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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 29 January · Morning Shift · Mathematics · Limits

_ x → (π)/(2) ((1)/( ( x - (π)/(2))2) ∫ _ x3 ^ ((π)/(2))3...

lim⁡x→π2\lim_{x \to \dfrac{π}{2}}\left(\dfrac{1}{ \left(x−x -\dfrac{π}{2}\right)2} {∫}_{x3}^{\left(\dfrac{π}{2}\right)3}\cos \left(\dfrac{1}{t3}\right)dt\right) is equal to

Options

  1. A3π8\dfrac{3\pi}{8}
  2. B3π24\dfrac{3{\pi}^{2}}{4}
  3. C3π28\dfrac{3{\pi}^{2}}{8}
  4. D3π4\dfrac{3\pi}{4}

Correct answer

C. 3π28\dfrac{3{\pi}^{2}}{8}

Step-by-step solution

Let, y=lim⁡x→π2∫x3(π2)3cost13dt(x−π2)2y = \lim_{x \to \dfrac{\pi}{2}}\dfrac{{∫}_{{x}^{3}}^{{\left(\dfrac{\pi}{2}\right)}^{3}}cos{t}^{\dfrac{1}{3}}dt}{{\left(x - \dfrac{\pi}{2}\right)}^{2}}

⇒y=lim⁡h→0∫(π2−h)3(π2)3cost13dt(π2−h−π2)2⇒ y = \lim_{h \to 0}\dfrac{{∫}_{{\left(\dfrac{\pi}{2} - h\right)}^{3}}^{{\left(\dfrac{\pi}{2}\right)}^{3}}cos{t}^{\dfrac{1}{3}}dt}{{\left(\dfrac{\pi}{2} - h - \dfrac{\pi}{2}\right)}^{2}}

⇒y=lim⁡h→0(∫(π2−h)3(π2)3cost13dth2)⇒ y = \lim_{h \to 0}\left(\dfrac{{∫}_{{\left(\dfrac{\pi}{2} - h\right)}^{3}}^{{\left(\dfrac{\pi}{2}\right)}^{3}}cos{t}^{\dfrac{1}{3}}dt}{{h}^{2}}\right)

Applying L-Hospital's rule and Newton Leibnitz Theorem we get,

⇒y=lim⁡h→0(0+cos⁡(π2−h)×3×(π2−h)22h)⇒ y = \lim_{h \to 0}\left(\dfrac{0 + \cos \left(\dfrac{\pi}{2} - h\right) \times 3 \times {\left(\dfrac{\pi}{2} - h\right)}^{2}}{2h}\right)

⇒y=lim⁡h→0(sinh×3×(π2−h)22h)⇒ y = \lim_{h \to 0}\left(\dfrac{sinh \times 3 \times {\left(\dfrac{\pi}{2} - h\right)}^{2}}{2h}\right)

We know that, lim⁡h→0(sinhh)=1\lim_{h \to 0}\left(\dfrac{sinh}{h}\right) = 1

⇒y=lim⁡h→0(3×(π2−h)22)⇒ y = \lim_{h \to 0}\left(\dfrac{3 \times {\left(\dfrac{\pi}{2} - h\right)}^{2}}{2}\right)

⇒y=3π28⇒ y = \dfrac{3{\pi}^{2}}{8}

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