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 \(\alpha = 2\), \(\lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0\)
(B) For \(\alpha = 2\), \(\lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=1\)
(C) For \(\alpha = 3\), \(\lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0\)
(D) For \(\alpha = 3\), \(\lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=\frac{2}{3}\)