Question

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

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

Answer 1

The correct option is D $a=11$

$f\left(x\right)=x^3-6x^2+ax+b$
$f^{\prime }\left(x\right)=3x^2-12x+a$
Since, Rolle's theorem holds in [1,3],
$\Rightarrow f\left(1\right)=f\left(3\right)$
$\Rightarrow -5+a+b=-27+3a+b$
$\Rightarrow a=11$