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Question

(C) The function $g(x) = |x|[x^2]$ is discontinuous where $[x^2]$ is discontinuous, provided $|x| 
eq 0$.
$[x^2]$ is discontinuous at $x^2 \in \math

Vedclass · Accepted Answer

$1-\sqrt{2}$

Vedclass · Answer

(C) The function $g(x) = |x|[x^2]$ is discontinuous where $[x^2]$ is discontinuous, provided $|x| 
eq 0$.
$[x^2]$ is discontinuous at $x^2 \in \mathbb{Z}$.
For $x \in (-2, 2)$, $x^2 \in [0, 4)$.
The integers in $[0, 4)$ are $0, 1, 2, 3$.
Thus, $x^2 = 1, 2, 3$ gives $x = \pm 1, \pm \sqrt{2}, \pm \sqrt{3}$.
At $x=0$, $g(x) = |x|[x^2] = 0 \cdot [0] = 0$, and $\lim_{x 	o 0} |x|[x^2] = 0$, so $g(x)$ is continuous at $x=0$.
Thus, $S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$.
Now, $f(x) = \min \{\sqrt{2}x, x^2\}$.
$f(-1) = \min \{-\sqrt{2}, 1\} = -\sqrt{2}$.
$f(1) = \min \{\sqrt{2}, 1\} = 1$.
$f(-\sqrt{2}) = \min \{-2, 2\} = -2$.
$f(\sqrt{2}) = \min \{2, 2\} = 2$.
$f(-\sqrt{3}) = \min \{-\sqrt{6}, 3\} = -\sqrt{6}$.
$f(\sqrt{3}) = \min \{\sqrt{6}, 3\} = \sqrt{6}$.
Summing these values: $\sum_{x \in S} f(x) = -\sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} = 1 - \sqrt{2}$.

Let $[\bullet]$ 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 $\sum_{x \in S} f(x)$ equals:

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