IIT JEE Advanced 2025, Paper-2, (+3, –1); Limits Question

Question

Let $x_0$ be the real number such that $e^{x_0}+x_0=0$ . For a given real number $\alpha$ , define

$g(x)=\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x +1)}$

for all real numbers $x$ .

Then which one of the following statements is TRUE?

(A) For $\alpha = 2$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0$

(B) For $\alpha = 2$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=1$

(C) For $\alpha = 3$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0$

(D) For $\alpha = 3$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=\frac{2}{3}$