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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना \(\mathrm{f}\) एक संतत फलन है तथा \(\int_0^{\mathrm{t}^2}\left(\mathrm{f}(\mathrm{x})+\mathrm{x}^2\right) \mathrm{dx}=\frac{4}{3} \mathrm{t}^3, \forall \mathrm{t}>0\) है, तो \(\mathrm{f}\left(\frac{\pi^2}{4}\right)\) बराबर है :

  1. A \(\pi\left(1-\frac{\pi^3}{16}\right)\)
  2. B \(-\pi^2\left(1+\frac{\pi^2}{16}\right)\)
  3. C \(-\pi\left(1+\frac{\pi^3}{16}\right)\)
  4. D \(\pi^2\left(1-\frac{\pi^2}{16}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\pi\left(1-\frac{\pi^3}{16}\right)\)

Step-by-step Solution

Detailed explanation

\(\int \limits_0^{t^2}\left( f ( x )+ x ^2\right) d x=\frac{4}{3} t ^3, \forall t > 0\) \(\left( f \left( t ^2\right)+ t ^4\right)=2 t\) \(f \left( t ^2\right)=2 t - t ^4\) \(t =\frac{\pi}{2} \Rightarrow f \left(\frac{\pi^2}{4}\right)=\frac{2 \pi}{2}-\frac{\pi^4}{16}\)…
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