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JEE Main 20211 SeptemberEvening Shift
Exam
JEE Main
Year
2021
Date
1 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2021 · 1 September · Evening Shift · Mathematics · Definite Integration

Let f: R → R be a continuous function. Then _ x → π / 4 (π)/(4) ∫ _...

Let f:R→Rf: R \to R be a continuous function. Then lim⁡x→π/4π4∫2sec⁡2xf(x)dxx2−π216\lim_{x \to \pi / 4}\dfrac{\dfrac{\pi}{4}{∫}_{2}^{\sec^{2}x}f ( x ) dx}{{x}^{2} - \dfrac{{\pi}^{2}}{16}} is equal to:

Options

  1. Af(2)f ( 2 )
  2. B2f(2)2f ( \sqrt{2} )
  3. C2f(2)2f ( 2 )
  4. D4f(2)4f ( 2 )

Correct answer

C. 2f(2)2f ( 2 )

Step-by-step solution

lim⁡x→π4π4∫2sec2xf(x)dxx2−π216\lim_{x \to \dfrac{\pi}{4}}\dfrac{\dfrac{\pi}{4}{∫}_{2}^{se{c}^{2}x}f\left(x\right) d x}{{x}^{2} - \dfrac{{\pi}^{2}}{16}}

Using L-Hospital's rule, we get

=lim⁡x→π4π4f(sec⁡2x)⋅2⋅sec⁡x⋅sec⁡x⋅tan⁡x−02x= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\dfrac{\pi}{4}f\left(\sec^{2}x\right) \cdot 2 \cdot \sec x \cdot \sec x \cdot \tan x - 0}{2x}

=π4⋅2(2)2⋅(1)⋅f(2)2⋅π4= \dfrac{\dfrac{\pi}{4} \cdot 2 ( \sqrt{2}{)}^{2} \cdot ( 1 ) \cdot f ( 2 )}{2 \cdot \dfrac{\pi}{4}}

=2f(2)= 2f\left(2\right)

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