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JEE Mains 26 July 2022 Shift 1 Paper

Section: Mathematics

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Question : 18 of 90

Marks: +1, -0

Let

S=\{\theta \in [0,2\pi]: 8^{2\sin^2\theta}+8^{2\cos^2\theta}=16\}

n(s)+\sum\limits_{\theta \in S}\left(\sec\left(\frac{\pi}{4}+2\theta\right)\csc\left(\frac{\pi}{4}+2\theta\right)\right)

[26-Jul-2022-Shift-1]

0
$-2$
$-4$
12

Solution: 👈: Video Solution

S=\{\theta \in [0,2\pi]: 8^{2\sin^2\theta}+8^{2\cos^2\theta}=16\}

Now apply AM

\geq

GM for

8^{2\sin^2\theta}, 8^{2\cos^2\theta}

\; \frac{8^{2\sin^2\theta}+8^{2\cos^2\theta}}{2} \geq (8^{2\sin^2\theta+2\cos^2\theta})^{\frac{1}{2}}

8 \geq 8

\Rightarrow 8^{2\sin^2\theta}=8^{2\cos^2\theta}

\; \text{ or } \; \sin^2\theta=\cos^2\theta

\therefore \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}

n(S)+\sum\limits_{\theta \in S}\sec\left(\frac{\pi}{4}+2\theta\right)\csc\left(\frac{\pi}{4}+2\theta\right)

4+\sum\limits_{\theta \in S}\frac{2}{2\sin\left(\frac{\pi}{4}+2\theta\right)\cos\left(\frac{\pi}{4}+2\theta\right)}

=4+\sum\limits_{\theta \in S}\frac{2}{\sin\left(\frac{\pi}{2}+4\theta\right)}=4+2\sum\limits_{\theta \in S}\csc\left(\frac{\pi}{2}+4\theta\right)

=4+2\left[\csc\left(\frac{\pi}{2}+\pi\right)\;\csc\left(\frac{\pi}{2}+3\pi\right)+\csc\left(\frac{\pi}{2}+5\pi\right)+\csc\left(\frac{\pi}{2}+7\pi\right)\right]

=4+2\left[-\csc\;\frac{\pi}{2}-\csc\;\frac{\pi}{2}-\csc\;\frac{\pi}{2}-\csc\;\frac{\pi}{2}\right]

=4-2(4)

=4-8

=-4

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