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MathMedAdv 2025 • Paper 2

Limit Evaluation of Function Involving Exponential Root

Let x0x_0 be the real number such that ex0+x0=0e^{x_0} + x_0 = 0. For a given real number α\alpha, define

g(x)=3xex+3x−αex−αx3(ex+1)g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}

for all real numbers xx.

Then which one of the following statements is TRUE?

Options

A

For α=2\alpha = 2, lim⁡x→x0∣g(x)+ex0x−x0∣=0\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0

B

For α=2\alpha = 2, lim⁡x→x0∣g(x)+ex0x−x0∣=1\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1

C

For α=3\alpha = 3, lim⁡x→x0∣g(x)+ex0x−x0∣=0\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0

Correct
D

For α=3\alpha = 3, lim⁡x→x0∣g(x)+ex0x−x0∣=23\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}

Topics & Concepts

Limits and DerivativesLimitsDifferentiation

Step-by-Step Solution

To determine which of the given statements is true, we first analyze the function g(x)g(x) and the given condition.

We are given that x0x_0 is the real root of the equation: ex0+x0=0  ⟹  ex0=−x0e^{x_0} + x_0 = 0 \implies e^{x_0} = -x_0

The function g(x)g(x) is defined as: g(x)=3xex+3x−αex−αx3(ex+1)g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}

We can rewrite the numerator by factoring terms: 3xex+3x−αex−αx=3x(ex+1)−α(ex+x)3xe^x + 3x - \alpha e^x - \alpha x = 3x(e^x + 1) - \alpha(e^x + x)

Thus, g(x)g(x) simplifies to: g(x)=3x(ex+1)−α(ex+x)3(ex+1)=x−α3(ex+xex+1)g(x) = \frac{3x(e^x + 1) - \alpha(e^x + x)}{3(e^x + 1)} = x - \frac{\alpha}{3}\left(\frac{e^x + x}{e^x + 1}\right)

Evaluating g(x)g(x) at x=x0x = x_0: g(x0)=x0−α3(ex0+x0ex0+1)g(x_0) = x_0 - \frac{\alpha}{3}\left(\frac{e^{x_0} + x_0}{e^{x_0} + 1}\right)

Since ex0+x0=0e^{x_0} + x_0 = 0, we have: g(x0)=x0−α3(0)=x0g(x_0) = x_0 - \frac{\alpha}{3}(0) = x_0

Since x0=−ex0x_0 = -e^{x_0}, we have g(x0)=−ex0g(x_0) = -e^{x_0}, which gives: g(x0)+ex0=0  ⟹  ex0=−g(x0)g(x_0) + e^{x_0} = 0 \implies e^{x_0} = -g(x_0)

Now, the given limit can be recognized as the absolute value of the derivative of g(x)g(x) at x=x0x = x_0: lim⁡x→x0∣g(x)+ex0x−x0∣=lim⁡x→x0∣g(x)−g(x0)x−x0∣=∣g′(x0)∣\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \lim_{x \to x_0} \left| \frac{g(x) - g(x_0)}{x - x_0} \right| = |g'(x_0)|

Next, we differentiate g(x)g(x) with respect to xx: g′(x)=1−α3⋅ddx(ex+xex+1)g'(x) = 1 - \frac{\alpha}{3} \cdot \frac{d}{dx}\left(\frac{e^x + x}{e^x + 1}\right)

Using the quotient rule: ddx(ex+xex+1)=(ex+1)ddx(ex+x)−(ex+x)ddx(ex+1)(ex+1)2\frac{d}{dx}\left(\frac{e^x + x}{e^x + 1}\right) = \frac{(e^x + 1)\frac{d}{dx}(e^x + x) - (e^x + x)\frac{d}{dx}(e^x + 1)}{(e^x + 1)^2} =(ex+1)(ex+1)−(ex+x)ex(ex+1)2= \frac{(e^x + 1)(e^x + 1) - (e^x + x)e^x}{(e^x + 1)^2}

Evaluating this derivative at x=x0x = x_0 using ex0+x0=0e^{x_0} + x_0 = 0: ddx(ex+xex+1)∣x=x0=(ex0+1)2−0⋅ex0(ex0+1)2=1\left. \frac{d}{dx}\left(\frac{e^x + x}{e^x + 1}\right) \right|_{x = x_0} = \frac{(e^{x_0} + 1)^2 - 0 \cdot e^{x_0}}{(e^{x_0} + 1)^2} = 1

Therefore: g′(x0)=1−α3(1)=1−α3g'(x_0) = 1 - \frac{\alpha}{3}(1) = 1 - \frac{\alpha}{3}

Hence, the limit is: lim⁡x→x0∣g(x)+ex0x−x0∣=∣1−α3∣\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \left| 1 - \frac{\alpha}{3} \right|

Evaluating for the values of α\alpha given in the options:

  1. For α=2\alpha = 2: ∣1−23∣=13\left| 1 - \frac{2}{3} \right| = \frac{1}{3}
  2. For α=3\alpha = 3: ∣1−33∣=0\left| 1 - \frac{3}{3} \right| = 0

Therefore, the statement in option (C) is TRUE.

Limit Evaluation of Function Involving Exponential Root | Mathematics PYQ Solution - JEE Challenger