Question

Find 'c' of the mean value theorem, if $f\left(x\right)=x(x-1)(x-2);a=0,b=1/2$

• A. $c=\frac{1+\sqrt{21}}{6}$
• B. $c=\frac{1-\sqrt{21}}{6}$
• C. $c=\frac{1+\sqrt{6}}{3}$
• D. $c=\frac{1-\sqrt{6}}{3}$

Answer 1

The correct option is B $c=\frac{1-\sqrt{21}}{6}$

We have $f\left(a\right)=f\left(0\right)=0$ and $f\left(b\right)=f\left(\frac{1}{2}\right)=\frac{3}{8}$
$\therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{\frac{3}{8}-0}{\frac{1}{2}-0}=\frac{3}{4}$
Now
$f\left(x\right)=x^3-3x^2+2x\Rightarrow f^{\prime }\left(x\right)={3x}^2-6x+2\Rightarrow f^{\prime }\left(c\right)={3c}^2-6c+2$
Putting all these value in lagrange's mean value theorem
$\frac{f\left(b\right)-f\left(a\right)}{b-a}=f^{\prime }\left(c\right),(a<c,b)$
We get $\frac{3}{4}={3c}^2-6c+2\Rightarrow c=1\pm \frac{\sqrt{21}}{6}$
Hence $c=\frac{1-\sqrt{21}}{6}$ lies in the open interval $(0,\frac{1}{2})$
Therefore it is the required value