Question

Let $f:R\to R$ be a differentiable function satisfying $f^{\prime }(3)+f^{\prime }(2)=0$.
Then $\underset{x\to 0}{\lim }{\left(\frac{1+f\left(3+x\right)-f\left(3\right)}{1+f\left(2-x\right)-f\left(2\right)}\right)}^{\frac{1}{x}}$ is equal to

• A. $e^2$
• B. e
• C. $e^{-1}$
• D. 1

Answer 1

Answer: (D)

${\lim }_{x\to 0}{\left(\frac{1+f\left(3+x\right)-f\left(3\right)}{1+f\left(2-x\right)-f\left(2\right)}\right)}^{\frac{1}{x}}\left(1^{\infty }\right)$
$\Rightarrow e^{{\lim }_{x\to 0}\frac{f\left(3+x\right)-f\left(2-x\right)-f\left(3\right)+f\left(2\right)}{x\left(1+f\left(2-x\right)-f\left(2\right)\right)}}$
$\Rightarrow e^{{\lim }_{x\to 0}\frac{f;\left(3+x\right)+f^{\prime }\left(2-x\right)}{f^{\prime }\left(2-x\right)+\left(1+f\left(2-x\right)-f\left(2\right)\right)}}$
$\Rightarrow e^{\frac{f^{\prime }\left(3\right)+f^{\prime }\left(2\right)}{1}}=1$