Question

If the function f(x) = x³ – 6x² + ax + b defined on [1, 3] satisfies Roll's theorem for c = $2+\frac{1}{\sqrt{3}},$ then a = ___________, b = __________.

Answer 1

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

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

$\therefore f\left(1\right)=f\left(3\right)$ and $f\text{'}\left(c\right)=0$

Now,

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

$\Rightarrow 1-6+a+b=27-54+3a+b$

$\Rightarrow -5+a=-27+3a$

$\Rightarrow 2a=22$

$\Rightarrow a=11$

Also,

f(x) = x³ − 6x² + ax + b

$\Rightarrow f\text{'}\left(x\right)=3x^2-12x+a$

$\therefore f\text{'}\left(c\right)=0$

$\Rightarrow 3c^2-12c+a=0$

$\Rightarrow 3c^2-12c+11=0$ (a = 11)

Since both equations f(1) = f(3) and $f\text{'}\left(c\right)=3c^2-12c+11=0$ are independent of b, so b can taken any real value.

∴ a = 11 and b ∈ R

Thus, if the function f(x) = x³ – 6x² + ax + b defined on [1, 3] satisfies Roll's theorem for $c=2+\frac{1}{\sqrt{3}}$, then a = 11 and b ∈ R.


If the function f(x) = x³ – 6x² + ax + b defined on [1, 3] satisfies Roll's theorem for c = $2+\frac{1}{\sqrt{3}},$ then a = ___11___, b = ___R___.