The function f x = x 3 - 6 x 2 + a x + b is such that f 2 = f 4 = 0 . Consider two statements: S 1 there…

Given:

$f\left(x\right)=x^3-6x^2+ax+b$

Given that:

$f\left(2\right)=0\Rightarrow 2a+b=16 ....\left(\mathrm{i}\right)$

$f\left(4\right)=0\Rightarrow 4a+b=32 ....\left(\mathrm{ii}\right)$

Solving both equations, we get

$a=8, b=0$

$\therefore f\left(x\right)=x^3-6x^2+8x$

$\Rightarrow f\left(x\right)=x\left(x-2\right)\left(x-4\right)$

Now,

$f^{\text{'}}\left(x\right)=3x^2-12x+8$

Now,

If $f^{\text{'}}\left(x\right)=-1$

$\Rightarrow 3x^2-12x+8=-1$

$\Rightarrow 3x^2-12x+9=0$

$\Rightarrow x^2-4x+3=0$

$\Rightarrow x^2-3x-x+3=0$

$\Rightarrow \left(x-1\right)\left(x-3\right)=0$

$\Rightarrow x=1, 3$

Since, $x_1\in \left(2,4\right)$, therefore $x_1=3$.

And,

$f^{\text{'}}\left(x\right)=0$

$\Rightarrow 3x^2-12x+8=0$

$\Rightarrow x=\frac{12\pm \sqrt{144-96}}{6}$

$\Rightarrow x=\frac{12\pm \sqrt{48}}{6}$

$\Rightarrow x=\frac{12\pm 4\sqrt{3}}{6}=\frac{6\pm 2\sqrt{3}}{3}$

$\Rightarrow x=2+\frac{2\sqrt{3}}{3}, 2-\frac{2\sqrt{3}}{3}$

Then, $\left(2+\frac{2\sqrt{3}}{3}\right)\in \left(2,4\right)$

Hence, $x_2=2+\frac{2\sqrt{3}}{3}$

So, $S_1$ is true.

Now,

$f^{\text{'}}\left(x\right)=\left(x-\left(2+\frac{2\sqrt{3}}{3}\right)\right)\left(x-\left(2-\frac{2\sqrt{3}}{3}\right)\right)$

Sign scheme for $f^{\text{'}}\left(x\right)$ is as follows:

$f^{\text{'}}\left(x\right)>0 \forall x\in \left(-\infty ,2-\frac{2\sqrt{3}}{3}\right)\cup \left(2+\frac{2\sqrt{3}}{3},\infty \right)$, hence it is increasing.

$f^{\text{'}}\left(x\right)<0 \forall x\in \left(2-\frac{2\sqrt{3}}{3},2+\frac{2\sqrt{3}}{3}\right)$, hence it is decreasing.

So, $x_4=\left(2+\frac{2\sqrt{3}}{3}\right)$.

Now,

$2f^{\text{'}}\left(x_3\right)=\sqrt{3}f\left(x_4\right)$

$\Rightarrow 2\left[3x_3^2-12x_3+8\right]=\sqrt{3}\left(2+\frac{2\sqrt{3}}{3}\right)\left(2+\frac{2\sqrt{3}}{3}-2\right)\left(2+\frac{2\sqrt{3}}{3}-4\right)$

$\Rightarrow 2\left[3x_3^2-12x_3+8\right]=\sqrt{3}\left(\frac{2\sqrt{3}}{3}+2\right)\left(\frac{2\sqrt{3}}{3}\right)\left(\frac{2\sqrt{3}}{3}-2\right)$

$\Rightarrow 2\left[3x_3^2-12x_3+8\right]=2\left(\frac{12}{9}-4\right)$

$\Rightarrow \left[3x_3^2-12x_3+8\right]=-\frac{8}{3}$

$\Rightarrow 9x_3^2-36x_3+32=0$

$\Rightarrow x_3=\frac{36\pm \sqrt{1296-1152}}{18}=\frac{36\pm 12}{18}$

$\Rightarrow x_3=\frac{8}{3}, \frac{4}{3}$

So, $S_2$ is true.