Let x 0 be the real number such that e x 0 + x 0 = 0 . For a given real number , define g ( x ) = 3 x e x + 3 x - e x - x 3 ( e x + 1 ) for all real numbers x . Then which one of the following statements is TRUE? [2025]

Question

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 [2025]

Smart Achievers · Accepted Answer

(3)

$Given that e^{x_{0}} + x_{0} = 0$

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

$= \lim_{h \to 0} \frac{x_{0} + h - \frac{α (e^{x_{0} + h} + x_{0} + h)}{3 (e^{x_{0} + h} + 1)} + e^{x_{0}}}{h}$ $[∵ e^{x_{0}} + x_{0} = 0]$

$= \lim_{h \to 0} 1 - \frac{α (e^{x_{0} + h} - e^{x_{0}} + h)}{3 h (e^{x_{0} + h} + 1)}$

$= \lim_{h \to 0} 1 - \frac{α}{3 (e^{x_{0} + h} + 1)} {\frac{e^{x_{0}} (e^{h} - 1)}{h} + 1}$ $[∵ \lim_{h \to 0} \frac{e^{h} - 1}{h} = 1]$

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

$So for α = 2, required limit is 1 - \frac{2}{3} = \frac{1}{3}$

$for α = 3, required limit is 1 - \frac{3}{3} = 0$

Solution

Q.
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 [2025]

1 For $α = 2$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= 0$

2 For $α = 2$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= 1$

3 For $α = 3$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= 0$

4 For $α = 3$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= \frac{2}{3}$

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Solution

Q. Let x0 be the real number such that ex0+x0=0. For a given real number α, define g(x)=3xex+3x-αex-αx3(ex+1) for all real numbers x. Then which one of the following statements is TRUE [2025]

1 For α=2, limx→x0|g(x)+ex0x-x0|=0

2 For α=2, limx→x0|g(x)+ex0x-x0|=1

3 For α=3, limx→x0|g(x)+ex0x-x0|=0