Question

Let $f\left(x\right)=x^3-x+p(0\ge x\ge 2),$ where $p$ is a constant. The value $c$ of mean value theorem is :

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{\sqrt{6}}{2}$
• C. $\frac{\sqrt{3}}{3}$
• D. $\frac{\sqrt{2}}{3}$
• E. $\frac{2\sqrt{3}}{3}$

Answer 1

The correct option is E $\frac{2\sqrt{3}}{3}$

$f\left(x\right)=x^3-x+p,f^{\prime }\left(x\right)=3x^2-1$

According to the mean value theorem, $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$⟹3c^2-1=\frac{p+6-p}{2-0}=3$

$⟹c^2=\frac{4}{3}⟹c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$