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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 23 of 160

Marks: +1, -0

The solution of the equation

[\sin x+\cos x]^{1+\sin 2x}=2

-\pi \leq x \leq \pi

is

$\frac{\pi}{2}$
$\pi$
$\frac{\pi}{4}$
$\frac{3\pi}{2}$

Solution:

(\sin x+\cos x)^{1+\sin 2x}=2

, where

-\pi \leq x \leq \pi

(\sin x+\cos x)^{(\sin x+\cos x)^{2}}=2

[\because (\sin x+\cos x)^{2}

=\sin^{2}x+\cos^{2}x+2\sin \cos x

=1+\sin 2x]

\because \sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}

\therefore -\sqrt{1^{2}+1^{2}} \leq \sin x+\cos x \leq \sqrt{1^{2}+1^{2}}

\Rightarrow -\sqrt{2} \leq \sin x+\cos x \leq \sqrt{2}

For

-\pi \leq x \leq \pi

\because \sin x+\cos x=-\sqrt{2}

at

x=-\frac{\pi}{4}

\sin x+\cos x=\sqrt{2}

at

x=\frac{\pi}{4}

Now, at

x=\frac{\pi}{4}

We get

(\sqrt{2})^{(\sqrt{2})^{2}}=2

=(\sqrt{2})^{2}=2=2=2

∴ It is a solution.

At

x=-\frac{\pi}{4}

We get

(-\sqrt{2})^{(-\sqrt{2})^{2}}=2

(-\sqrt{2})^{2}=2

\Rightarrow 2=2

It is also a solution.

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