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. $a=11,bϵR$
• D. none of these

Answer 1

The correct option is C $a=11,bϵR$

$f\left(x\right)=x^3-6x^2+ax+b$
$f^{\prime }\left(x\right)=3x^2-12x+a$
Since, Rolle's theorem holds for $c=\frac{2\sqrt{3}+1}{\sqrt{3}}$
$\Rightarrow f^{\prime }\left(c\right)=0$
$3(2+\frac{1}{\sqrt{3}}{)}^2-12(2+\frac{1}{\sqrt{3}})+a=0$
$\Rightarrow a=11$
Also, f'(x) is independent of b, so $b\in R$