Question

If the function $f\left(x\right)=a{x}^3+b{x}^2+11x-6,a,b\in R$ satisfies Rolle's theorem in $[1,3]$ and $f^{\prime }(2+\frac{1}{\sqrt{3}})=0,$ then the value of $(a-b)$ is

Answer 1

By Rolle’s theorem,
$f\left(1\right)=f\left(3\right)$
$\Rightarrow a+b+11-6=27a+9b+33-6$
$\Rightarrow 13a+4b+11=0$

Now, $f^{\prime }\left(x\right)=3a{x}^2+2bx+11$
Given, $x=2+\frac{1}{\sqrt{3}}$ is a root of $f^{\prime }\left(x\right)=0$
Since coefficients are real, $x=2-\frac{1}{\sqrt{3}}$ is another root.
Product of roots $=\frac{11}{3a}=\frac{11}{3}$
$\Rightarrow a=1$
Sum of roots $=-\frac{2b}{3}=4$
$\Rightarrow b=-6$
$\therefore a-b=1+6=7$