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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 65 of 160

Marks: +1, -0

Find the value of '

k

\frac{d}{dx}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2\cos(4x)}}}}\right\}

= k \sec\left(\frac{x}{2}\right) \tan\left(\frac{x}{2}\right)

$\frac{1}{2}$
2
1
$\frac{1}{8}$

Solution:

Let

y=\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2\cos 4x}}}}

\because 1+\cos 2A = 2\cos^2 A

∴

y=\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4x)}}}}

=\frac{2}{\sqrt{2+\sqrt{2+\sqrt{4\cos^2 2x}}}}=\frac{2}{\sqrt{2+\sqrt{2+2\cos 2x}}}

=\frac{2}{\sqrt{2+\sqrt{2(1+\cos 2x)}}}=\frac{2}{\sqrt{2+\sqrt{4\cos^2 x}}}

=\frac{2}{\sqrt{2+2\cos x}}=\frac{2}{\sqrt{2(1+\cos x)}}

\Rightarrow y=\frac{2}{\sqrt{4\cos^2\frac{x}{2}}}

y=\sec\frac{x}{2}

Differentiate w.r.t. x,

\frac{dy}{dx}=\sec\frac{x}{2}\tan\frac{x}{2}\left(\frac{1}{2}\right)

[using chain rule]

Now, according to the question,

\frac{dy}{dx}=k\sec\frac{x}{2}\tan\frac{x}{2}

\sec\frac{x}{2}\tan\frac{x}{2}\frac{1}{2}=k\sec\frac{x}{2}\tan\frac{x}{2}

∴

k = \frac{1}{2}

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