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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Mathematics
Chapter
Limits

JEE Advanced 2020 · Paper 2 · Mathematics · Limits

The value of the limit _ x → (π)/(2) 4 2 ( sin 3x + x ) (2...

The value of the limit \lim_{x \to \dfrac{\pi}{2}}\dfrac{4\sqrt{2}\left(\text{ sin }3x +\text{ \sin x }\right)}{\left(2\sin 2xsin\dfrac{3x}{2} + cos\dfrac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2}\cos 2x + cos\dfrac{3x}{2}\right)} is _________

Correct answer

0

Step-by-step solution

lim⁡x→π282sin⁡2x⋅cos⁡xcosx2−cos7x2+cos5x2−(2⋅2cos⁡2x+cos3x2)\lim_{x \to \dfrac{\pi}{2}}\dfrac{8\sqrt{2}\sin 2x \cdot \cos x}{cos\dfrac{x}{2} - cos\dfrac{7x}{2} + cos\dfrac{5x}{2} - \left(\sqrt{2} \cdot 2\cos^{2}x + cos\dfrac{3x}{2}\right)}

=lim⁡x→π282sin⁡2x⋅cos⁡x(cosx2−cos3x2)+(cos5x2−cos7x2)−(22cos⁡2x)= \lim_{x \to \dfrac{\pi}{2}}\dfrac{8\sqrt{2}\sin 2x \cdot \cos x}{\left(cos\dfrac{x}{2} - cos\dfrac{3x}{2}\right) + \left(cos\dfrac{5x}{2} - cos\dfrac{7x}{2}\right) - \left(2\sqrt{2}\cos^{2}x\right)}

=lim⁡x→π2162sin⁡xcos⁡x⋅cos⁡x2sin⁡xsin⁡x2+2sin⁡3xsinx2−22cos⁡2x= \lim_{x \to \dfrac{\pi}{2}}\dfrac{16\sqrt{2}\sin x \cos x \cdot \cos x}{2\sin x \sin\dfrac{x}{2} + 2\sin 3xsin\dfrac{x}{2} - 2\sqrt{2}\cos^{2}x}

=lim⁡x→π2162sin⁡xcos⁡x⋅cos⁡x2sinx2(sin⁡x+sin⁡3x)−22cos⁡2x= \lim_{x \to \dfrac{\pi}{2}}\dfrac{16\sqrt{2}\sin x \cos x \cdot \cos x}{2sin\dfrac{x}{2}\left(\sin x + \sin 3x\right) - 2\sqrt{2}\cos^{2}x}

=lim⁡x→π2162sin⁡xcos⁡2x2sinx2(2sin⁡2x⋅cos⁡x)−22cos⁡2x= \lim_{x \to \dfrac{\pi}{2}}\dfrac{16\sqrt{2}\sin x\cos^{2}x}{2sin\dfrac{x}{2}\left(2\sin 2x \cdot \cos x\right) - 2\sqrt{2}\cos^{2}x}

=lim⁡x→π2162sin⁡x2sinx2.4sin⁡x−22= \lim_{x \to \dfrac{\pi}{2}}\dfrac{16\sqrt{2}\sin x}{2sin\dfrac{x}{2} . 4\sin x - 2\sqrt{2}}

=16242−22=16222=8= \dfrac{16\sqrt{2}}{4\sqrt{2} - 2\sqrt{2}} = \dfrac{16\sqrt{2}}{2\sqrt{2}} = 8.

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