Question

It is given that for the function f(x) = x³ - 6x² + ax + b on [1, 3], Rolle's theorem holds with c = $2+\frac{1}{\sqrt{3}}.$ If f(1) = f(3) = 0, then a =_______, b =________.

Answer 1

The given function is f(x) = x³ − 6x² + ax + b.

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

$f\left(1\right)=f\left(3\right)=0$ (Given)

$\therefore f\left(1\right)=0$

$\Rightarrow 1-6+a+b=0$

$\Rightarrow a+b=5$ .....(1)

Also,

$f\left(3\right)=0$

$\Rightarrow 27-54+3a+b=0$

$\Rightarrow 3a+b=27$ .....(2)

Solving (1) and (2), we get

a = 11 and b = −6

It can be verified that for a = 11 and b = −6, $f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=0$.

Thus, the values of a and b are 11 and −6, respectively.


It is given that for the function f(x) = x³ − 6x² + ax + 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___.