JEE Advanced 2025 Paper 1 · Q11 · Definite Integrals

Question

Let $\alpha$ and $\beta$ be real numbers such that $$\lim_{x 	o 0} \frac{1}{x^3} \left( \frac{\alpha}{2} \int_{0}^{x} \frac{1}{1 - t^2} \, dt + \beta x \cos(x) ight) = 2.$$ Find $\alpha + \beta$.

SolveKar AI · Accepted Answer

Expand each piece in Taylor series near $x = 0$. $\int_{0}^{x} \frac{1}{1 - t^2} \, dt = x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots$ (since $\frac{1}{1-t^2} = 1 + t^2 + t^4 + \ldots$). So $\frac{\alpha}{2} \cdot \int = \frac{\alpha}{2}\left(x + \frac{x^3}{3} + O(x^5)\right) = \frac{\alpha}{2} x + \frac{\alpha}{6} x^3 + O(x^5)$. $\beta x \cos(x) = \beta x \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots\right) = \beta x - \frac{\beta}{2} x^3 + O(x^5)$. Sum $= \left(\frac{\alpha}{2} + \beta\right) x + \left(\frac{\alpha}{6} - \frac{\beta}{2}\right) x^3 + O(x^5)$. Divide by $x^3$ and take the limit. For the limit to be finite ($= 2$), the $x$ term must vanish: $\frac{\alpha}{2} + \beta = 0$ $\Rightarrow$ $\alpha = -2\beta$. Then the $x^3$ coefficient: $\frac{\alpha}{6} - \frac{\beta}{2} = \frac{-2\beta}{6} - \frac{\beta}{2} = -\frac{\beta}{3} - \frac{\beta}{2} = -\frac{5\beta}{6} = 2$. So $\beta = -\frac{12}{5} = -2.4$ and $\alpha = \frac{24}{5} = 4.8$. $\alpha + \beta = 4.8 - 2.4 = 2.4$.

JEE Advanced 2025 Paper 1 · Q11 · Definite Integrals

Solution