JEE Advanced 2025 — Paper 1 — Question 25

$\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_{0}^{x} \frac{1}{1-t^{2}} d t+\beta x \cos x}{x^{3}}$

$=\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2}\left(\frac{1}{1-x^{2}}\right)+\beta \cos x-\beta x \sin x}{3 x^{2}}$

$\frac{=\frac{\alpha}{2}\left(1-x^{2}\right)^{-1}+\beta\left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!} \ldots\right)-\beta x\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} \ldots\right)}{3 x^{2}}$ $=\frac{\frac{\alpha}{2}\left(1+x^{2}+x^{4} \ldots\right)+\beta\left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!} \ldots\right)-\beta\left(x^{2}-\frac{x^{4}}{3!} \ldots\right)}{3 x^{2}}$ $=\frac{\left(\frac{\alpha}{2}+\beta\right)+\mathrm{x}^{2}\left(\frac{\alpha}{2}-\frac{\beta}{2}-\beta\right)+\mathrm{x}^{4}() \ldots .}{3 \mathrm{x}^{2}}=2($ Given $)$

$\therefore \frac{\alpha}{2}+\beta=0$ and $\frac{\alpha-3 \beta}{6}=2$

$\Rightarrow \alpha=-2 \beta$ and $\alpha=12+3 \beta$

$\Rightarrow \beta=-\frac{12}{5}$ and $\alpha=\frac{24}{5}$

$\therefore \alpha+\beta=\frac{12}{5}=2.40$