JEE Advanced 2025 — Paper 2 — Question 17

Given $e^{x_0}+x_0=0$, so $e^{x_0} = -x_0$. Simplify $g(x)=\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x+1)} = \frac{3x(e^x+1)-\alpha(e^x+x)}{3(e^x+1)} = x - \frac{\alpha(e^x+x)}{3(e^x+1)}$. Thus $g(x)+e^{x_0} = x + e^{x_0} - \frac{\alpha(e^x+x)}{3(e^x+1)} = (x-x_0) - \frac{\alpha(e^x+x)}{3(e^x+1)}$. As $x\to x_0$, numerator $g(x)+e^{x_0}\to 0$ and denominator $x-x_0\to 0$, so we use L'Hospital's rule. $\lim_{x\to x_0} \left|\frac{g(x)+e^{x_0}}{x-x_0}\right| = \lim_{x\to x_0} \left|g'(x)\right| = |g'(x_0)|$. Compute $g'(x) = 1 - \frac{\alpha}{3} \cdot \frac{(e^x+1)(e^x+1) - (e^x+x)e^x}{(e^x+1)^2} = 1 - \frac{\alpha}{3} \cdot \frac{(e^x+1)^2 - e^x(e^x+x)}{(e^x+1)^2}$. At $x=x_0$, $e^{x_0}+1 = -x_0+1$ and $e^{x_0}+x_0 = 0$, so $(e^{x_0}+1)^2 - e^{x_0}(e^{x_0}+x_0) = (e^{x_0}+1)^2$.

Hence $g'(x_0) = 1 - \frac{\alpha}{3} \cdot \frac{(e^{x_0}+1)^2}{(e^{x_0}+1)^2} = 1 - \frac{\alpha}{3}$. Therefore the limit is $\left|1 - \frac{\alpha}{3}\right|$.

For $\alpha=3$, the limit is 0, which corresponds to option C.