Question

If the function $f\left(x\right)=x^3-6x^2+ax+b$ defined on $[1,3]$ satisfies the rolle's theorem for $c=\frac{2\sqrt{3}+1}{\sqrt{3}}$ then

• A. $a=11,b=6$
• B. $a=-11,b=6$
• C. $a=11,b\in R$
• D. None of these

Answer 1

The correct option is C $a=11,b\in R$

Given that

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

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

$\Rightarrow f^{\prime }\left(c\right)=3c^2-12c+a$

According to Rolle's theorem

$\Rightarrow f^{\prime }\left(c\right)=0$

$\Rightarrow 3c^2-12c+a=0$

Substitute the given value of $c$

$\Rightarrow 3(\frac{2\sqrt{3}+1}{\sqrt{3}}{)}^2-12(\frac{2\sqrt{3}+1}{\sqrt{3}})+a=0$

$\Rightarrow 13+4\sqrt{3}-24-4\sqrt{3}=-a$

$\Rightarrow a=11$ and $b$ belongs to $R$