JEE Advanced 2025 Paper 2 Q01 Mathematics Differentiation & Applications Limits & Continuity Hard

JEE Advanced 2025 Paper 2 · Q01 · Limits & Continuity

Let $x_0$ be the real number such that $e^{x_0} + x_0 = 0$. For a given real number $\alpha$, define $$g(x) = \frac{3x e^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$$ for all real numbers $x$.

Then which one of the following statements is TRUE?

  1. A. For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$
  2. B. For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1$
  3. C. For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$
  4. D. For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Rewrite $g(x) = x - \frac{\alpha}{3} \cdot \frac{e^x + x}{e^x + 1}$. At $x_0$, $e^{x_0} + x_0 = 0$, so the second term vanishes and $g(x_0) = x_0$. Hence $g(x_0) + e^{x_0} = x_0 + e^{x_0} = 0$, so the limit equals $|g'(x_0)|$. Differentiate: $g'(x) = 1 - \frac{\alpha}{3} \cdot \frac{d}{dx}\left[\frac{e^x+x}{e^x+1}\right]$. $\frac{d}{dx}\left[\frac{e^x+x}{e^x+1}\right] = \frac{(e^x+1)(e^x+1) - (e^x+x) e^x}{(e^x+1)^2} = \frac{(e^x+1)^2 - e^x(e^x+x)}{(e^x+1)^2}$. At $x = x_0$ with $e^{x_0} = -x_0$: $(e^{x_0}+1)^2 = (1-x_0)^2$, and $e^{x_0}(e^{x_0}+x_0) = (-x_0)(0) = 0$. So the bracket $= (1-x_0)^2$ and the derivative of the inner ratio at $x_0$ equals $1$. Therefore $g'(x_0) = 1 - \frac{\alpha}{3}$. For $\alpha = 2$: $|g'(x_0)| = \frac{1}{3}$ (neither $0$ nor $1$). For $\alpha = 3$: $|g'(x_0)| = 0$. So (C) is TRUE.

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