Question

It is given that for the function f given by $f\left(x\right)=x^3+b{x}^2+ax,x\in [1,3]$, Rolle's theorem holds with $c=2+\frac{1}{\sqrt{3}}$. Find the values of a and b.

Answer 1

Here Rolle's theorem holds for $f\left(x\right)$ defined on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$.

Given : $f\left(x\right)=x^3+b{x}^2+ax$

Then, $f^{\prime }\left(x\right)=3x^2+2bx+a$

Therefore,

$f\left(1\right)=f\left(3\right)$ and $f^{\prime }\left(c\right)=0$

$⟹1+b+a=27+9b+3a$ and $3c^2+2bc+a=0$

$⟹2a+8b+26=0$ and $3(2+\frac{1}{\sqrt{3}}{)}^2+2b(2+\frac{1}{\sqrt{3}})+a=0$

$⟹a+4b+13=0$ and $a+4b+13+\frac{2}{\sqrt{3}}(b+6)=0$

$⟹a+4b+13=0$ and $0+\frac{2}{\sqrt{3}}(b+6)=0$

$⟹a+4b+13=0$ and $b=-6$

Therefore, $a=11$ and $b=-6$