Question

Let $S$ be the set of all functions $f:[0,1]\to R$, which are continuous on $[0,1]$ and differentiable on $(0,1)$. Then for every $f$ in $S$, there exists a $c\in (0,1)$, depending on $f$, such that:

• A. $\frac{f\left(1\right)-f\left(c\right)}{1-c}=f^{\prime }\left(c\right)$
• B. $\left|f\right(c)-f(1\left)\right|<\left|f^{\prime }\right(c\left)\right|$
• C. $\left|f\right(c)+f(1\left)\right|<(1+c)\left|f^{\prime }\right(c\left)\right|$
• D. $\left|f\right(c)-f(1\left)\right|<(1-c)\left|f^{\prime }\right(c\left)\right|$

Answer 1

The correct option is C $\left|f\right(c)+f(1\left)\right|<(1+c)\left|f^{\prime }\right(c\left)\right|$

Note : This is a BONUS question, as none of the options are correct.
$S$ is set of all functions.
If we consider a constant function, then option $2,3$ and $4$ are incorrect.
For option 1:
$\frac{f\left(1\right)-f\left(c\right)}{1-c}=f^{\prime }\left(c\right)$
This may not be true for $f\left(x\right)=x^2$
None of the option are correct.