JEE Advanced 2025 · Paper 2 · Mathematics Q1

MathematicsSingle correct+3 / -1Question 1 of 48

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?

For

\alpha = 2

\lim_{x\to x_0}\left|\frac{g(x) + e^{x_0}}{x - x_0}\right| = 0

For

\alpha = 2

\lim_{x\to x_0}\left|\frac{g(x) + e^{x_0}}{x - x_0}\right| = 1

For

\alpha = 3

\lim_{x\to x_0}\left|\frac{g(x) + e^{x_0}}{x - x_0}\right| = 0

For

\alpha = 3

\lim_{x\to x_0}\left|\frac{g(x) + e^{x_0}}{x - x_0}\right| = \dfrac23

Answer and worked solution

Official answer: C

Simplify g.
$g(x)=x-\frac{\alpha}{3}\cdot\frac{e^x+x}{e^x+1}.$

Value at $x_0$ . Since $e^{x_0}+x_0=0$ , we get $g(x_0)=x_0=-e^{x_0}$ . So the limit is $|g'(x_0)|$ .

Derivative.
$g'(x)=1-\frac\alpha3\cdot\frac{(e^x+1)^2-(e^x+x)e^x}{(e^x+1)^2}\implies g'(x_0)=1-\frac\alpha3.$

JEE Advanced 2025 · Paper 2, Mathematics question 1