When is the greatest integer function [x] discontinuous?

[x] is discontinuous at all integers. So [x²] is discontinuous where x² is an integer, i.e., at x = 0, ±1, ±√2, ±√3, ±2, ...

Is g(x) = |x|[x²] discontinuous at x=0?

No. Near x=0, [x²]=0 so g(x)=|x|·0=0 from both sides. g(0)=0. Hence g is continuous at x=0.

How to check discontinuity of a product?

A product f·g can be continuous even if g has a jump, if f=0 at that point. At x=0: |x|→0, so the jump in [x²] is suppressed.

What is S for this problem?

S = {-√3, -√2, -1, 1, √2, √3} — all six points in (-2,2) where [x²] jumps and |x|≠0.

How to evaluate f(x) = min{√2·x, x²} for negative x?

For x 0, so min = √2·x (the negative value). For example, f(-1) = min{-√2, 1} = -√2.

How to find min{√2·x, x²} for x=1?

f(1) = min{√2, 1} = 1 since √2≈1.414 > 1.

How to find min{√2·x, x²} for x=√3?

f(√3) = min{√2·√3, 3} = min{√6, 3} = √6 ≈ 2.449 < 3.

Why is the answer negative?

For negative x in S, f(x) = √2·x < 0. The negative contributions (-√2 from x=-1) outweigh after cancellations, giving final sum 1-√2 < 0.

Let f(x) = min{√2·x, x²} and g(x) = |x|[x²] discontinuous set S in (-2,2), then Σf(x)

Q: Why do √6 terms cancel?

f(-√3) = -√6 and f(√3) = √6. They add to zero. Similarly the -2 and +2 cancel. Only 1-√2 remains.

Q: Why is the answer negative?

For negative x in S, f(x) = √2·x < 0. The negative contributions (-√2 from x=-1) outweigh after cancellations, giving final sum 1-√2 < 0.

Let f(x) = min{√2·x, x²} and g(x) = |x|[x²] discontinuous set S in (-2,2), then Σf(x) | JEE Main Mathematics

NEETJEE

Math Functions GIF

Q MCQ GIF

Let $[\cdot]$ 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 $\displaystyle\sum_{x \in S} f(x)$ equals

A) $2 - \sqrt{2}$

B) $2\sqrt{6} - 3\sqrt{2}$

C) $1 - \sqrt{2}$ ✓

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

✅ Correct Answer

C) 1 − √2

✓

Solution Steps

Identify where [x²] is discontinuous

$[t]$ (GIF) is discontinuous at every integer. So $[x^2]$ jumps where $x^2$ is an integer.

In $(-2,2)$: $x^2 \in \{0,1,2,3\}$ gives $x \in \{0, \pm1, \pm\sqrt{2}, \pm\sqrt{3}\}$.

Check x = 0 separately

$g(x) = |x|\cdot[x^2]$. Near $x=0$: $[x^2] = 0$, so $g(x) = |x| \cdot 0 = 0$.

Both limits = 0 = g(0). So $g$ is continuous at x = 0. Exclude from S.

Check ±1, ±√2, ±√3 for discontinuity

At each $x = \pm\sqrt{n}$ (n = 1,2,3), $|x| = \sqrt{n} \neq 0$, so the jump in $[x^2]$ is not suppressed.

Example at $x=1$: left limit $= 1 \cdot 0 = 0$, right limit $= 1 \cdot 1 = 1$. Jump exists. ✗

All six points are discontinuities. Therefore:

$$S = \{-\sqrt{3},\; -\sqrt{2},\; -1,\; 1,\; \sqrt{2},\; \sqrt{3}\}$$

Evaluate f(x) = min{√2·x, x²} at each point

For $x < 0$: $\sqrt{2}\,x < 0 < x^2$, so $f(x) = \sqrt{2}\,x$.

For $x > 0$: Compare $\sqrt{2}\,x$ vs $x^2$. They're equal when $x = \sqrt{2}$.

For $0 < x < \sqrt{2}$: $x^2 < \sqrt{2}\,x$, so $f(x) = x^2$.

For $x > \sqrt{2}$: $x^2 > \sqrt{2}\,x$, so $f(x) = \sqrt{2}\,x$.

Compute f at each element of S

x	√2·x	x²	f(x) = min
$-\sqrt{3}$	$-\sqrt{6}$	$3$	$-\sqrt{6}$
$-\sqrt{2}$	$-2$	$2$	$-2$
$-1$	$-\sqrt{2}$	$1$	$-\sqrt{2}$
$1$	$\sqrt{2}$	$1$	$1$
$\sqrt{2}$	$2$	$2$	$2$
$\sqrt{3}$	$\sqrt{6}$	$3$	$\sqrt{6}$

Sum all f(x) values

$$\sum_{x \in S} f(x) = -\sqrt{6} + (-2) + (-\sqrt{2}) + 1 + 2 + \sqrt{6}$$

$= \underbrace{(-\sqrt{6} + \sqrt{6})}_{0} + \underbrace{(-2 + 2)}_{0} + 1 - \sqrt{2}$

$$= \boxed{1 - \sqrt{2}}$$

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Key Insight

At x=0, |x|=0 kills the jump in [x²] → continuous. Symmetric ±√3 and ±√2 cancel. Only −√2 from x=−1 survives, giving 1−√2.

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Theory

1. Greatest Integer Function Discontinuity

$[x]$ is discontinuous at every integer $n$: the right limit is $n$ while the left limit is $n-1$. So $[x^2]$ is discontinuous wherever $x^2$ is an integer, i.e., at $x = 0, \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2, \ldots$ In any interval $(-a,a)$, these are finitely many points. Always list them and check each individually.

2. Product Continuity at Jump Points

If $h(x) = p(x) \cdot q(x)$ and $q(x)$ has a jump discontinuity at $x_0$ but $p(x_0) = 0$, then $h$ may still be continuous. At $x=0$: $|x| \to 0$ and $[x^2]$ jumps from $-1$ to $0$, but $|0| = 0$ suppresses the jump. Hence $g(x) = |x|[x^2]$ is continuous at $x=0$ even though $[x^2]$ jumps there.

3. Min of Two Functions

$f(x) = \min\{u(x), v(x)\}$ equals whichever is smaller. Find where $u(x) = v(x)$ (crossover point), then determine which is smaller on each side. Here $\sqrt{2}\,x = x^2 \Rightarrow x = 0$ or $x = \sqrt{2}$. For $0 < x < \sqrt{2}$: $x^2 < \sqrt{2}\,x$ so $f=x^2$. For $x > \sqrt{2}$: $\sqrt{2}\,x < x^2$ so $f = \sqrt{2}\,x$. For $x < 0$: $\sqrt{2}\,x < 0 < x^2$ so $f = \sqrt{2}\,x$.

4. Symmetric Cancellation Strategy

When $S$ has symmetric pairs $\{a, -a\}$, check if $f(a) + f(-a)$ simplifies. Here $f(\sqrt{3}) + f(-\sqrt{3}) = \sqrt{6} + (-\sqrt{6}) = 0$ and $f(\sqrt{2}) + f(-\sqrt{2}) = 2 + (-2) = 0$. Only the non-cancelling pair $f(1) + f(-1) = 1 + (-\sqrt{2})$ remains.

5. Counting Discontinuities in Open Intervals

Always work on the open interval carefully. $x = \pm2$ satisfy $x^2 = 4$ (integer), but $\pm2 \notin (-2,2)$ (open interval), so exclude them. $x=0$ is in the interval but g is continuous there. So $|S| = 6$: $\{-\sqrt{3}, -\sqrt{2}, -1, 1, \sqrt{2}, \sqrt{3}\}$. Careful boundary analysis prevents errors.

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FAQs

When is [x²] discontinuous?

⌄

Whenever x² is an integer. In (-2,2): x = 0, ±1, ±√2, ±√3. At x=±2, not in open interval.

Why is g continuous at x=0?

⌄

g(x) = |x|·[x²]. Near 0, [x²]=0 so g=0 from both sides. The factor |x|→0 kills the GIF jump.

What is S?

⌄

S = {-√3, -√2, -1, 1, √2, √3}. Six points where g has a genuine jump (|x|≠0 at each point).

How is f(x) = min{√2·x, x²} evaluated for x<0?

⌄

For x<0: √2·x is negative, x²>0. So min = √2·x. e.g., f(-1) = min{-√2, 1} = -√2.

Why do √6 terms cancel?

⌄

f(√3)=√6 and f(-√3)=-√6. Sum = 0. Similarly f(√2)=2 and f(-√2)=-2. Sum = 0. Cancellation leaves 1-√2.

How to evaluate f(1)?

⌄

f(1) = min{√2·1, 1²} = min{√2, 1} = 1 since √2≈1.414 > 1. So the parabola x² wins.

How to evaluate f(√2)?

⌄

f(√2) = min{√2·√2, (√2)²} = min{2, 2} = 2. Both are equal at x=√2 (crossover point).

Why is the final answer negative?

⌄

After cancellations, only f(1)+f(-1) = 1+(-√2) = 1-√2 ≈ -0.414 remains. Negative x values contribute negative f values.

How to verify the answer is C and not A?

⌄

Option A: 2-√2 ≈ 0.586. Option C: 1-√2 ≈ -0.414. Our sum = 0+0+(1-√2) = 1-√2. Option C confirmed.

What is the crossover between √2·x and x²?

⌄

Set √2·x = x²: x(x-√2) = 0, so x=0 or x=√2. For 0 < x < √2, x² < √2·x. For x > √2, x² > √2·x.

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