Let [·] denote the greatest integer function, and let f(x) = min{√2x, x²}. Let S = {x ∈ (-2, 2) : th

Q: Wait, why is $x=0$ not in the set $S$ if $[x^2]$ is discontinuous at $0$?

That is a brilliant observation! You are right that $[x^2]$ has a jump at $x=0$. However, $g(x)$ is the product $|x| [x^2]$. As $x \to 0$, $|x| \to 0$ and $[x^2]$ is bounded (specifically, it is $0$ for $x^2 \in [0, 1)$). The product of a function approaching zero and a bounded function is always zero. Since the limit matches the function value $g(0)=0$, the function is continuous there. The $|x|$ factor 'heals' the jump!

Q: How do I handle the $\min$ function when summing $f(x)$?

Think of $f(x) = \min\{\sqrt{2}x, x^2\}$ as a path that follows the lower of the two curves. For $x 0$, the parabola starts below the line but intersects it at $x = \sqrt{2}$. So, for $x \in (0, \sqrt{2})$, the parabola is the minimum, and for $x > \sqrt{2}$, the line is the minimum. Just check which curve is lower at each specific point in $S$.

Q: Do I really need to check all six points in $S$?

Yes, but don't let the number of points intimidate you! Once you identify the set $S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$, the evaluation becomes a simple arithmetic exercise. You will notice that many terms cancel out beautifully, which is a hallmark of well-designed JEE problems. Stay patient and methodical!

Question

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

Options:

(A) $\sqrt{6} - 2\sqrt{2}$

(B) $1 - \sqrt{2}$

(C) $2 - \sqrt{2}$

(D) $2\sqrt{6} - 3\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the functions $f(x) = \min\{\sqrt{2}x, x^2\}$ and $g(x) = |x| [x^2]$. Our objective is to find the sum of $f(x)$ over the set $S$, where $S$ is the set of points where $g(x)$ is discontinuous within the domain $x \in (-2, 2)$.

### The Staircase Trap

The function $g(x) = |x| [x^2]$ is governed by the greatest integer function $[x^2]$. This function exhibits jump discontinuities whenever the argument $x^2$ is an integer.

Given $x \in (-2, 2)$, the range of $x^2$ is $[0, 4)$. The integers within this range are $0, 1, 2,$ and $3$. Solving $x^2 = k$ for $k \in \{0, 1, 2, 3\}$ yields the candidate points:
$$x \in \{0, \pm 1, \pm \sqrt{2}, \pm \sqrt{3}\}$$

### The Zero-Product Savior

We must verify if these points are truly discontinuities. Consider the point $x = 0$:
$$\lim_{x 	o 0} |x| [x^2] = 0 \cdot [0] = 0$$
Since $g(0) = |0|[0] = 0$, the function is **continuous** at $x = 0$ because the vanishing factor $|x|$ suppresses the jump in $[x^2]$.

For all other points in our set, $|x| 
eq 0$. Consequently, the jump in the greatest integer function $[x^2]$ is preserved. Thus, the set of discontinuities is:
$$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$$

### Evaluating the Minimum Function

The function $f(x) = \min\{\sqrt{2}x, x^2\}$ selects the smaller value between the line $y = \sqrt{2}x$ and the parabola $y = x^2$. We evaluate $f(x)$ for each $x \in S$:

1. For $x = -1$: $f(-1) = \min\{-\sqrt{2}, 1\} = -\sqrt{2}$
2. For $x = 1$: $f(1) = \min\{\sqrt{2}, 1\} = 1$
3. For $x = -\sqrt{2}$: $f(-\sqrt{2}) = \min\{-2, 2\} = -2$
4. For $x = \sqrt{2}$: $f(\sqrt{2}) = \min\{2, 2\} = 2$
5. For $x = -\sqrt{3}$: $f(-\sqrt{3}) = \min\{-\sqrt{6}, 3\} = -\sqrt{6}$
6. For $x = \sqrt{3}$: $f(\sqrt{3}) = \min\{\sqrt{6}, 3\} = \sqrt{6}$

### The Grand Summation

We now calculate the sum of these values:
$$\sum_{x \in S} f(x) = (-\sqrt{2}) + (1) + (-2) + (2) + (-\sqrt{6}) + (\sqrt{6})$$

Observing the terms, we see that $-2$ cancels with $2$, and $-\sqrt{6}$ cancels with $\sqrt{6}$. This leaves us with the final result:
$$\sum_{x \in S} f(x) = 1 - \sqrt{2}$$

The final answer is **$1 - \sqrt{2}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[\cdot]$ denote the greatest integer function, and let $f (x) = min {2 x, x^{2}}$ . 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

Select Answer:

Visualized Solution

Analyze the function $g (x)$

Identify Potential Discontinuity Points

Check Continuity at $x = 0$

Define the Set $S$

Introduce $f (x) = min {2 x, x^{2}}$

Visualize the Minimum Function

Evaluate $f (\pm 1)$

Evaluate $f (\pm 2)$

Evaluate $f (\pm 3)$

Calculate the Final Sum

Final Answer

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Staircase Trap

The Zero-Product Savior

Evaluating the Minimum Function

The Grand Summation

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Select Answer:

Visualized Solution

Analyze the function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Identify Potential Discontinuity Points

Check Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Define the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Introduce .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=min{2​x,x2}

Visualize the Minimum Function

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(±1)

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(±2​)

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(±3​)