MCQ
Let $f$ be a continuous function satisfying $\int \limits_0^{t^2}\left( f ( x )+ x ^2\right) dx =\frac{4}{3} t ^3, \forall t > 0 . \quad$ Then $f \left(\frac{\pi^2}{4}\right)$ equal to :
  • $\pi\left(1-\frac{\pi^3}{16}\right)$
  • B
    $-\pi^2\left(1+\frac{\pi^2}{16}\right)$
  • C
    $-\pi\left(1+\frac{\pi^3}{16}\right)$
  • D
    $\pi^2\left(1-\frac{\pi^2}{16}\right)$

Answer

Correct option: A.
$\pi\left(1-\frac{\pi^3}{16}\right)$
a
$\int \limits_0^{t^2}\left( f ( x )+ x ^2\right) d x=\frac{4}{3} t ^3, \forall t > 0$

$\left( f \left( t ^2\right)+ t ^4\right)=2 t$

$f \left( t ^2\right)=2 t - t ^4$

$t =\frac{\pi}{2} \Rightarrow f \left(\frac{\pi^2}{4}\right)=\frac{2 \pi}{2}-\frac{\pi^4}{16}$

$=\pi-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right)$

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