Consider the function f: [(1)/(2), 1] → R defined by f(x) = 4 2 x ^...

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

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

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

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

So, the function is increasing in $x \in \left(\dfrac{1}{2}, 1\right)$.

Thus, the function will intersect the $x -$axis only at one point.

Hence, statement $I$ is correct.

Now, putting $x = \dfrac{\pi}{12}$in the equation $\cos 3x = 4\cos^{3}\dfrac{\pi}{12} - 3cos\dfrac{\pi}{12}$.

$⇒ cos\dfrac{\pi}{4} = 4\cos^{3}\dfrac{\pi}{12} - 3cos\dfrac{\pi}{12}$

$⇒ 4\sqrt{2}\cos^{3}\dfrac{\pi}{12} - 3\sqrt{2}cos\dfrac{\pi}{12} - 1 = 0$

Now, comparing with $f\left(x\right) = 4\sqrt{2}{x}^{3} - 3\sqrt{2}x - 1$ we get.

$cos\dfrac{\pi}{12}$ is a solution of $f\left(x\right)$.

So, statement-$II$ is also correct.

Hence, both the statements are correct.