Consider the function $f:[\frac{1}{2},1]\to <!--\; \to \; -->\mathrm{R}$ defined by $f(x)=4\sqrt{2}{x}^3-<!--\; -\; -->3\sqrt{2}x-<!--\; -\; -->1$.  Consider the statements :

(I) The curve y = f(x) intersects the x-axis is exactly at one point.

(II) The curve y = f(x) intersects the x-axis at $x=\cos <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{12}$

Then

The correct answer is Both (I) and (II) are correct.

We are given the function

$f\left(x\right)=4\sqrt{2}{x}^3-3\sqrt{2}x-1$

defined on the interval $[\frac{1}{2},1]$. We need to find where, if anywhere, this curve crosses the x-axis (i.e., where f(x) = 0).

Step 1: Rewrite the equation and recall the triple-angle identity.

Setting f(x) = 0:

$4\sqrt{2}{x}^3-3\sqrt{2}x-1=0$

Factor out $\sqrt{2}$ from the first two terms:

$\sqrt{2}(4x^3-3x)=1$

Now recall the key trigonometric identity for the triple angle of cosine:

$\cos \left(3\theta \right)=4{\cos }^3\theta -3\cos \theta$

This matches the expression inside the parentheses exactly! So if we let $x=\cos \theta$, the equation becomes:

$\sqrt{2}·\cos \left(3\theta \right)=1$

$\cos \left(3\theta \right)=\frac{1}{\sqrt{2}}$

Step 2: Solve for θ.

We know that $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$, so the general solutions for $3\theta$ in $[0,2\pi ]$ are:

$3\theta =\frac{\pi }{4}\text{or}3\theta =\frac{7}{\pi }·\frac{1}{4}=\frac{7}{\pi }/4$

More precisely, the two principal solutions in [0, 2π] are $3\theta =\frac{\pi }{4}$ and $3\theta =\frac{7}{\pi }/4$, giving:

$\theta =\frac{\pi }{12}\Rightarrow x=\cos \frac{\pi }{12}$

$\theta =\frac{7}{\pi }/12\Rightarrow x=\cos \frac{7}{\pi }/12$

Step 3: Check which solutions lie in the domain $[\frac{1}{2},1]$.

Numerically evaluating:

• $\cos \frac{\pi }{12}=\cos (15°)\approx 0.966$ — this lies in $[\frac{1}{2},1]$ ✓

• $\cos \frac{7}{\pi }/12=\cos (105°)\approx -0.259$ — this is NOT in $[\frac{1}{2},1]$ ✗

So the only root within the domain is $x=\cos \frac{\pi }{12}$.

Step 4: Confirm uniqueness using monotonicity (verifying Statement I).

Let's compute the derivative:

$f^{\text{'}}\left(x\right)=12\sqrt{2}{x}^2-3\sqrt{2}=3\sqrt{2}(4x^2-1)$

Setting f'(x) = 0 gives $x=\frac{1}{2}$, which is the left endpoint of our domain. For all $x>\frac{1}{2}$, we have $4x^2>1$, so $f^{\text{'}}\left(x\right)>0$. This means f is strictly increasing on $(\frac{1}{2},1]$.

Also note:

• $f\left(\frac{1}{2}\right)=4\sqrt{2}·\frac{1}{8}-3\sqrt{2}·\frac{1}{2}-1=\frac{\sqrt{2}}{2}-\frac{3}{\sqrt{2}}/2-1=-\sqrt{2}-1<0$

• $f\left(1\right)=4\sqrt{2}-3\sqrt{2}-1=\sqrt{2}-1>0$

Since f is continuous and strictly increasing on the entire interval (with only one critical point, at the endpoint x = 1/2), and it goes from a negative value to a positive value, by the Intermediate Value Theorem, it crosses zero exactly once. A strictly monotone function can only cross zero at most once, confirming Statement (I).

Conclusion:

Statement (I) is correct — the curve intersects the x-axis at exactly one point in $[\frac{1}{2},1]$.
Statement (II) is correct — that unique intersection point is at $x=\cos \frac{\pi }{12}$.

Therefore, Both (I) and (II) are correct.