Let f:R→R be a differentiable function satisfying f'(2)+f'(1)=0 . Then limx→01+f(2+x)-f(2)1+f(1-x)-f(1)1/x is equal to

limx→01+f(2+x)−f(2)1+f(1−x)−f(1)1/x=limx→01+f(2+x)−f(2)−f(1−x)+f(1)1+f(1−x)−f(1)1/x=elimx→01xf(2+x)−f(2)−f(1−x)+f(1)1+f(1−x)−f(1)=elimx→0f(2+x)−f(2)xelimx→0f(1)−f(1−x)x=ef'(2)ef'(1)=ef'(2)+f'(1)=e0=1

Let f:R→R be a differentiable function satisfying f'(2)+f'(1)=0 . Then limx→01+f(2+x)-f(2)1+f(1-x)-f(1)1/x is equal to

limx→01+f(2+x)−f(2)1+f(1−x)−f(1)1/x=limx→01+f(2+x)−f(2)−f(1−x)+f(1)1+f(1−x)−f(1)1/x=elimx→01xf(2+x)−f(2)−f(1−x)+f(1)1+f(1−x)−f(1)=elimx→0f(2+x)−f(2)xelimx→0f(1)−f(1−x)x=ef'(2)ef'(1)=ef'(2)+f'(1)=e0=1

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$Let f : R \to R be a differentiable function satisfying f^{'} (2) + f^{'} (1) = 0 . Then$ $\lim_{x \to 0} {(\frac{1 + f (2 + x) - f (2)}{1 + f (1 - x) - f (1)})}^{1 / x} is equal to$

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$e^{2}$

$e^{3}$

answer is A.

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Detailed Solution

$\begin{matrix} \lim_{x \to 0} {(\frac{1 + f (2 + x) - f (2)}{1 + f (1 - x) - f (1)})}^{1 / x} = \lim_{x \to 0} {(1 + \frac{f (2 + x) - f (2) - f (1 - x) + f (1)}{1 + f (1 - x) - f (1)})}^{1 / x} \\ = e^{\lim_{x \to 0} \frac{1}{x} (\frac{f (2 + x) - f (2) - f (1 - x) + f (1)}{1 + f (1 - x) - f (1)})} \\ = e^{\lim_{x \to 0} (\frac{f (2 + x) - f (2)}{x})} e^{\lim_{x \to 0} (\frac{f (1) - f (1 - x)}{x})} \\ = e^{f' (2)} e^{f' (1)} \\ = e^{f' (2) + f' (1)} \\ = e^{0} \\ = 1 \end{matrix}$