Let $$y = fx$$ represent a parabola with focus $$(-\dfrac{1}{2}, 0)$$ and directrix $$y = -\dfrac{1}{2}$$. Then
$$S = \{x \in \mathbb{R}: \tan^{-1}(\sqrt{fx}) + \sin^{-1}(\sqrt{fx+1}) = \dfrac{\pi}{2}\}$$:

We need to find the parabola $$y = f(x)$$ with focus $$(-1/2, 0)$$ and directrix $$y = -1/2$$, then determine the set $$S$$.

The vertex of a parabola is the midpoint between the focus and the nearest point on the directrix. The focus is at $$(-1/2, 0)$$ and the directrix is $$y = -1/2$$. The point on the directrix closest to the focus (dropping a perpendicular) is $$(-1/2, -1/2)$$. The vertex is:

$$V = \left(-\frac{1}{2}, -\frac{1}{4}\right)$$

Since the focus is above the directrix, the parabola opens upward. The distance from vertex to focus is $$a = \bigl|0 - (-\tfrac{1}{4})\bigr| = \tfrac{1}{4}$$. The standard form (with vertex at $$(h,k)$$) is:

$$ (x-h)^2 = 4a(y-k) \implies \left(x+\frac{1}{2}\right)^2 = 4 \times \frac{1}{4}\!\left(y+\frac{1}{4}\right) = y + \frac{1}{4}$$

So $$y = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} = x^2 + x + \frac{1}{4} - \frac{1}{4} = x^2 + x$$. Therefore $$f(x) = x^2 + x = x(x+1)$$.

$$S = \{x \in \mathbb{R} : \tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x)+1}) = \tfrac{\pi}{2}\}$$

For $$\tan^{-1}(\sqrt{f(x)})$$ to be defined, we need $$f(x) \ge 0$$, i.e., $$x(x+1) \ge 0$$. For $$\sin^{-1}(\sqrt{f(x)+1})$$ to be defined, we need $$0 \le f(x) + 1 \le 1$$, i.e., $$-1 \le f(x) \le 0$$.

Combining these conditions gives $$f(x) = 0$$.

When $$f(x) = 0$$ we have $$\tan^{-1}(0) + \sin^{-1}(1) = 0 + \tfrac{\pi}{2} = \tfrac{\pi}{2}$$, which satisfies the equation.

Setting $$x(x+1)=0$$ yields $$x = 0$$ or $$x = -1$$, so $$S = \{0, -1\}$$, which contains exactly two elements.

The correct answer is Option 1: contains exactly two elements.