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Limits Continuity and Differentiability question

2025 · Shift 2 · Q17

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JEE Advanced
Mathematics
Limits Continuity and Differentiability
2025 · Shift 2 · Q17

Limits Continuity and Differentiability question

2025 · Shift 2 · Q17

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMCQ+3 / −1

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?

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

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