Key Concepts and Formulas

• Intermediate Value Theorem (IVT): If a function $f(x)$ is continuous on a closed interval $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs (i.e., $f(a)f(b) < 0$), then there exists at least one point $c \in (a, b)$ such that $f(c) = 0$.
• Monotonicity of a Function: A function $f(x)$ is monotonically increasing on an interval if $f'(x) \ge 0$ for all $x$ in that interval.
• Trigonometric Identity: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Also, $\cos(\pi/4) = \frac{\sqrt{2}}{2}$.

Step-by-Step Solution

Part 1: Analyzing Statement (I) - Uniqueness of Intersection Point

Statement (I) says: "The curve $y=f(x)$ intersects the $x$-axis exactly at one point."

Step 1: Check for Monotonicity

We compute the derivative of $f(x)$ to analyze its monotonicity. $f(x) = 4 \sqrt{2} x^3 - 3 \sqrt{2} x - 1$ $f'(x) = \frac{d}{dx} (4 \sqrt{2} x^3 - 3 \sqrt{2} x - 1)$ $f'(x) = 12 \sqrt{2} x^2 - 3 \sqrt{2} = 3\sqrt{2}(4x^2 - 1)$

Now, let's analyze the sign of $f'(x)$ on the given domain $\left[\frac{1}{2}, 1\right]$. For $x \in \left(\frac{1}{2}, 1\right]$, $4x^2 > 1$, so $4x^2 - 1 > 0$. At $x = \frac{1}{2}$, $f'(\frac{1}{2}) = 0$. Therefore, $f'(x) \ge 0$ on $[\frac{1}{2}, 1]$.

Explanation: Since $f'(x) \ge 0$ for all $x \in \left[\frac{1}{2}, 1\right]$, the function $f(x)$ is monotonically increasing on this interval. A monotonically increasing function can intersect the x-axis at most once.

Step 2: Apply the Intermediate Value Theorem (IVT)

To prove the existence of at least one root, we evaluate $f(x)$ at the endpoints of the interval $\left[\frac{1}{2}, 1\right]$.

• Evaluate $f\left(\frac{1}{2}\right)$: $f\left(\frac{1}{2}\right) = 4 \sqrt{2} \left(\frac{1}{2}\right)^3 - 3 \sqrt{2} \left(\frac{1}{2}\right) - 1$ $f\left(\frac{1}{2}\right) = \frac{\sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} - 1$ $f\left(\frac{1}{2}\right) = -\frac{2 \sqrt{2}}{2} - 1 = -\sqrt{2} - 1 < 0$

• Evaluate $f(1)$: $f(1) = 4 \sqrt{2} (1)^3 - 3 \sqrt{2} (1) - 1$ $f(1) = 4 \sqrt{2} - 3 \sqrt{2} - 1$ $f(1) = \sqrt{2} - 1 > 0$

Explanation: Since $f\left(\frac{1}{2}\right) < 0$ and $f(1) > 0$, and $f(x)$ is continuous on $\left[\frac{1}{2}, 1\right]$, by the Intermediate Value Theorem, there must exist at least one value $c \in \left(\frac{1}{2}, 1\right)$ such that $f(c) = 0$.

Conclusion for Statement (I): Combining the monotonicity (at most one root) and the IVT (at least one root), we conclude that there is exactly one point $x \in \left(\frac{1}{2}, 1\right)$ where $f(x) = 0$. Therefore, statement (I) is correct.

Part 2: Analyzing Statement (II) - Specific Intersection Point

Statement (II) says: "The curve $y=f(x)$ intersects the $x$-axis at $x=\cos \frac{\pi}{12}$."

Step 1: Trigonometric Substitution

Let $x = \cos \theta$. Then $f(x) = 4\sqrt{2}\cos^3\theta - 3\sqrt{2}\cos\theta - 1$. We want to show that $f(\cos(\pi/12)) = 0$.

$f(\cos\theta) = \sqrt{2}(4\cos^3\theta - 3\cos\theta) - 1$ Using the triple angle formula, $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, we have: $f(\cos\theta) = \sqrt{2}\cos(3\theta) - 1$

Step 2: Evaluate at $\theta = \frac{\pi}{12}$

Now, substitute $\theta = \frac{\pi}{12}$: $f\left(\cos\frac{\pi}{12}\right) = \sqrt{2}\cos\left(3 \cdot \frac{\pi}{12}\right) - 1$ $f\left(\cos\frac{\pi}{12}\right) = \sqrt{2}\cos\left(\frac{\pi}{4}\right) - 1$ Since $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, $f\left(\cos\frac{\pi}{12}\right) = \sqrt{2} \cdot \frac{\sqrt{2}}{2} - 1$ $f\left(\cos\frac{\pi}{12}\right) = \frac{2}{2} - 1 = 1 - 1 = 0$

Explanation: Since $f\left(\cos\frac{\pi}{12}\right) = 0$, the curve $y=f(x)$ intersects the $x$-axis at $x=\cos \frac{\pi}{12}$. Therefore, statement (II) is correct. Also, note that $\cos(\pi/12)$ is between 1/2 and 1, so it is in the interval.

Conclusion for Statement (II): Statement (II) is correct.

Since both statements (I) and (II) are correct, the answer is (A).

Common Mistakes & Tips

• Forgetting to check the interval when applying the Intermediate Value Theorem.
• Not recognizing the triple angle formula.
• Making algebraic errors when simplifying the function or its derivative.

Summary

We first proved that statement (I) is correct by showing that the function is monotonically increasing on the interval $[\frac{1}{2}, 1]$ and applying the Intermediate Value Theorem to show that there exists at least one root in the interval. We then proved that statement (II) is correct by using the triple angle formula and evaluating the function at $x = \cos \frac{\pi}{12}$. Both statements are correct.

Final Answer

The final answer is \boxed{Both (I) and (II) are correct}, which corresponds to option (A).