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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Limits

_ x arrow 0 cosec x ( 2 ^2 x+3 x - ^2 x+ x+4 ) is:

\lim _{x \rightarrow 0} \operatorname{cosec} x(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right) is:

Options

  1. A00
  2. B115\frac{1}{\sqrt{15}}
  3. C125\frac{1}{2 \sqrt{5}}
  4. D−125-\frac{1}{2 \sqrt{5}}

Correct answer

D. −125-\frac{1}{2 \sqrt{5}}

Step-by-step solution

 lim⁡x→0cosec⁡(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)lim⁡x→0cos⁡ec⁡x(cos⁡2x+3cos⁡x−sin⁡x−4)(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→01sin⁡x(cos⁡2x+3cos⁡x−4)−sin⁡x(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→0(cos⁡x+4)(cos⁡x−1)−sin⁡xsin⁡x(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→0−2sin⁡2x2(cos⁡x+4)−2sin⁡x2cos⁡x22sin⁡x2cos⁡x2(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→0−(sin⁡x(cos⁡x+4)+cos⁡x2)cos⁡x2(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)−125\begin{aligned} & \text{ } \lim _{x \rightarrow 0} \operatorname{cosec}\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right) \\ & \lim _{x \rightarrow 0} \frac{\cos \operatorname{ec} x\left(\cos ^2 x+3 \cos x-\sin x-4\right)}{\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{1}{\sin x} \frac{\left(\cos ^2 x+3 \cos x-4\right)-\sin x}{\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{\left(\cos ^x+4\right)(\cos x-1)-\sin x}{\sin x\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{-2 \sin ^2 \frac{x}{2}(\cos x+4)-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{-\left(\sin ^x(\cos x+4)+\cos \frac{x}{2}\right)}{\cos \frac{x}{2}\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & -\frac{1}{2 \sqrt{5}}\end{aligned}
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