1. Tardigrade
  2. Question
  3. Mathematics
  4. Let f: R → R be a differentiable function satisfying f '(3) + f'(2) = 0. Then displaystyle limx→0 ((1+f(3+x)-f(3)/1+f(2-x)-f(2)))(1/x) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Let $f : R \to R$ be a differentiable function satisfying $f '(3) + f'(2) = 0$.
Then $\displaystyle\lim_{x\to0} \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

JEE MainJEE Main 2019Limits and Derivatives

Solution:

$\lim_{x\to0} \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\to0} \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\to0} \frac{f;\left(3+x\right)+f'\left(2-x\right)}{f'\left(2-x\right)+\left(1+f\left(2-x\right)-f\left(2\right)\right)}} $
$ \Rightarrow e^{\frac{f'\left(3\right)+f'\left(2\right)}{1}} = 1 $