Key Concepts and Formulas

• Rolle's Theorem: If a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
• Intermediate Value Theorem (IVT): If a function $f(x)$ is continuous on $[a, b]$, and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that $f(c) = k$. A version exists for derivatives called Darboux's Theorem.
• Monotonicity: If $f'(x) > 0$ on an interval, then $f(x)$ is increasing on that interval. If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing on that interval.

Step-by-Step Solution

Step 1: Determine the Function $f(x)$

We are given $f(x) = x^3 - 6x^2 + ax + b$, $f(2) = 0$, and $f(4) = 0$. We will use these conditions to find $a$ and $b$.

• Using $f(2) = 0$: Substitute $x = 2$ into $f(x)$: $f(2) = (2)^3 - 6(2)^2 + a(2) + b = 8 - 24 + 2a + b = 0$ $2a + b = 16 \hspace{1cm} (1)$

• Using $f(4) = 0$: Substitute $x = 4$ into $f(x)$: $f(4) = (4)^3 - 6(4)^2 + a(4) + b = 64 - 96 + 4a + b = 0$ $4a + b = 32 \hspace{1cm} (2)$

• Solve for $a$ and $b$: Subtract equation (1) from equation (2): $(4a + b) - (2a + b) = 32 - 16$ $2a = 16$ $a = 8$ Substitute $a = 8$ into equation (1): $2(8) + b = 16$ $16 + b = 16$ $b = 0$

Therefore, the function is $f(x) = x^3 - 6x^2 + 8x$.

Step 2: Find the Derivative $f'(x)$

Differentiate $f(x)$ with respect to $x$: $f'(x) = 3x^2 - 12x + 8$

Step 3: Analyze Statement 1

Statement 1: There exists $x_1, x_2 \in (2, 4)$, $x_1 < x_2$, such that $f'(x_1) = -1$ and $f'(x_2) = 0$.

• Check if $f'(x) = -1$ has a solution in $(2, 4)$: We need to solve $f'(x) = -1$: $3x^2 - 12x + 8 = -1$ $3x^2 - 12x + 9 = 0$ $x^2 - 4x + 3 = 0$ $(x - 1)(x - 3) = 0$ $x = 1, 3$ Since $x = 3 \in (2, 4)$, we can take $x_1 = 3$, and thus $f'(x_1) = f'(3) = -1$.

• Check if $f'(x) = 0$ has a solution in $(2, 4)$: We need to solve $f'(x) = 0$: $3x^2 - 12x + 8 = 0$ Using the quadratic formula: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3}$ Let $x_2 = 2 + \frac{2\sqrt{3}}{3}$. Since $\sqrt{3} \approx 1.732$, we have $\frac{2\sqrt{3}}{3} \approx \frac{2(1.732)}{3} \approx 1.155$. Thus, $x_2 \approx 2 + 1.155 = 3.155$, which is in the interval $(2, 4)$. Since $1 < \sqrt{3} < 2$, we have $2 < 2 + \frac{2\sqrt{3}}{3} < 2 + \frac{4}{3} = \frac{10}{3} < 4$. Therefore, $x_2 \in (2, 4)$. We also need $x_1 < x_2$, so $3 < 2 + \frac{2\sqrt{3}}{3}$, which means $1 < \frac{2\sqrt{3}}{3}$, or $3 < 2\sqrt{3}$, or $9 < 12$, which is true. Thus, we can choose $x_1 = 3$ and $x_2 = 2 + \frac{2\sqrt{3}}{3}$, and Statement 1 is true.

Step 4: Analyze Statement 2

Statement 2: There exists $x_3, x_4 \in (2, 4)$, $x_3 < x_4$, such that $f$ is decreasing in $(2, x_4)$, increasing in $(x_4, 4)$ and $2f'(x_3) = \sqrt{3} f(x_4)$.

The function $f(x)$ is decreasing when $f'(x) < 0$ and increasing when $f'(x) > 0$. From the analysis in Step 3, $f'(x) = 3x^2 - 12x + 8 = 0$ when $x = 2 \pm \frac{2\sqrt{3}}{3}$. We found that $x_2 = 2 + \frac{2\sqrt{3}}{3} \approx 3.155$, and $f'(x) = 0$ at $x_2$. Since the coefficient of $x^2$ in $f'(x)$ is positive, $f'(x) < 0$ between the roots and $f'(x) > 0$ outside the roots. Therefore, $f$ is decreasing in $(2, 2 + \frac{2\sqrt{3}}{3})$ and increasing in $(2 + \frac{2\sqrt{3}}{3}, 4)$. So, we can take $x_4 = 2 + \frac{2\sqrt{3}}{3}$.

Now, we need to find $x_3 \in (2, 4)$ such that $x_3 < x_4$ and $2f'(x_3) = \sqrt{3}f(x_4)$. Since $f'(x_4) = 0$, $x_4$ is a local minimum. However, $f(x_4) \neq 0$ since $x_4$ is not 2 or 4. We have $f(x_4) = f(2 + \frac{2\sqrt{3}}{3}) = (2 + \frac{2\sqrt{3}}{3})^3 - 6(2 + \frac{2\sqrt{3}}{3})^2 + 8(2 + \frac{2\sqrt{3}}{3})$. We need to find $x_3$ such that $2f'(x_3) = \sqrt{3} f(x_4)$. This is difficult to solve analytically.

However, we know that $f(2) = f(4) = 0$. By Rolle's Theorem, there exists a $c \in (2, 4)$ such that $f'(c) = 0$. We found that $c = 2 + \frac{2\sqrt{3}}{3} = x_4$. Since $f$ is decreasing in $(2, x_4)$ and increasing in $(x_4, 4)$, $x_4$ is the point of local minimum. Consider the interval $(2, x_4)$. Since $f'(2) = 3(4) - 12(2) + 8 = 12 - 24 + 8 = -4$ and $f'(x_4) = 0$. By Darboux's Theorem, $f'$ attains all values between -4 and 0 in $(2, x_4)$. Since $x_4 = 2 + \frac{2\sqrt{3}}{3}$, $f(x_4)$ is negative. Therefore, $\sqrt{3}f(x_4)$ is negative. We need to find $x_3$ such that $2f'(x_3) = \sqrt{3}f(x_4)$, so $f'(x_3)$ must be negative.

Since $f'(2) = -4$ and $f'(x_4) = 0$, we can find an $x_3 \in (2, x_4)$ such that $f'(x_3) = \frac{\sqrt{3}}{2}f(x_4)$. Therefore, Statement 2 is true.

Common Mistakes & Tips

• Be careful with the arithmetic and algebraic manipulations when solving for the coefficients and roots.
• Remember to check if the solutions you obtain lie within the specified intervals.
• Don't forget to apply the correct theorems (Rolle's Theorem, IVT, etc.) when needed.
• The condition $x_1 < x_2$ and $x_3 < x_4$ must be satisfied.

Summary

We first determined the function $f(x)$ by using the given conditions $f(2) = 0$ and $f(4) = 0$. Then we found the derivative $f'(x)$ and analyzed Statement 1, showing that there exist $x_1, x_2 \in (2, 4)$ such that $f'(x_1) = -1$ and $f'(x_2) = 0$. Finally, we analyzed Statement 2, showing that there exist $x_3, x_4 \in (2, 4)$ such that $f$ is decreasing in $(2, x_4)$, increasing in $(x_4, 4)$ and $2f'(x_3) = \sqrt{3} f(x_4)$. Therefore, both statements are true.

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).