JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Step 1: Differentiation

Dono sides ko $t$ ke respect mein differentiate karne ke liye hum Leibniz Rule ka use karenge:

\frac{d}{dt} \left[ \int_{0}^{t^{2}} (f(x) + x^{2})dx \right] = \frac{d}{dt} \left( \frac{4}{3}t^{3} \right)

Rule apply karne par:

(f(t^{2}) + (t^{2})^{2}) \cdot \frac{d}{dt}(t^{2}) = \frac{4}{3} \cdot 3t^{2}

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

Step 2: Simplify karein

Dono taraf $2t$ se divide karne par ( $t > 0$ ):

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

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

Step 3: Value find karein

Humein $f\left(\frac{\pi^{2}}{4}\right)$ nikaalna hai, toh hum $t^{2} = \frac{\pi^{2}}{4}$ rakhenge, jiska matlab hai $t = \frac{\pi}{2}$ .

f\left(\frac{\pi^{2}}{4}\right) = 2\left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^{4}

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

Common factor $\pi$ lene par:

f\left(\frac{\pi^{2}}{4}\right) = \pi \left( 1 - \frac{\pi^{3}}{16} \right)

Choose the correct answer:

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