 4 = 48$.

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