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

Answer: C

Solution:

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

First put value of $g(x)$ in the given limit, in the option

$\lim\limits_{x \to x_0}\left|\Large \frac{\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x +1)}+e^{x_0}}{{{x-x_0}}} \right|=0$

$\because$ $e^{x_0}+x_0=0$ , so putting its value in the above equation, we get

$\lim\limits_{x \to x_0}\left|\Large \frac{\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x +1)}-{x_0}}{{{x-x_0}}} \right|$

now taking LCM, we get

$\lim\limits_{x \to x_0}\left|\Large \frac{{3xe^x+3x-\alpha e^x-\alpha x}-{3{x_0}e^x}-3{x_0}}{{{3(e^x+1)}({x-x_0})}} \right|$

Now, replacing ${3(e^{x}+1)}$ with ${3(e^{x_0}+1)}$ as it is not taking part in making indeterminate form, so we get

$\lim\limits_{x \to x_0}\left|\Large \frac{{3xe^x+3x-\alpha e^x-\alpha x}-{3{x_0}e^x}-3{x_0}}{{{3(e^{x_0}+1)}({x-x_0})}} \right|$

Because it is in $\frac {0}{0}$ form, therefore applying L’ Hospital Rule, we get

$\lim\limits_{x \to x_0}\left|\Large \frac{{3e^x+3xe^x+3-\alpha e^x-\alpha}-{3{x_0}e^x}-0}{{{3(e^{x_0}+1)}(1)}} \right|$

now, putting $x=x_0$ in the above limit, we get the desired value of the limit

$= \left|\Large \frac{{3e^{x_0}+3x_0 e^{x_0}+3-\alpha e^{x_0}-\alpha}-{3{x_0}e^x}-0}{{{3(e^{x_0}+1)}}} \right|$

$=\left|\Large \frac{{e^{x_0}(3-\alpha )}+(3-\alpha))}{{{3(e^{x_0}+1)}}} \right|$

$= \left|\Large \frac{{(e^{x_0}+1)(3-\alpha )}}{{{3(e^{x_0}+1)}}} \right|$

$= \left|\Large \frac{{(3-\alpha )}}{{{3}}} \right|$

$\therefore$ If $\alpha = 2$ then limit $=\frac{1}{3}$

Similarly, If $\alpha = 3$ then limit $=0$

Hence, option (C) is the correct option.

This is the final answer given by IIT JEE Advanced.

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