What is {d{1-sin2x}}{dx} equal to, where π /4

Correct Answer is : cos x + sin x

Correct Answer is : cos x + sin x

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UPSC NDA II 2018 Mathematics Solved Paper

Question : 89 of 120

Marks: +1, -0

What is

\frac{d\sqrt{1-sin2x}}{dx}

\frac{\pi }{4} < x < \frac{\pi }{2}?

Solution:

Concept:

Simplify

1 - \sin 2x

using the identity

\sin^2 x + \cos^2 x = 1

and

\sin 2x = 2\sin x \cos x

, then evaluate the square root considering the sign in the given interval.

Explanation:

Given

\frac{d}{dx} \sqrt{1 - \sin 2x}

for

\frac{\pi}{4} < x < \frac{\pi}{2}

Write

1 - \sin 2x = \sin^2 x + \cos^2 x - 2\sin x \cos x = (\sin x - \cos x)^2

Thus

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

For

x

(\frac{\pi}{4}, \frac{\pi}{2})

\sin x > \cos x

, so

\sin x - \cos x > 0

Therefore

\sqrt{(\sin x - \cos x)^2} = \sin x - \cos x

Differentiate:

\frac{\mathrm{d}}{\mathrm{d}x}(\sin x - \cos x) = \cos x + \sin x

Hence the derivative equals

\cos x + \sin x

Answer:

Option A –

\cos x + \sin x

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