MHT-CET Maths · Limits

Discontinuities of [x], |x| and sgn x — Counting the Points

The greatest-integer, modulus and sign functions carry built-in jumps — the question is where they land, whether another factor cancels them, and how many there are in a given interval.

All Limits notes MHT-CET Maths strategy

Why this matters

The smallest page in the chapter — 6 PYQs at 50% HARD — but a distinct question type with no coverage anywhere else: 'how many points of discontinuity' and 'discontinuous at which set'. It reuses the one-sided habit from the first page and the three-part test from the continuity pages; what is new is the counting, and the one case where a vanishing factor swallows a jump. One of the six carries an exam key that contradicts the mathematics, and knowing which is part of the preparation.

Concept 1 of 5

[x] Is Discontinuous at Every Integer — Counting Them

Intuition

The staircase

[x]

steps up by

1

at each integer and is flat in between. So its discontinuities are exactly the integers, and 'how many in

(a, b)

' is a counting question: how many integers lie strictly inside.

Definition

At an integer $n$ : left limit $n - 1$ , right limit $n$ , value $n$ . The left limit disagrees, so $[x]$ is discontinuous at every integer and continuous everywhere else.
$[x]$ is right-continuous at integers (right limit $=$ value). At a closed left endpoint that is an integer, there is no left approach, so no discontinuity there; at a closed right endpoint that is an integer, the left approach fails, so it counts.
Counting integers in an open interval $(a, b)$ : from $\lceil a \rceil$ (or $a + 1$ if $a$ is an integer) to $\lfloor b \rfloor$ (or $b - 1$ if $b$ is an integer). In $\left(-\frac{7}{2}, 100\right)$ : $-3, -2, \dots, 99$ — that is $3 + 1 + 99 = 103$ .
$[kx]$ jumps where $kx$ is an integer, i.e. at multiples of $\dfrac{1}{k}$ ; $\left[\dfrac{x}{2}\right]$ jumps at even integers.

Jumps of the greatest-integer function

\lim_{x\to n^-}[x] = n - 1 \ne n = [n] \quad (n \in \mathbb{Z}) \qquad [kx] \text{ jumps at } x = \frac{m}{k},\ m \in \mathbb{Z}

Worked example

How many points of discontinuity does

f(x) = [x]

have on

(-2.5, 6)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1LimitsMODERATE

The number of discontinuities of the greatest integer function

f(x)=[x]

x\in\left(-\frac{7}{2},100\right)

[Q142 · 9th May Shift 1 · 2024]

Miscounting the negative side

\left(-\frac{7}{2}, 100\right)

the integers are

-3, -2, -1

(three of them), then

0

, then

1

99

. Forgetting

0

, or including

100

, gives

102

104

— both offered.

Concept 2 of 5

Signum-Type Jumps: (x − a)/|x − a| and Products with It

Intuition

\dfrac{x - a}{|x - a|}

+1

on one side of

a

and

-1

on the other: a jump of size

2

that no choice of

f(a)

can bridge. Anything multiplied by it inherits the jump unless the other factor is

0

a

Definition

$\dfrac{x - a}{|x - a|} = \dfrac{|x - a|}{x - a} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases}$ ; it is undefined at $a$ and discontinuous there whatever value is assigned.
$\dfrac{p(x)}{|(x - 1)(x - 2)|}$ with $p$ divisible by $(x - 1)(x - 2)$ : the polynomial parts cancel, leaving sign factors at $1$ and $2$ . Each gives a jump, so $f$ is continuous on $\mathbb{R} - \{1, 2\}$ and nowhere else can be repaired.
A piecewise function built from $\dfrac{x - 3}{|x - 3|} + a$ on the left and $\dfrac{|x - 3|}{x - 3} + b$ on the right can be continuous — the sign factors evaluate to $-1$ and $+1$ , giving ordinary equations $-1 + a = 1 + b = f(3)$ .
$|x|$ itself is continuous everywhere; only $|x|$ divided by something vanishing at the same point produces a jump.

The sign factor

\frac{x - a}{|x - a|} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases} \qquad \text{jump of size } 2 \text{ at } a

Worked example

f(x) = \dfrac{x^2 - 1}{|x - 1|}

for

x \neq 1

. Can

f(1)

be chosen so that

f

is continuous at

1

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2LimitsMODERATE

Let

f(x) =\left\{ \begin{matrix} \frac{x^{4}- 5x^{2}+ 4}{|(x- 1)(x- 2)|} & ,x\neq 1,2 \\ 6 & ,x= 1 \\ 12 & ,x= 2 \end{matrix} \right.

.Then

f(x)

is continuous on the set

[Q102 · 26 April Shift II · 2025]

Cancelling |x − 1| against (x − 1)

\dfrac{x - 1}{|x - 1|}

is not

1

; it is

\pm 1

depending on the side. Cancelling as if the modulus were absent loses the jump and produces 'continuous on

\mathbb{R}

', which is always an option.

Concept 3 of 5

Piecewise with [x] and |x|: Check Every Join and Both Endpoints

Intuition

A function built from

[x]

and

|x|

on several intervals can fail at the seams between pieces AND at the integers inside each piece. List every candidate, test each with left/right/value, and count only the failures.

Definition

Candidates: every seam between pieces, every integer inside a piece that contains $[x]$ , and both endpoints of a closed domain.
At each candidate compute left limit, right limit and value using the correct piece for each side.
Endpoints: at a closed left endpoint only the right limit exists, so continuity there means right limit $=$ value; symmetrically at a closed right endpoint. $x + [x]$ at the right endpoint $3$ has left limit $3 + 2 = 5$ but value $6$ — discontinuous.
Two pieces that happen to agree at a seam ( $x + |x|$ and $x + [x]$ both give $4$ at $2$ ) make that seam continuous even though the formulas differ.
Report the count or the set, as asked; 'only three points' means exactly three.

Continuity at a closed endpoint

\text{left endpoint } a:\ \lim_{x\to a^+} f(x) = f(a) \qquad \text{right endpoint } b:\ \lim_{x\to b^-} f(x) = f(b)

Worked example

f(x) = [x] + |x|

[-1, 2)

. At how many points is

f

discontinuous?

Practice this conceptself-check

From the bank · past-year question

Example 3LimitsHARD

Let

f:[-1,3]\to\mathbb{R}

be defined as

f(x) = \begin{cases}|x|+[x],&-1\leq x<1\\x+|x|,&1\leq x<2\\x+[x],&2\leq x\leq3\end{cases}

where

[t]

denotes the greatest integer function. Then

f

is discontinuous at

[Q107 · 15th May Shift 1 · 2023]

Forgetting the closed right endpoint

[-1, 3]

with

f(x) = x + [x]

near

3

, the value

f(3) = 6

is not the left limit

5

. A student who only checks interior seams reports two points; the answer is three.

Concept 4 of 5

Composites of [x]: Where Does the Inner Function Cross an Integer?

Intuition

[g(x)]

jumps whenever

g(x)

crosses an integer. For

[x^2]

that happens at

x = \pm\sqrt{n}

; for

\left[x\left[\frac{x}{2}\right]\right]

you have to track both the inner staircase and the product it feeds into.

Definition

$[g(x)]$ is discontinuous at each $x$ where $g(x)$ passes through an integer value, provided $g$ is continuous and strictly monotone there. Where $g$ merely touches an integer without crossing, check separately.
$[x^2]$ on $(0, 2)$ : $x^2$ crosses $1, 2, 3$ at $x = 1, \sqrt2, \sqrt3$ — three jumps.
$[-x^2]$ uses $[-u] = -[u] - 1$ for non-integer $u$ : as $x^2 \to 9^-$ , $[x^2] = 8$ and $[-x^2] = -9$ ; as $x^2 \to 9^+$ , $[x^2] = 9$ and $[-x^2] = -10$ . So $[x^2] - [-x^2]$ has left limit $17$ and right limit $19$ at $x = 3$ — no $k$ makes it continuous.
$\left[x\left[\frac{x}{2}\right]\right]$ on $(-10, 10)$ : the inner $\left[\frac{x}{2}\right]$ is constant on each $[2m, 2m + 2)$ , so the product is $mx$ there, and $[mx]$ jumps at multiples of $\frac{1}{|m|}$ inside that block; the seams $x = 2m$ must also be tested. The bank's answer is $8$ points.

Jumps of a composite

[g(x)] \text{ jumps where } g(x) \in \mathbb{Z} \text{ and } g \text{ crosses it} \qquad [-u] = -[u] - 1 \ (u \notin \mathbb{Z})

Worked example

How many points of discontinuity does

f(x) = [x^2]

have on

(0, 2)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4LimitsHARD

Let

f(x) = \left[x\left[\frac{x}{2}\right]\right]

, for

-10 < x < 10

, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to

[Q147 · 9th May Shift 2 · 2023]

Answering the 'find k' question when no k exists

[x^2] - [-x^2]

3

has one-sided limits

17

and

19

. The exam key takes the right-hand value

19

; mathematically no

k

works. On the paper choose

19

; in your notes, know why the question is flawed.

Drill 1 more on composites of [x]: where does the inner function cross an integer?

Concept 5 of 5

When the Other Factor Vanishes at the Jump

Intuition

[x]

jumps by

1

n

. Multiply it by something that is

0

n

and continuous there, and the jump is multiplied by

0

: both one-sided limits become

0

, the value is

0

, and the product is continuous.

Definition

If $g$ is continuous at $n$ with $g(n) = 0$ , and $h$ is bounded near $n$ (any $[\cdot]$ or sign function is), then $g\cdot h \to 0 = g(n)h(n)$ : the product is continuous at $n$ .
$[x]\sin\pi x$ : $\sin\pi n = 0$ at every integer, so the product is continuous on all of $\mathbb{R}$ .
$[x]\cos\dfrac{(2x - 1)\pi}{2}$ : at $x = n$ , $\cos\dfrac{(2n - 1)\pi}{2} = 0$ — the cosine of an odd multiple of $\dfrac{\pi}{2}$ — so this too is continuous at every integer, hence everywhere.
If $g(n) \neq 0$ the jump survives: $[x](x - 1)(x - 2)$ is discontinuous at every integer except $1$ and $2$ .
The test is at each integer separately: which integers make the other factor zero?

A zero swallows a bounded jump

g \text{ continuous at } n,\ g(n) = 0,\ |h| \le M \text{ near } n \ \Rightarrow\ \lim_{x\to n} g(x)h(x) = 0 = g(n)h(n)

Worked example

Where is

f(x) = [x]\sin\pi x

discontinuous?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5LimitsMODERATE

The function

f(x)=[x]\cdot\cos\left(\dfrac{2x-1}{2}\pi\right)

, where

[\cdot]

denotes the greatest integer function, is discontinuous at

[Q119 · 14th May Shift 1 · 2024]

The exam key that contradicts the mathematics

For

[x]\cos\dfrac{(2x - 1)\pi}{2}

the official MHT-CET key marks 'all integer points', reasoning only that

[x]

jumps there. The cosine is

0

at every integer, so the function is in fact continuous everywhere. Know the correct mathematics; on that particular paper, the marked answer was the key's.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

[x] Is Discontinuous at Every Integer — Counting Them
Jumps of the greatest-integer function
$\lim_{x\to n^-}[x] = n - 1 \ne n = [n] \quad (n \in \mathbb{Z}) \qquad [kx] \text{ jumps at } x = \frac{m}{k},\ m \in \mathbb{Z}$
Signum-Type Jumps: (x − a)/|x − a| and Products with It
The sign factor
$\frac{x - a}{|x - a|} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases} \qquad \text{jump of size } 2 \text{ at } a$
Piecewise with [x] and |x|: Check Every Join and Both Endpoints
Continuity at a closed endpoint
$\text{left endpoint } a:\ \lim_{x\to a^+} f(x) = f(a) \qquad \text{right endpoint } b:\ \lim_{x\to b^-} f(x) = f(b)$
Composites of [x]: Where Does the Inner Function Cross an Integer?
Jumps of a composite
$[g(x)] \text{ jumps where } g(x) \in \mathbb{Z} \text{ and } g \text{ crosses it} \qquad [-u] = -[u] - 1 \ (u \notin \mathbb{Z})$
When the Other Factor Vanishes at the Jump
A zero swallows a bounded jump
$g \text{ continuous at } n,\ g(n) = 0,\ |h| \le M \text{ near } n \ \Rightarrow\ \lim_{x\to n} g(x)h(x) = 0 = g(n)h(n)$

Watch out for (5)

Miscounting the negative side
→ [x] Is Discontinuous at Every Integer — Counting Them
Cancelling |x − 1| against (x − 1)
→ Signum-Type Jumps: (x − a)/|x − a| and Products with It
Forgetting the closed right endpoint
→ Piecewise with [x] and |x|: Check Every Join and Both Endpoints
Answering the 'find k' question when no k exists
→ Composites of [x]: Where Does the Inner Function Cross an Integer?
The exam key that contradicts the mathematics
→ When the Other Factor Vanishes at the Jump

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