Let $x_{0}$ be the real number such that $e^{x_{0}}+x_{0}=0$. For a given real number $\alpha$, define \[ g(x)=\frac{3 x e^{x}+3 x-\alpha e^{x}-\alpha x}{3\left(e^{x}+1\right)} \] for all real numbers $x$. Then which one of the following statements is TRUE?

(C) For $\alpha=3, \lim _{x \rightarrow x_{0}}\left|\frac{g(x)+e^{x_{0}}}{x-x_{0}}\right|=0$

(A) For $\alpha=2, \lim _{x \rightarrow x_{0}}\left|\frac{g(x)+e^{x_{0}}}{x-x_{0}}\right|=0$

(B) For $\alpha=2, \lim _{x \rightarrow x_{0}}\left|\frac{g(x)+e^{x_{0}}}{x-x_{0}}\right|=1$

(D) For $\alpha=3, \lim _{x \rightarrow x_{0}}\left|\frac{g(x)+e^{x_{0}}}{x-x_{0}}\right|=\frac{2}{3}$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Continuity and Differentiability

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?

(A)For $α = 2, lim_{x \to x_{0}} \frac{g ( x ) + e ^{x_{0}}}{x - x _{0}} = 0$
(B)For $α = 2, lim_{x \to x_{0}} \frac{g ( x ) + e ^{x_{0}}}{x - x _{0}} = 1$
(C)For $α = 3, lim_{x \to x_{0}} \frac{g ( x ) + e ^{x_{0}}}{x - x _{0}} = 0$ ✓ Correct
(D)For $α = 3, lim_{x \to x_{0}} \frac{g ( x ) + e ^{x_{0}}}{x - x _{0}} = \frac{2}{3}$

Answer: (C)

Given

e^{x_{0}} + x_{0} = 0

, we have

e^{x_{0}} = - x_{0}

.

Rewrite

g (x) = x - \frac{α}{3} (\frac{e ^{x} + x}{e ^{x} + 1})

.

Then

g (x) + e^{x_{0}} = (x - x_{0}) - \frac{α}{3} (\frac{e ^{x} + x}{e ^{x} + 1})

.

Note that

g (x_{0}) + e^{x_{0}} = 0

(can be verified by substitution).

Applying L'Hôpital's rule:

x \to x_{0} lim \frac{g ( x ) + e ^{x_{0}}}{x - x _{0}} = ∣ g^{'} (x_{0}) ∣

Compute

g^{'} (x) = 1 - \frac{α}{3} (\frac{( e ^{x} + 1 ) ^{2} - ( e ^{x} + x ) e ^{x}}{( e ^{x} + 1 ) ^{2}}) = 1 - \frac{α}{3} (\frac{e ^{x} + 1 - x e ^{x}}{( e ^{x} + 1 ) ^{2}})

.

At

x = x_{0}

g^{'} (x_{0}) = 1 - \frac{α}{3}

.

Therefore

lim_{x \to x_{0}} \frac{g ( x ) + e ^{x_{0}}}{x - x _{0}} = 1 - \frac{α}{3}

.

For

α = 3

∣ 1 - 1 ∣ = 0

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option