JEE Advanced 2025 Paper 2, Mathematics Section 1 Q1: Differentiability Of A Function

JEE Advanced2025Paper 2Mathematics Section 1

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

Solution

Step 1: Simplify $g (x)$ .

Factor the numerator as $3 x (e^{x} + 1) - α (e^{x} + x)$ , giving $g (x) = x - \frac{α}{3} \cdot \frac{e ^{x} + x}{e ^{x} + 1}$ .

Step 2: Use the defining relation.

Since $e^{x_{0}} + x_{0} = 0$ , we have $e^{x_{0}} = - x_{0}$ , so $x + e^{x_{0}} = x - x_{0}$ . Therefore

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

Step 3: Recognise the limit as $∣ g^{'} (x_{0}) ∣$ .

Dividing by $(x - x_{0})$ and taking the limit gives $∣ g^{'} (x_{0}) ∣$ (since the value of $g$ at $x_{0}$ is $- e^{x_{0}}$ , the quotient is a difference-quotient form).

Step 4: Compute $g^{'} (x_{0})$ .

Differentiating, $g^{'} (x) = 1 - \frac{α}{3} \cdot \frac{( e ^{x} + 1 ) ^{2} - ( e ^{x} + x ) e ^{x}}{( e ^{x} + 1 ) ^{2}}$ . At $x = x_{0}$ the term $e^{x_{0}} + x_{0}$ vanishes, so the bracket collapses to $(e^{x_{0}} + 1)^{2}$ , leaving $g^{'} (x_{0}) = 1 - \frac{α}{3}$ .

Step 5: Evaluate for each $α$ .

For $α = 3$ : $g^{'} (x_{0}) = 0$ , hence the limit is $0$ . For $α = 2$ : the limit equals $\frac{1}{3}$ , not $0$ or $1$ .

$∴$ The correct option is (C).

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