Let $$\left[\cdot\right]$$ denote the greatest integer function, and let f (x) = $$\min \left\{\sqrt{2x},x^{2}\right\}$$. Let S = $$\left\{x \in (-2,2): \text{the function,} g(x)= |x|\left[x^{2}\right]\text{is discontinuous at x} \right\}.$$ Then $$\sum_{x\in S}f(x)$$ equals

We need to find the set $$S$$ of points in $$(-2, 2)$$ where $$g(x) = |x|[x^2]$$ is discontinuous, where $$[\cdot]$$ is the greatest integer function.

The function $$|x|$$ is continuous everywhere. The function $$[x^2]$$ is discontinuous where $$x^2$$ takes integer values. In the interval $$(-2, 2)$$, $$x^2$$ ranges from $$0$$ to $$4$$ (not including 4).

So $$[x^2]$$ has potential discontinuities where $$x^2 = 0, 1, 2, 3$$, i.e., at $$x = 0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}$$.

The product $$g(x) = |x| \cdot [x^2]$$ is discontinuous at points where $$[x^2]$$ is discontinuous, UNLESS the discontinuity is "absorbed" by $$|x| = 0$$.

At $$x = 0$$: $$|x| = 0$$, and $$[x^2]$$ jumps from $$0$$ to $$0$$ (for small $$|x|$$, $$x^2$$ is between 0 and 1, so $$[x^2] = 0$$). Actually, for $$x \to 0$$, $$x^2 \to 0^+$$, so $$[x^2] = 0$$. And $$g(0) = 0 \cdot 0 = 0$$. So $$g$$ is continuous at $$x = 0$$.

At $$x = \pm 1$$: $$|x| = 1 \neq 0$$, and $$[x^2]$$ has a jump. Let us check $$x = 1$$:

For $$x \to 1^-$$: $$x^2 \to 1^-$$, so $$[x^2] = 0$$, giving $$g(x) \to 1 \cdot 0 = 0$$.

For $$x \to 1^+$$: $$x^2 \to 1^+$$, so $$[x^2] = 1$$, giving $$g(x) \to 1 \cdot 1 = 1$$.

$$g(1) = 1 \cdot [1] = 1 \cdot 1 = 1$$.

Since left limit $$\neq$$ right limit, $$g$$ is discontinuous at $$x = 1$$. Similarly at $$x = -1$$.

At $$x = \pm\sqrt{2}$$: $$|x| = \sqrt{2} \neq 0$$. Check $$x = \sqrt{2}$$:

For $$x \to \sqrt{2}^-$$: $$x^2 \to 2^-$$, $$[x^2] = 1$$, $$g(x) \to \sqrt{2} \cdot 1 = \sqrt{2}$$.

For $$x \to \sqrt{2}^+$$: $$x^2 \to 2^+$$, $$[x^2] = 2$$, $$g(x) \to \sqrt{2} \cdot 2 = 2\sqrt{2}$$.

Left limit $$\neq$$ right limit, so $$g$$ is discontinuous at $$x = \sqrt{2}$$. Similarly at $$x = -\sqrt{2}$$.

At $$x = \pm\sqrt{3}$$: $$|x| = \sqrt{3} \neq 0$$. Check $$x = \sqrt{3}$$:

For $$x \to \sqrt{3}^-$$: $$[x^2] = 2$$, $$g(x) \to 2\sqrt{3}$$.

For $$x \to \sqrt{3}^+$$: $$[x^2] = 3$$, $$g(x) \to 3\sqrt{3}$$.

Discontinuous at $$x = \pm\sqrt{3}$$.

So $$S = \{-\sqrt{3}, -\sqrt{2}, -1, 1, \sqrt{2}, \sqrt{3}\}$$.

Now we compute $$f(x) = \min\{\sqrt{2x}, x^2\}$$ for each $$x \in S$$. Note that $$\sqrt{2x}$$ requires $$x \geq 0$$, so for negative $$x$$, $$\sqrt{2x}$$ is not real. Let us re-examine the problem.

Actually, reading more carefully: $$f(x) = \min\{\sqrt{2x}, x^2\}$$. For negative values of $$x$$, $$\sqrt{2x}$$ is undefined (in reals). This likely means the problem intends $$f(x) = \min\{|2x|^{1/2}, x^2\}$$ or similar, but let us check with just $$f(x) = x^2$$ for negative $$x$$ (since $$\sqrt{2x}$$ is not real, the min defaults to $$x^2$$).

Alternatively, perhaps $$f(x) = \min\{\sqrt{2}x, x^2\}$$ (i.e., $$\sqrt{2} \cdot x$$). Let us check this interpretation since it works for all real $$x$$.

With $$f(x) = \min\{\sqrt{2} \cdot x, x^2\}$$:

Finding where $$\sqrt{2} \cdot x = x^2$$: $$x^2 - \sqrt{2}x = 0$$, so $$x(x - \sqrt{2}) = 0$$, giving $$x = 0$$ or $$x = \sqrt{2}$$.

For $$x < 0$$: $$\sqrt{2}x < 0$$ and $$x^2 > 0$$, so $$\min = \sqrt{2}x$$.

For $$0 < x < \sqrt{2}$$: $$\sqrt{2}x > x^2$$ (since $$x < \sqrt{2}$$), so $$\min = x^2$$.

For $$x > \sqrt{2}$$: $$x^2 > \sqrt{2}x$$ (since $$x > \sqrt{2}$$), so $$\min = \sqrt{2}x$$.

Now computing $$f(x)$$ for each $$x \in S$$:

$$f(-\sqrt{3}) = \sqrt{2} \cdot (-\sqrt{3}) = -\sqrt{6}$$

$$f(-\sqrt{2}) = \sqrt{2} \cdot (-\sqrt{2}) = -2$$

$$f(-1) = \sqrt{2} \cdot (-1) = -\sqrt{2}$$

$$f(1) = 1^2 = 1$$ (since $$0 < 1 < \sqrt{2}$$)

$$f(\sqrt{2}) = (\sqrt{2})^2 = 2$$ (at the boundary, both are equal: $$\sqrt{2} \cdot \sqrt{2} = 2$$)

$$f(\sqrt{3}) = \sqrt{2} \cdot \sqrt{3} = \sqrt{6}$$ (since $$\sqrt{3} > \sqrt{2}$$)

$$\sum_{x \in S} f(x) = -\sqrt{6} - 2 - \sqrt{2} + 1 + 2 + \sqrt{6} = 1 - \sqrt{2}$$

The answer is Option B: $$1 - \sqrt{2}$$.