Question

The value of $c$ prescribed by Lagrange's mean value theorem (where $f^{\prime }\left(c\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$ for some $c$ in $[a,b]$), 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

Answer: (B)

$\sqrt{5}$

$f\left(x\right)=\sqrt{x^2-4}$
$f^{\prime }\left(x\right)=\frac{x}{\sqrt{x^2-4}}$
By Lagrange's mean value theorem in [2,3], there exists c $\in$ (2,3)
$f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$
$c^2=5(c^2-4)$
$\Rightarrow c=\pm \sqrt{5}$
As $c=-\sqrt{5}\notin (2,3)$, so $c=\sqrt{5}$