Crey | JEE Advanced -2025 Paper-2

# Q1 of 48

$Let x_{0} be the real number such that e^{x_{0}} + x_{0} = 0 .$

$For a given real number α, define$
$g (x) = \frac{3 x e^{x} + 3 x - α e^{x} - α x}{3 (e^{x} + 1)}$

for all real numbers x. Then which one of the following statements is TRUE ?

Options

$For α = 2, lim_{x \to x_{0}} | \frac{g (x) + e^{x_{0}}}{x - x_{0}} | = 0$

$For α = 2, lim_{x \to x_{0}} | \frac{g (x) + e^{x_{0}}}{x - x_{0}} | = 1$

$For α = 3, lim_{x \to x_{0}} | \frac{g (x) + e^{x_{0}}}{x - x_{0}} | = 0$

$For α = 3, lim_{x \to x_{0}} | \frac{g (x) + e^{x_{0}}}{x - x_{0}} | = \frac{2}{3}$

Show Answer

Correct Answer

$For α = 3, lim_{x \to x_{0}} | \frac{g (x) + e^{x_{0}}}{x - x_{0}} | = 0$

Solution

$α = 3$

$lim_{x \to x_{0}} | \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$

$lim_{x \to x_{0}} | \frac{\frac{3 x e^{x} + 3 x - 3 e^{x} - 3 x}{3 (e^{x} + 1)} + e^{x_{0}}}{x - x_{0}} |$

$lim_{x \to x_{0}} | \frac{3 e^{x} (x - x_{0}) - 3 e^{x_{0}} (e^{x} - e^{x_{0}})}{3 (e^{x} + 1) (x - x_{0})} |$

$lim_{x \to x_{0}} | \frac{e^{x} - e^{x_{0}}}{e^{x} + 1} | = 0$