Let f : R → R be a differentiable function


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. $e^{2}$

  2. $\mathrm{e}$

  3. $\mathrm{e}^{-1}$

  4. 1

Correct Option: , 4


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

$\Rightarrow \mathrm{e}^{\lim _{x \rightarrow 0} \frac{f(3+\mathrm{x})-f(2-\mathrm{x})-f(3)+f(2)}{\mathrm{x}(1+f(2-\mathrm{x})-f(2))}}$

using L'Hopital

$\Rightarrow \mathrm{e}^{\lim _{\mathrm{x} \rightarrow 0-\mathrm{x}} \frac{f^{\prime}(2-\mathrm{x}+\mathrm{x})+f^{\prime}(2-\mathrm{x})}{(1+f(2-\mathrm{x})-f(2))}}$

$\Rightarrow \mathrm{e}^{\frac{f^{\prime}(3)+f^{\prime}(2)}{1}}=1$


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