1. Use the given conditions to determine $a,b$

We have $f(x)=x^3-6x^2+ax+b,$ with $f(2)=0,\, f(4)=0.$

Compute: $f(2)=8-24+2a+b=0 \implies 2a+b=16 \,\,\,(1)$ $f(4)=64-96+4a+b=0 \implies 4a+b=32 \,\,\,(2)$

Subtracting (1) from (2): $2a=16 \implies a=8.$ Then from (1): $16+b=16 \implies b=0.$

So $f(x)=x^3-6x^2+8x=x(x-2)(x-4).$

2. Find the derivative

$f'(x)=3x^2-12x+8.$

Also, $f''(x)=6x-12=6(x-2).$ Thus for $x\in(2,4)$, $f''(x)>0,$ so $f'(x)$ is strictly increasing on $(2,4)$.

3. Check Statement 1

We need $x_1,x_2\in(2,4)$ with $x_1<x_2$ such that $f'(x_1)=-1, \qquad f'(x_2)=0.$

First evaluate $f'$ at the endpoints of the interval: $f'(2)=12-24+8=-4,$ $f'(4)=48-48+8=8.$

Since $f'(x)$ is continuous and strictly increasing on $(2,4)$, it takes every value between $-4$ and $8$ exactly once in $(2,4)$.

Both $-1$ and $0$ lie in $(-4,8)$, so there exist unique points $x_1,x_2\in(2,4)$ such that $f'(x_1)=-1, \qquad f'(x_2)=0.$ Because $f'$ is increasing and $-1<0$, we get $x_1<x_2.$

Hence, Statement 1 is true.

4. Check Statement 2: monotonicity part

We need $x_4\in(2,4)$ such that $f$ is decreasing in $(2,x_4)$ and increasing in $(x_4,4)$. This means $x_4$ should be a local minimum point, so $f'(x_4)=0.$

Solve: $3x^2-12x+8=0$ $x=\frac{12\pm\sqrt{144-96}}{6}=\frac{12\pm4\sqrt3}{6}=2\pm\frac{2}{\sqrt3}.$

Among these, only $x_4=2+\frac{2}{\sqrt3}$ lies in $(2,4)$.

Since $f'$ is increasing, we have:

• $f'(x)<0$ for $2<x<x_4$
• $f'(x)>0$ for $x_4<x<4$

So $f$ is decreasing on $(2,x_4)$ and increasing on $(x_4,4)$. Thus the monotonicity part is true.

5. Compute $f(x_4)$

Let $t=x_4=2+\frac{2}{\sqrt3}.$ Since $f(x)=x(x-2)(x-4),$ we get $t-2=\frac{2}{\sqrt3}, \qquad t-4=-2+\frac{2}{\sqrt3}=\frac{2}{\sqrt3}-2.$

So $f(t)=t(t-2)(t-4)=\left(2+\frac{2}{\sqrt3}\right)\left(\frac{2}{\sqrt3}\right)\left(\frac{2}{\sqrt3}-2\right).$

A simpler way: let $u=\frac{2}{\sqrt3}.$ Then $t=2+u,$ and $f(t)=(2+u)(u)(u-2)=u(u^2-4).$ Now $u^2=\frac{4}{3},$ so $u^2-4=\frac{4}{3}-4=-\frac{8}{3}.$ Hence $f(x_4)=u\left(-\frac{8}{3}\right)=-\frac{16}{3\sqrt3}.$ Rationalizing, $f(x_4)=-\frac{16\sqrt3}{9}.$

6. Find $x_3$ such that $2f'(x_3)=\sqrt3\,f(x_4)$

We just found $\sqrt3\,f(x_4)=\sqrt3\left(-\frac{16}{3\sqrt3}\right)=-\frac{16}{3}.$ Therefore the condition becomes $2f'(x_3)=-\frac{16}{3} \implies f'(x_3)=-\frac{8}{3}.$

Now on $(2,4)$, $f'(x)$ is continuous and strictly increasing from $f'(2)=-4 \quad \text{to} \quad f'(4)=8.$ Since $-\frac{8}{3}\in(-4,8),$ there exists a unique $x_3\in(2,4)$ such that $f'(x_3)=-\frac{8}{3}.$

Also, because on $(2,x_4)$ we have $f'(x)<0$ and $f'(x_4)=0$, the value $-\frac{8}{3}$ occurs for some point strictly before $x_4$. Hence $x_3<x_4.$ So the required $x_3,x_4$ exist.

Hence, Statement 2 is true.

7. Conclusion

• Statement 1: True
• Statement 2: True

Therefore, the correct option is $\boxed{\text{A}}$

8. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.