Suppose y = f(x) is a differentiable function of interval [a, b], such that $f\text{'}\left(a\right)=f\text{'}\left(b\right)$ . Then, consider the following statement P, 
P : There exists atleast one $c\in (a,b)suchthatf\text{'}\left(c\right)=\frac{f\left(c\right)-f\left(a\right)}{c-a}$ 
Then which of the following options is/are correct?  
 

 
$f\text{'}\left(a\right)=f\text{'}\left(b\right)$ means tangents at A, B are par 
Hence AC such that AC is tangent at C
 $m_{AC}=\frac{f\left(c\right)-f\left(a\right)}{c-a}=m_{T@c}=f\text{'}\left(c\right)$
$g\left(x\right)=\{\begin{array}{l}\frac{f\left(x\right)-f\left(a\right)}{x-a};x\ne a\\ f\text{'}\left(a\right);x=a\end{array}$

 if g(x) is not strictly monotonic, then 
$\exists c\in (a,b):g\text{'}\left(c\right)=0\Rightarrow f\text{'}\left(c\right)=\frac{f\left(c\right)-f\left(a\right)}{c-a}$
Otherwise, we assume  $g\left(a\right)\le g\left(x\right)\le g\left(b\right)$

$\therefore g\left(x\right)\le g\left(b\right)\Rightarrow f\left(x\right)\le f\left(a\right)+(x-a)g\left(b\right)$

$nowf\text{'}\left(b\right)=\underset{t\to b}{\lim }\frac{f\left(t\right)-f\left(b\right)}{t-b}\ge \underset{t\to b}{\lim }\frac{f\left(a\right)-f\left(b\right)+(t-a)g\left(b\right)}{t-b}$

$\Rightarrow f\left(b\right)\ge g\left(b\right)\ge g\left(x\right)\ge g\left(a\right)=f\text{'}\left(a\right)$

This is only possible when g(x) = g(a) = g(b) = constant