Question ID: #766
Let [.] denote the greatest integer function, and let $f(x)=\min\{\sqrt{2}x,x^{2}\}$. Let $S=\{x\in(-2,2)$: the function $g(x)=|x|[x^{2}]$ is discontinuous at x}. Then $\sum_{x\in S}f(x)$ equals:
- (1) $2-\sqrt{2}$
- (2) $2\sqrt{6}-3\sqrt{2}$
- (3) $1-\sqrt{2}$
- (4) $\sqrt{6}-2\sqrt{2}$
Solution:
The function is $g(x) = |x|[x^2]$.
Possible points of discontinuity in $(-2, 2)$ are where $x^2$ is an integer.
$x^2 \in \{1, 2, 3\}$.
So, $x \in \{\pm 1, \pm \sqrt{2}, \pm \sqrt{3}\}$.
Check continuity at $x=0$: $\lim_{x \to 0} |x|[x^2] = 0 \times 0 = 0 = g(0)$. Continuous.
Let’s check the other points:
At $x = 1, -1$: $[x^2]$ jumps from 0 to 1. $g(x)$ jumps. Discontinuous.
At $x = \sqrt{2}, -\sqrt{2}$: $[x^2]$ jumps from 1 to 2. $g(x)$ jumps. Discontinuous.
At $x = \sqrt{3}, -\sqrt{3}$: $[x^2]$ jumps from 2 to 3. $g(x)$ jumps. Discontinuous.
So, $S = \{-\sqrt{3}, -\sqrt{2}, -1, 1, \sqrt{2}, \sqrt{3}\}$.
We need to sum $f(x) = \min\{\sqrt{2}x, x^2\}$ for these values.
Since $x^2$ is always positive, and $\sqrt{2}x$ is negative for negative $x$, for negative $x$, $\min$ is $\sqrt{2}x$.
For positive $x$, we compare $\sqrt{2}x$ and $x^2$.
$\sqrt{2}x < x^2 \Rightarrow \sqrt{2} < x \Rightarrow x > 1.414$.
Calculating $f(x)$ for each point:
1. $x = -\sqrt{3}: f(x) = \min(-\sqrt{6}, 3) = -\sqrt{6}$
2. $x = -\sqrt{2}: f(x) = \min(-2, 2) = -2$
3. $x = -1: f(x) = \min(-\sqrt{2}, 1) = -\sqrt{2}$
4. $x = 1: f(x) = \min(\sqrt{2}, 1) = 1$ (since $1 < 1.414$) 5. $x = \sqrt{2}: f(x) = \min(2, 2) = 2$ 6. $x = \sqrt{3}: f(x) = \min(\sqrt{6}, 3) = \sqrt{6}$ (since $\sqrt{6} \approx 2.45 < 3$)
Sum $= (-\sqrt{6} – 2 – \sqrt{2}) + (1 + 2 + \sqrt{6})$
Sum $= -\sqrt{6} – 2 – \sqrt{2} + 3 + \sqrt{6}$
Sum $= 1 – \sqrt{2}$
Ans. (3)
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