Question

The value of $c$ in the Lagrange's mean - value theorem for $f\left(x\right)=\sqrt{x-2}$ in the interval $[2,6]$ is

• A. $\frac{9}{2}$
• B. $\frac{5}{2}$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

We have, $f\left(x\right)=\sqrt{x-2}$

$\therefore f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x-2}}$

Now using LMVT:

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$\Rightarrow \frac{1}{2\sqrt{c-2}}=\frac{f\left(6\right)-f\left(2\right)}{4}=\frac{2-0}{4}=\frac{1}{2}$
$\Rightarrow$ $\sqrt{c-2}=1$
$\Rightarrow c-2=1$
$\Rightarrow c=3$