Question

If the function $f\left(x\right)=x^3-6x^2+ax+b$ satisfies Rolle's theorem in the interval $\left[1,3\right]$ and $f\text{'}\left[\frac{\left(2\sqrt{3}+1\right)}{\sqrt{3}}\right]=0$, then:

• A. $a=-11$
• B. $a=-6$
• C. $a=6$
• D. $a=11$

Answer 1

The correct option is D

$a=11$

Explanation for correct option:

Rolle's Theorem:

Given function, $f\left(x\right)=x^3-6x^2+ax+b$

According to Rolle's theorem, if a function $f$ is continuous on the closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$ such that $f\left(a\right)=f\left(b\right)$, then $f\text{'}\left(x\right)=0$ for some $x$ with $a\le x\le b$.

So, upon differentiating the given function, we get,

$f\text{'}\left(x\right)=3x^2-12x+a$

Now, as it is given that the function satisfies the Rolle's theorem, so, $f\text{'}\left(c\right)=0$

$\Rightarrow f\text{'}\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right)=0$

We can replace this in $f\text{'}\left(x\right)=3x^2-12x+a$,

$$\begin{gathered} \Rightarrow 0=3{\left(2+\frac{1}{\sqrt{3}}\right)}^2-12\left(2+\frac{1}{\sqrt{3}}\right)+a \\ \Rightarrow 0=3\left(4+\frac{1}{3}+\frac{4}{\sqrt{3}}\right)-24-\frac{12}{\sqrt{3}}+a \\ \Rightarrow 0=12+1+\cancel{4\sqrt{3}}-24-\cancel{4\sqrt{3}}+a \\ \Rightarrow 0=-11+a \\ \Rightarrow a=11 \end{gathered}$$

Therefore, option (D) is the correct answer.