Question

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

• A. \(2 - \sqrt{2}\)
• B. \(1 - \sqrt{2}\)
• C. \(2\sqrt{6} - 3\sqrt{2}\)
• D. \(\sqrt{6} - 2\sqrt{2}\)

Hint:

Discontinuities of expressions involving greatest integer function occur at integer values of the inner expression.

Answer 1

The Correct Option is A

To solve this problem, we must identify where the function \(g(x) = x[x^2]\) is discontinuous in the interval \((-2,2)\) and evaluate \(f(x) = \min\{\sqrt{2}x, x^2\}\) at those points.

1. First, determine when \(x[x^2]\) is discontinuous. The greatest integer function \([\,\cdot\,]\) returns the largest integer less than or equal to its argument. Discontinuities occur at points where \(x^2\) transitions from one integer to another.
2. Find the critical points in the range \(0 \leq x^2 < 4\) that cause changes in \([x^2]\). These occur at \(x^2 = 1, 2, 3\), corresponding to \(x = \pm1, \pm\sqrt{2}, \pm\sqrt{3}\).
3. Since \(x\) is restricted in \((-2,2)\), consider f(x) = \min\{\sqrt{2}x, x^2\} at each point:
   • \(f(-\sqrt{3}) = \min(-\sqrt{6}, 3) = -\sqrt{6}\)
   • \(f(-\sqrt{2}) = \min(-2, 2) = -2\)
   • \(f(-1) = \min(-\sqrt{2}, 1) = -\sqrt{2}\)
   • \(f(1) = \min(\sqrt{2}, 1) = 1\)
   • \(f(\sqrt{2}) = \min(2, 2) = 2\)
   • \(f(\sqrt{3}) = \min(\sqrt{6}, 3) = \sqrt{6}\)
4. Sum the contributions from all points: \(S = -\sqrt{6} - 2 - \sqrt{2} + 1 + 2 + \sqrt{6}\)
5. Simplify the expression: \(S = 1 - \sqrt{2}\)

Thus, the sum \(\sum_{x \in S} f(x)\) is \(2 - \sqrt{2}\).