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UPSC NDA Math Model Paper 6

Question : 17 of 120

Marks: +1, -0

\lim\limits_{x \to \frac{\pi}{4}} \left[ \frac{1-\sin 2x}{1+\cos 4x} \right] ?

Solution:

\lim\limits_{x \to \frac{\pi}{4}} \left[ \frac{1-\sin 2x}{1+\cos 4x} \right] = \frac{0}{0}

Since this is an indeterminate form, we can apply the L-Hospital Rule as:

\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}

\lim\limits_{x \to \frac{\pi}{4}} \left[ \frac{1-\sin 2x}{1+\cos 4x} \right]

= \lim\limits_{x \to \frac{\pi}{4}} \left[ \frac{-2 \cos 2x}{-4 \sin 4x} \right] = \frac{0}{0}

Since this is also 0/0 form, we again apply the L-Hospital Rule as:

\lim\limits_{x \to \frac{\pi}{4}} \left[ \frac{-2 \cos 2x}{-4 \sin 4x} \right]

= \lim\limits_{x \to \frac{\pi}{4}} \left[ \frac{4 \sin 2x}{-16 \cos 4x} \right] = \frac{1}{4}

Hence, option A is the correct answer.

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