Let $f$ be a continuous function satisfying $\int_{0}^{\mathrm{t}^{2}}\left(\mathrm{f}(\mathrm{x})+\mathrm{x}^{2}\right) \mathrm{dx}=\frac{4}{3} \mathrm{t}^{3}, \forall \mathrm{t}>0$. Then $\mathrm{f}\left(\frac{\pi^{2}}{4}\right)$ is equal to :

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

(B) $-\pi^{2}\left(1+\frac{\pi^{2}}{16}\right)$

(C) $-\pi\left(1+\frac{\pi^{3}}{16}\right)$

(D) $\pi^{2}\left(1-\frac{\pi^{2}}{16}\right)$

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Definite Integrals

Let

f

be a continuous function satisfying

\int_{0}^{t^{2}} (f (x) + x^{2}) dx = \frac{4}{3} t^{3}, \forall t > 0

. Then

f (\frac{π ^{2}}{4})

is equal to :

Answer: (A)

Differentiating

\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}

with respect to

t

gives

(f (t^{2}) + t^{4}) \cdot 2 t = 4 t^{2}

, so

f (t^{2}) = 2 t - t^{4}

. Taking

t = \frac{π}{2}

f (\frac{π ^{2}}{4}) = π - \frac{π ^{4}}{16} = π (1 - \frac{π ^{3}}{16})

. The required value is

π (1 - \frac{π ^{3}}{16})

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option