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

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