Question

If $f\left(x\right)=(x-1)(x-2)$ and interval given was (0,4), find 'c' using Langrange's mean value theorem.

• A. $2+\frac{2}{\sqrt{3}}$
• B. $2-\frac{2}{\sqrt{3}}$
• C. $2$
• D. Both A and B

Answer 1

The correct option is C $2$

Here $f\left(x\right)=(x-1)(x-2)$.
Now $f\left(x\right)$ is continuous on $[0,4]$ and differentiable on $(0,4)$.
Thus applying LMVT, we get
$\frac{f\left(4\right)-f\left(0\right)}{4-0}=f^{\prime }\left(c\right)$
$\Rightarrow \frac{6-2}{4}=c-2+c-1$
$\Rightarrow 1=2c-3$
$\Rightarrow c=2$