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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Limits

_ x arrow (π)/(2) ( ∫_ x^3 ^ (π / 2)^3 ( (2 t^ 1 / 3 )+ (t^ 1 / 3...

lim⁡x→π2\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right) is equal to

Options

  1. A5π29\frac{5 \pi^2}{9}
  2. B9π28\frac{9 \pi^2}{8}
  3. C11π210\frac{11 \pi^2}{10}
  4. D3π22\frac{3 \pi^2}{2}

Correct answer

B. 9π28\frac{9 \pi^2}{8}

Step-by-step solution

lim⁡x→π20−{sin⁡(2x)+cos⁡(x)}⋅3x22(x−π2)=lim⁡x→π2−{2sin⁡xcos⁡x+cos⁡x}3x22(x−π2)=lim⁡x→π2{2sin⁡xsin⁡(π2−x)2(x−π2)+sin⁡(π2−x)2(π2−x)}3x2=(1(1)+12)3(π2)2=9π28\begin{aligned} & \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 \rightarrow \frac{\pi}{2}} \frac{-\{2 \sin x \cos x+\cos x\} 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\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 \\ & =\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^2 \\ & =\frac{9 \pi^2}{8}\end{aligned}
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