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Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2020 — Paper 2 — Question 39

The value of the limit lim⁡x→π242(sin⁡3x+sin⁡x)(2sin⁡2xsin⁡3x2+cos⁡5x2)−(2+2cos⁡2x+cos⁡3x2)\lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3 x+\sin x)}{\left(2 \sin 2 x \sin \frac{3 x}{2}+\cos \frac{5 x}{2}\right)-\left(\sqrt{2}+\sqrt{2} \cos 2 x+\cos \frac{3 x}{2}\right)} is

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Answer: 8

Numerical answer — enter this value.

Step-by-step solution

lim⁡x→π2(42⋅2sin⁡2xcos⁡x2sin⁡2xsin⁡3x2+(cos⁡5x2−cos⁡3x2)−2(1+cos⁡2x))⇒lim⁡x→π2(162sin⁡xcos⁡2x8sin⁡xsin⁡x2⋅cos⁡2x−22cos⁡2x)⇒lim⁡x→π2162sin⁡x8sin⁡xsin⁡x2−22=8\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4 \sqrt{2} \cdot 2 \sin 2 x \cos x}{2 \sin 2 x \sin \frac{3 x}{2}+\left(\cos \frac{5 x}{2}-\cos \frac{3 x}{2}\right)-\sqrt{2}(1+\cos 2 x)}\right) \\& \Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{16 \sqrt{2} \sin x \cos ^{2} x}{8 \sin x \sin \frac{x}{2} \cdot \cos ^{2} x-2 \sqrt{2} \cos ^{2} x}\right) \\& \Rightarrow \lim _{x \rightarrow \frac{\pi}{2}} \frac{16 \sqrt{2} \sin x}{8 \sin x \sin \frac{x}{2}-2 \sqrt{2}}=8 \end{aligned}

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Application of L'Hospital rule, series expansion.

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