Question

In the mean value theorem $\frac{f \left(b\right)-f \left(a\right)}{b-a}=f '\left(c\right),$ if $a = 0, b = 1/2$ and $f (x) = x (x - 1) (x - 2)$, the value of $c$ is -

• A. $1-\frac{\sqrt{15}}{6}$
• B. $1+\sqrt{15}$
• C. $1-\frac{\sqrt{21}}{6}$
• D. $1+\sqrt{21}$

Answer 1

Answer: (C)

$f '\left(c\right)=\frac{f \left(b\right)-f \left(a\right)}{b-a}$
$\Rightarrow 3c^{2}-6c+2=\frac{3/8-0}{1/2-0}=\frac{3}{4}$
$\Rightarrow c=1\pm\frac{\sqrt{21}}{6}$
$ \Rightarrow c=1+\frac{\sqrt{21}}{6} \notin \left(0, \frac{1}{2}\right)$
$ \Rightarrow c=1-\frac{\sqrt{21}}{6}$