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Mathematics · Definite Integration

JEE Main 2024 — 9 April, Shift 2 — Question 11

lim⁡x→x2(∫x3(π/2)3(sin⁡(2t1/3)+cos⁡(t1/3))dt(x−π2)2)\lim _{x \rightarrow \frac{x}{2}}\left(\frac{\int_{x^{3}}^{(\pi / 2)^{3}}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) \mathrm{dt}}{\left(x-\frac{\pi}{2}\right)^{2}}\right) is equal to :

  1. Option A:

    9π28\frac{9 \pi^{2}}{8}

    Correct
  2. Option B:

    11π210\frac{11 \pi^{2}}{10}

  3. Option C:

    3π22\frac{3 \pi^{2}}{2}

  4. Option D:

    5π29\frac{5 \pi^{2}}{9}

Answer: A

Step-by-step solution

lim⁡x→π20−{sin⁡(2x)+cos⁡(x)}⋅3x22(x−π2)\lim _{x \rightarrow \frac{\pi}{2}} \frac{0-\{\sin (2 x)+\cos (x)\} \cdot 3 x^{2}}{2\left(x-\frac{\pi}{2}\right)}

=lim⁡x→π2−{2sin⁡xcos⁡x+cos⁡x}3x22(x−π2)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\{2 \sin x \cos x+\cos x\} 3 x^{2}}{2\left(x-\frac{\pi}{2}\right)}

=lim⁡x→π2{2sin⁡xsin⁡(π2−x)2(x−π2)+sin⁡(π2−x)2(π2−x)}3x2=\lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{2 \sin x \sin \left(\frac{\pi}{2}-x\right)}{2\left(x-\frac{\pi}{2}\right)}+\frac{\sin \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}\right\} 3 x^{2}

=(1(1)+12)3(π2)2=\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^{2}

=9π28=\frac{9 \pi^{2}}{8}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Mathematics
Chapter
Definite Integration
Topic
Leibnitz rule & its application in limits

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