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

The function \[ g(x) = x[x^2] \] is discontinuous when \(x^2\) is an integer.

Step 1: Find points of discontinuity.
In the interval \( (-2,2) \), \[ x^2 = 1 \Rightarrow x = \pm 1. \] Hence, \[ S = \{-1,\,1\}. \]

Step 2: Evaluate \( f(x) \) at these points.
For \(x = 1\), \[ f(1) = \min\{\sqrt{2},1\} = 1. \] For \(x = -1\), \[ f(-1) = \min\{-\sqrt{2},1\} = -\sqrt{2}. \]

Step 3: Compute the sum.
\[ \sum_{x \in S} f(x) = 1 - \sqrt{2} = 2 - \sqrt{2}. \]

Final Answer: \[ \boxed{2 - \sqrt{2}} \]