JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 6)
Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then
$$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to :
$$-\pi\left(1+\frac{\pi^{3}}{16}\right)$$
$$\pi\left(1-\frac{\pi^{3}}{16}\right)$$
$$-\pi^{2}\left(1+\frac{\pi^{2}}{16}\right)$$
$$\pi^{2}\left(1-\frac{\pi^{2}}{16}\right)$$
Explanation
Given that
$$ \int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0 $$
On differentiating using Leibnitz rule, we get
$$ \begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned} $$
On substituting $\frac{\pi}{2}$ for $t$, we get
$$ f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right) $$
$$ \int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0 $$
On differentiating using Leibnitz rule, we get
$$ \begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned} $$
On substituting $\frac{\pi}{2}$ for $t$, we get
$$ f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right) $$
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