JEE MAIN - Mathematics Hindi (2023 - 10th April Evening Shift - No. 6)
माना $$f$$ एक संतत फलन है तथा $$\int_\limits0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t > 0$$ है। तो $$f\left(\frac{\pi^2}{4}\right)$$ बराबर हे
$$-\pi\left(1+\frac{\pi^3}{16}\right)$$
$$\pi\left(1-\frac{\pi^3}{16}\right)$$
$$-\pi^2\left(1+\frac{\pi^2}{16}\right)$$
$$\pi^2\left(1-\frac{\pi^2}{16}\right)$$
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