The number of distinct real roots of the equation $$x^7 - 7x - 2 = 0$$ is

We need to find the number of distinct real roots of $$x^7 - 7x - 2 = 0$$.

Let $$f(x) = x^7 - 7x - 2$$.

Finding critical points: $$f'(x) = 7x^6 - 7 = 7(x^6 - 1) = 0$$

This gives $$x^6 = 1$$, so $$x = 1$$ and $$x = -1$$ are the only real critical points.

$$f''(x) = 42x^5$$

At $$x = -1$$: $$f''(-1) = -42 < 0$$, so $$x = -1$$ is a local maximum.

At $$x = 1$$: $$f''(1) = 42 > 0$$, so $$x = 1$$ is a local minimum.

Evaluating $$f$$ at the critical points:

$$f(-1) = (-1)^7 - 7(-1) - 2 = -1 + 7 - 2 = 4 > 0$$

$$f(1) = 1 - 7 - 2 = -8 < 0$$

Since $$f$$ is a degree 7 polynomial with positive leading coefficient:

As $$x \to -\infty$$, $$f(x) \to -\infty$$ and as $$x \to +\infty$$, $$f(x) \to +\infty$$.

Analyzing sign changes:

From $$-\infty$$ to $$x = -1$$: $$f$$ goes from $$-\infty$$ to $$f(-1) = 4 > 0$$. So there is exactly one root in $$(-\infty, -1)$$.

From $$x = -1$$ to $$x = 1$$: $$f$$ goes from $$4 > 0$$ to $$f(1) = -8 < 0$$. So there is exactly one root in $$(-1, 1)$$.

From $$x = 1$$ to $$+\infty$$: $$f$$ goes from $$-8 < 0$$ to $$+\infty$$. So there is exactly one root in $$(1, +\infty)$$.

Since the function is monotonically decreasing between the local max and local min, and monotonically increasing outside, there are exactly 3 distinct real roots.

The correct answer is Option D.