JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

$\lim_{x \to 0} \frac{48}{x^4} \int_{0}^{x} \frac{t^3}{t^6 + 1} dt$ is equal to

Choose the correct answer:

Explanation

Solution

1. Form Recognition ( $\frac{0}{0}$ ):

As $x \to 0$ , the integral $\int_{0}^{0} \dots dt = 0$ and the denominator $x^4 = 0$ .

2. Applying L'Hôpital's Rule & Leibniz Rule:

\lim_{x \to 0} \frac{48 \cdot \frac{d}{dx} \left[ \int_{0}^{x} \frac{t^3}{t^6 + 1} dt \right]}{\frac{d}{dx} [x^4]}

3. Differentiation:

Using the Leibniz Rule for the numerator:

\frac{d}{dx} \int_{0}^{x} \frac{t^3}{t^6 + 1} dt = \frac{x^3}{x^6 + 1}

Substituting this back into the limit:

\lim_{x \to 0} \frac{48 \cdot \left( \frac{x^3}{x^6 + 1} \right)}{4x^3}

4. Simplification:

Cancel $x^3$ from the numerator and denominator:

\lim_{x \to 0} \frac{48}{4(x^6 + 1)}

\lim_{x \to 0} \frac{12}{x^6 + 1}

5. Final Substitution:

Put $x = 0$ :

\frac{12}{0^6 + 1} = 12

Answer: 12

Choose the correct answer:

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