Question

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

• A. $2.5$
• B. $\sqrt{5}$
• C. $\sqrt{3}$
• D. $\sqrt{3}+1$

Answer 1

The correct option is B $\sqrt{5}$

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 $\epsilon$(2, 3) such that
$f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$
$\Rightarrow \frac{c}{\sqrt{c^2-4}}=\sqrt{5}-0$
$[\because f(x)=\sqrt{x^2-4}\Rightarrow f^{\prime }(x)=\frac{x}{\sqrt{x^2-4}}]$
$\Rightarrow c^2=5(c^2-4)$
$\Rightarrow 4c^2=20$
$\Rightarrow c=\sqrt{5}$