The function f(x) = x ^ 3 - 6 x ^ 2 + ax + b is such that f(2) =...

Given:

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

Given that:

$f\left(2\right) = 0 ⇒ 2a + b = 16 . . . . \left(i\right)$

$f\left(4\right) = 0 ⇒ 4a + b = 32 . . . . \left(ii\right)$

Solving both equations, we get

$a = 8, b = 0$

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

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

Now,

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

Now,

If ${f}^{'}\left(x\right) = - 1$

$⇒ 3{x}^{2} - 12x + 8 = - 1$

$⇒ 3{x}^{2} - 12x + 9 = 0$

$⇒ {x}^{2} - 4x + 3 = 0$

$⇒ {x}^{2} - 3x - x + 3 = 0$

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

$⇒ x = 1, 3$

Since, ${x}_{1} \in \left(2, 4\right)$, therefore ${x}_{1} = 3$.

And,

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

$⇒ 3{x}^{2} - 12x + 8 = 0$

$⇒ x = \dfrac{12 \pm \sqrt{144 - 96}}{6}$

$⇒ x = \dfrac{12 \pm \sqrt{48}}{6}$

$⇒ x = \dfrac{12 \pm 4\sqrt{3}}{6} = \dfrac{6 \pm 2\sqrt{3}}{3}$

$⇒ x = 2 + \dfrac{2\sqrt{3}}{3}, 2 - \dfrac{2\sqrt{3}}{3}$

Then, \left(2 + \dfrac{2\sqrt{3}}{3}\right) \in \left(2, 4\right)$</p><p>Hence,${x}_{2} = 2 + \dfrac{2\sqrt{3}}{3}$</p><p>So,${S}_{1}$is true.</p><p>Now,</p><p>${f}^{'}\left(x\right) = $\left(x - \left(2 + \dfrac{2\sqrt{3}}{3}\right)\right)\left(x - \left(2 - \dfrac{2\sqrt{3}}{3}\right)\right)$

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

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

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

So, ${x}_{4} = \left(2 + \dfrac{2\sqrt{3}}{3}\right)$.

Now,

$2{f}^{'}\left({x}_{3}\right) = \sqrt{3}f\left({x}_{4}\right)$

$⇒ 2\left[3{x}_{3}^{2} - 12{x}_{3} + 8\right] = \sqrt{3}\left(2 + \dfrac{2\sqrt{3}}{3}\right)\left(2 + \dfrac{2\sqrt{3}}{3} - 2\right)\left(2 + \dfrac{2\sqrt{3}}{3} - 4\right)$

$⇒ 2\left[3{x}_{3}^{2} - 12{x}_{3} + 8\right] = \sqrt{3}\left(\dfrac{2\sqrt{3}}{3} + 2\right)\left(\dfrac{2\sqrt{3}}{3}\right)\left(\dfrac{2\sqrt{3}}{3} - 2\right)$

$⇒ 2\left[3{x}_{3}^{2} - 12{x}_{3} + 8\right] = 2\left(\dfrac{12}{9} - 4\right)$

$⇒ \left[3{x}_{3}^{2} - 12{x}_{3} + 8\right] = - \dfrac{8}{3}$

$⇒ 9{x}_{3}^{2} - 36{x}_{3} + 32 = 0$

$⇒ {x}_{3} = \dfrac{36 \pm \sqrt{1296 - 1152}}{18} = \dfrac{36 \pm 12}{18}$

$⇒ {x}_{3} = \dfrac{8}{3}, \dfrac{4}{3}$

So, ${S}_{2}$ is true.