Let f


Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be a differentiable function satisfying

$f^{\prime}(3)+f^{\prime}(2)=0 .$ Then $\lim _{x \rightarrow 0}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac{1}{x}}$ is equal to :

  1. (1) 1

  2. (2) $e^{-1}$

  3. (3) $e$

  4. (4) $e^{2}$

Correct Option: 1


$I=\lim _{x \rightarrow 0}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac{1}{x}} \quad\left[1^{\infty}\right.$ form $]$

$\Rightarrow I=e^{\ell} 1$, where

$I_{1}=\lim _{x \rightarrow 0}\left(\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}-1\right)\right)\left(\frac{1}{x}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\left(\frac{f(3+x)-f(3)-f(2-x)+f(2)}{1+f(2-x)-f(2)}\right)$

$\left(\frac{0}{0}\right.$ form $)$

By L. Hospital Rule,

$I_{1}=\lim _{x \rightarrow 0}\left(\frac{f^{\prime}(3+x)+f^{\prime}(2-x)}{1}\right) \lim _{x \rightarrow 0}\left(\frac{1}{1+f(2-x)-f(2)}\right)$


$\Rightarrow I=e=e=$

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