It is given that for the function $f(x)=x^{3}-6 x^{2}+a x+b$ on $[1,3]$, Rolle's theorem holds with $c=2+\frac{1}{\sqrt{3}} .$ If $f(1)=f(3)=0$, then a =_______, b =________.
The given function is $f(x)=x^{3}-6 x^{2}+a x+b$.
It is given that Rolle's theorem holds for $f(x)$ defined on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$.
$f(1)=f(3)=0 \quad$ (Given)
$\therefore f(1)=0$
$\Rightarrow 1-6+a+b=0$
$\Rightarrow a+b=5$ ......(1)
Also,
$\Rightarrow 27-54+3 a+b=0$
$\Rightarrow 3 a+b=27$ ......(2)
Solving (1) and (2), we get
$a=11$ and $b=-6$
It can be verified that for $a=11$ and $b=-6, f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$.
Thus, the values of a and b are 11 and −6, respectively.
It is given that for the function $f(x)=x^{3}-6 x^{2}+a x+b$ on $[1,3]$, Rolle's theorem holds with $c=2+\frac{1}{\sqrt{3}} .$ If $f(1)=f(3)=0$, then a = ___11___, b =___−6___.