The function $$f(x) = |\sin 4x| + |\cos 2x|$$, is a periodic function with a fundamental period:

The function is $$ f(x) = |\sin 4x| + |\cos 2x| $$. To find its fundamental period, we need the smallest positive $$ T $$ such that $$ f(x + T) = f(x) $$ for all $$ x $$.

First, consider the individual terms. The function $$ \sin 4x $$ has a period of $$ \frac{2\pi}{4} = \frac{\pi}{2} $$ because the period of $$ \sin(bx) $$ is $$ \frac{2\pi}{b} $$. Taking the absolute value, $$ |\sin 4x| $$, the period becomes half of that, which is $$ \frac{\pi}{4} $$, since the absolute value reflects the negative part and makes the function repeat more frequently.

Similarly, the function $$ \cos 2x $$ has a period of $$ \frac{2\pi}{2} = \pi $$. Taking the absolute value, $$ |\cos 2x| $$, the period becomes half, which is $$ \frac{\pi}{2} $$.

So, we have:

• Period of $$ |\sin 4x| $$ is $$ \frac{\pi}{4} $$.
• Period of $$ |\cos 2x| $$ is $$ \frac{\pi}{2} $$.

The sum $$ f(x) $$ will be periodic with a period that is a common multiple of these two periods. The fundamental period is the least common multiple (LCM) of $$ \frac{\pi}{4} $$ and $$ \frac{\pi}{2} $$.

To find the LCM, set $$ T = k \cdot \frac{\pi}{4} = m \cdot \frac{\pi}{2} $$ for some integers $$ k $$ and $$ m $$. Dividing both sides by $$ \pi $$, we get $$ \frac{k}{4} = \frac{m}{2} $$. Multiplying both sides by 4 gives $$ k = 2m $$. The smallest positive integers satisfying this are $$ m = 1 $$ and $$ k = 2 $$. Substituting back, $$ T = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} $$ (or $$ T = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2} $$). Thus, the LCM is $$ \frac{\pi}{2} $$, suggesting that $$ \frac{\pi}{2} $$ might be the fundamental period.

Now, verify if $$ f(x + \frac{\pi}{2}) = f(x) $$:

Compute $$ f\left(x + \frac{\pi}{2}\right) = \left| \sin 4\left(x + \frac{\pi}{2}\right) \right| + \left| \cos 2\left(x + \frac{\pi}{2}\right) \right| $$.

Simplify the arguments:

• $$ 4\left(x + \frac{\pi}{2}\right) = 4x + 2\pi $$, so $$ \sin(4x + 2\pi) = \sin 4x $$ (since sine has period $$ 2\pi $$), and $$ \left| \sin(4x + 2\pi) \right| = |\sin 4x| $$.
• $$ 2\left(x + \frac{\pi}{2}\right) = 2x + \pi $$, so $$ \cos(2x + \pi) = -\cos 2x $$ (since $$ \cos(\theta + \pi) = -\cos \theta $$), and $$ \left| \cos(2x + \pi) \right| = |-\cos 2x| = |\cos 2x| $$.

Thus, $$ f\left(x + \frac{\pi}{2}\right) = |\sin 4x| + |\cos 2x| = f(x) $$, confirming that $$ \frac{\pi}{2} $$ is a period.

Next, check if it is the fundamental period by testing smaller options. First, test $$ T = \frac{\pi}{4} $$:

Compute $$ f\left(x + \frac{\pi}{4}\right) = \left| \sin 4\left(x + \frac{\pi}{4}\right) \right| + \left| \cos 2\left(x + \frac{\pi}{4}\right) \right| $$.

Simplify the arguments:

• $$ 4\left(x + \frac{\pi}{4}\right) = 4x + \pi $$, so $$ \sin(4x + \pi) = -\sin 4x $$, and $$ \left| \sin(4x + \pi) \right| = |-\sin 4x| = |\sin 4x| $$.
• $$ 2\left(x + \frac{\pi}{4}\right) = 2x + \frac{\pi}{2} $$, so $$ \cos\left(2x + \frac{\pi}{2}\right) = -\sin 2x $$ (since $$ \cos(\theta + \frac{\pi}{2}) = -\sin \theta $$), and $$ \left| \cos\left(2x + \frac{\pi}{2}\right) \right| = |-\sin 2x| = |\sin 2x| $$.

Thus, $$ f\left(x + \frac{\pi}{4}\right) = |\sin 4x| + |\sin 2x| $$, which is not equal to $$ f(x) = |\sin 4x| + |\cos 2x| $$ for all $$ x $$. For example, at $$ x = 0 $$:

• $$ f(0) = |\sin 0| + |\cos 0| = 0 + 1 = 1 $$.
• $$ f\left(\frac{\pi}{4}\right) = \left| \sin \pi \right| + \left| \cos \frac{\pi}{2} \right| = |0| + |0| = 0 $$ (which is $$ f(0 + \frac{\pi}{4}) $$ at $$ x = 0 $$).

Since $$ 1 \neq 0 $$, $$ T = \frac{\pi}{4} $$ is not a period.

Now, test $$ T = \frac{\pi}{8} $$:

Compute $$ f\left(x + \frac{\pi}{8}\right) = \left| \sin 4\left(x + \frac{\pi}{8}\right) \right| + \left| \cos 2\left(x + \frac{\pi}{8}\right) \right| $$.

Simplify the arguments:

• $$ 4\left(x + \frac{\pi}{8}\right) = 4x + \frac{\pi}{2} $$, so $$ \sin\left(4x + \frac{\pi}{2}\right) = \cos 4x $$, and $$ \left| \sin\left(4x + \frac{\pi}{2}\right) \right| = |\cos 4x| $$.
• $$ 2\left(x + \frac{\pi}{8}\right) = 2x + \frac{\pi}{4} $$, so $$ \left| \cos\left(2x + \frac{\pi}{4}\right) \right| $$ remains as is.

Thus, $$ f\left(x + \frac{\pi}{8}\right) = |\cos 4x| + \left| \cos\left(2x + \frac{\pi}{4}\right) \right| $$, which is not equal to $$ f(x) $$. For example, at $$ x = 0 $$:

• $$ f(0) = 1 $$.
• $$ f\left(\frac{\pi}{8}\right) = |\cos 0| + \left| \cos \frac{\pi}{4} \right| = |1| + \left| \frac{\sqrt{2}}{2} \right| = 1 + \frac{\sqrt{2}}{2} \approx 1.707 $$.

Since $$ 1 \neq 1.707 $$, $$ T = \frac{\pi}{8} $$ is not a period.

Similarly, test $$ T = \frac{\pi}{6} $$ at $$ x = 0 $$:

• $$ f(0) = 1 $$.
• $$ f\left(\frac{\pi}{6}\right) = \left| \sin \frac{2\pi}{3} \right| + \left| \cos \frac{\pi}{3} \right| = \left| \frac{\sqrt{3}}{2} \right| + \left| \frac{1}{2} \right| = \frac{\sqrt{3}}{2} + \frac{1}{2} \approx 1.366 $$.

Since $$ 1 \neq 1.366 $$, $$ T = \frac{\pi}{6} $$ is not a period.

Now, consider the behavior of $$ f(x) $$ in $$ [0, \frac{\pi}{2}] $$:

• At $$ x = 0 $$, $$ f(0) = |\sin 0| + |\cos 0| = 0 + 1 = 1 $$.
• At $$ x = \frac{\pi}{8} $$, $$ f\left(\frac{\pi}{8}\right) = \left| \sin \frac{\pi}{2} \right| + \left| \cos \frac{\pi}{4} \right| = |1| + \left| \frac{\sqrt{2}}{2} \right| = 1 + \frac{\sqrt{2}}{2} \approx 1.707 $$.
• At $$ x = \frac{\pi}{4} $$, $$ f\left(\frac{\pi}{4}\right) = \left| \sin \pi \right| + \left| \cos \frac{\pi}{2} \right| = |0| + |0| = 0 $$.
• At $$ x = \frac{3\pi}{8} $$, $$ f\left(\frac{3\pi}{8}\right) = \left| \sin \frac{3\pi}{2} \right| + \left| \cos \frac{3\pi}{4} \right| = |-1| + \left| -\frac{\sqrt{2}}{2} \right| = 1 + \frac{\sqrt{2}}{2} \approx 1.707 $$.
• At $$ x = \frac{\pi}{2} $$, $$ f\left(\frac{\pi}{2}\right) = \left| \sin 2\pi \right| + \left| \cos \pi \right| = |0| + |-1| = 1 $$.

The function starts at 1, increases to approximately 1.707 at $$ \frac{\pi}{8} $$, decreases to 0 at $$ \frac{\pi}{4} $$, increases to approximately 1.707 at $$ \frac{3\pi}{8} $$, and returns to 1 at $$ \frac{\pi}{2} $$. This pattern repeats every $$ \frac{\pi}{2} $$, and no smaller period matches the values at all points.

Therefore, the fundamental period is $$ \frac{\pi}{2} $$.

Hence, the correct answer is Option D.