The value of c prescribed by Lagrange's mean value theorem, when $f\left(x\right)=\sqrt{x^2-4} ,a=2$ and $b=3$ is  

Solution:

Clearly $f\left(x\right)=\sqrt{x^2-4}$ is continuous on $[2,3]$ and 

differentiable on (2, 3). So, by Lagrange's mean value 

theorem there exists $c\in (2,3)$  such that 

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}\\ \Rightarrow \frac{c}{\sqrt{c^2-4}}=\sqrt{5}-0\left[\because f\left(x\right)=\sqrt{x^2-4}\Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{x}{\sqrt{x^2-4}}\right]\\ \Rightarrow c^2=5\left(c^2-4\right)\\ \Rightarrow 4c^2=20\end{array}$

$\Rightarrow c=\sqrt{5}$