The function f(x)=x^3-6x^2+ax+b is such that f(2)=f(4)=0. Consider two statements: (S_1) there exists x_1,x_2∈ (2,4),x_1<x_2, such that f^'(x_1)=-1…

Given:

$$f(x)={x}^{3}-6{x}^{2}+ax+b$$f(x)=x3−6x2+ax+b

Given that:

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

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

Solving both equations, we get

$$a=8,b=0$$a=8,b=0

$$\therefore f(x)={x}^{3}-6{x}^{2}+8x$$∴f(x)=x3−6x2+8x

$$\Rightarrow f(x)=x(x-2)(x-4)$$⇒f(x)=x(x−2)(x−4)

Now,

$${f}^{'}(x)=3{x}^{2}-12x+8$$f′(x)=3x2−12x+8

Now,

If ${f}^{'}(x)=-1$f′(x)=−1

$$\Rightarrow 3{x}^{2}-12x+8=-1$$⇒3x2−12x+8=−1

$$\Rightarrow 3{x}^{2}-12x+9=0$$⇒3x2−12x+9=0

$$\Rightarrow {x}^{2}-4x+3=0$$⇒x2−4x+3=0

$$\Rightarrow {x}^{2}-3x-x+3=0$$⇒x2−3x−x+3=0

$$\Rightarrow (x-1)(x-3)=0$$⇒(x−1)(x−3)=0

$$\Rightarrow x=1,3$$⇒x=1,3

Since, ${x}_{1}\in (2,4)$x1​∈(2,4), therefore ${x}_{1}=3$x1​=3.

And,

$${f}^{'}(x)=0$$f′(x)=0

$$\Rightarrow 3{x}^{2}-12x+8=0$$⇒3x2−12x+8=0

$$\Rightarrow x=\frac{12\pm \sqrt{144-96}}{6}$$⇒x=612±144−96​​

$$\Rightarrow x=\frac{12\pm \sqrt{48}}{6}$$⇒x=612±48​​

$$\Rightarrow x=\frac{12\pm 4\sqrt{3}}{6}=\frac{6\pm 2\sqrt{3}}{3}$$⇒x=612±43​​=36±23​​

$$\Rightarrow x=2+\frac{2\sqrt{3}}{3},2-\frac{2\sqrt{3}}{3}$$⇒x=2+323​​,2−323​​

Then, $(2+\frac{2\sqrt{3}}{3})\in (2,4)$(2+323​​)∈(2,4)

Hence, ${x}_{2}=2+\frac{2\sqrt{3}}{3}$x2​=2+323​​

So, ${S}_{1}$S1​ is true.

Now,

$${f}^{'}(x)=(x-(2+\frac{2\sqrt{3}}{3}))(x-(2-\frac{2\sqrt{3}}{3}))$$f′(x)=(x−(2+323​​))(x−(2−323​​))

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

${f}^{'}(x)>0\forall x\in (-\infty ,2-\frac{2\sqrt{3}}{3})\cup (2+\frac{2\sqrt{3}}{3},\infty )$f′(x)>0∀x∈(−∞,2−323​​)∪(2+323​​,∞), hence it is increasing.

${f}^{'}(x)<0\forall x\in (2-\frac{2\sqrt{3}}{3},2+\frac{2\sqrt{3}}{3})$f′(x)<0∀x∈(2−323​​,2+323​​), hence it is decreasing.

So, ${x}_{4}=(2+\frac{2\sqrt{3}}{3})$x4​=(2+323​​).

Now,

$$2{f}^{'}({x}_{3})=\sqrt{3}f({x}_{4})$$2f′(x3​)=3​f(x4​)

$$\Rightarrow 2[3{x}_{3}^{2}-12{x}_{3}+8]=\sqrt{3}(2+\frac{2\sqrt{3}}{3})(2+\frac{2\sqrt{3}}{3}-2)(2+\frac{2\sqrt{3}}{3}-4)$$⇒2[3x32​−12x3​+8]=3​(2+323​​)(2+323​​−2)(2+323​​−4)

$$\Rightarrow 2[3{x}_{3}^{2}-12{x}_{3}+8]=\sqrt{3}(\frac{2\sqrt{3}}{3}+2)(\frac{2\sqrt{3}}{3})(\frac{2\sqrt{3}}{3}-2)$$⇒2[3x32​−12x3​+8]=3​(323​​+2)(323​​)(323​​−2)

$$\Rightarrow 2[3{x}_{3}^{2}-12{x}_{3}+8]=2(\frac{12}{9}-4)$$⇒2[3x32​−12x3​+8]=2(912​−4)

$$\Rightarrow [3{x}_{3}^{2}-12{x}_{3}+8]=-\frac{8}{3}$$⇒[3x32​−12x3​+8]=−38​

$$\Rightarrow 9{x}_{3}^{2}-36{x}_{3}+32=0$$⇒9x32​−36x3​+32=0

$$\Rightarrow {x}_{3}=\frac{36\pm \sqrt{1296-1152}}{18}=\frac{36\pm 12}{18}$$⇒x3​=1836±1296−1152​​=1836±12​

$$\Rightarrow {x}_{3}=\frac{8}{3},\frac{4}{3}$$⇒x3​=38​,34​

So, ${S}_{2}$S2​ is true.