MathematicsRelations and FunctionsJEE

Relations and Functions · Part 16 of 20

Greatest Integer, Fractional Part, Signum and Max-Min

Functions with jumps and corners: the greatest integer [x], the fractional part {x}, the signum function and max-min envelopes, with their graphs, properties and equations.

Builds on: Part 11 · Modulus Function.

Video coming soon

Greatest integer

[x]: a staircase

[x] = n on [n, n + 1): filled dot left, hollow right. [2.7] = 2 but [−2.3] = −3. x − 1 < [x] ≤ x, and [x] + [−x] = 0 or −1.

Sawtooth graph. — Each tooth: closed at the left, open at the right.

Fractional part

{x} = x − [x]

A sawtooth in [0, 1) with period 1. {−2.3} = 0.7, not 0.3. {x} + {−x} = 0 or 1.

Signum

sgn(x)

1 for x > 0, 0 at x = 0, −1 for x < 0; sgn(x) = |x|/x for x ≠ 0. Range {−1, 0, 1}.

Max and min

Envelopes

max{x, x²} is the upper envelope, min{x, x²} the lower. $max (a, b) = \frac{a + b + ∣ a - b ∣}{2}$ , $min (a, b) = \frac{a + b - ∣ a - b ∣}{2}$ .

Summary

Key formulas

x - 1 < [x] \leq x

Greatest integer function

[x] \geq n ⟺ x \geq n

Greatest integer function

[x] < n ⟺ x < n

Greatest integer function

sgn x = 1, 0, - 1 for x > 0, x = 0, x < 0

Signum function

sgn x = \frac{∣ x ∣}{x}, x \neq = 0

Signum function

max (a, b) = \frac{a + b + ∣ a - b ∣}{2}

Max-min of functions

min (a, b) = \frac{a + b - ∣ a - b ∣}{2}

Max-min of functions

Worked examples

From the video

1. Solve [x]² − 5[x] + 6 = 0.

[x] = 2 or 3, so x ∈ [2, 4).

2. Domain of 1/√([x]² − [x] − 6).

(−∞, −2) ∪ [4, ∞).

JEE-style question

Your turn

The solution set of [x]² + [x] − 2 = 0 ([·] = greatest integer function) is:

(a){−2, 1}

(b)[−2, −1) ∪ [1, 2)

(c)[−2, 2)

(d)(−2, −1] ∪ (1, 2]

Show the answer and the traps

([x] + 2)([x] − 1) = 0, so [x] = −2 or 1: x ∈ [−2, −1) ∪ [1, 2). Option (b).

(a) solves for x instead of [x]. (c) joins the two pieces, but [x] = 0 or −1 doesn't work.

(d) puts the filled and hollow ends the wrong way round.

Watch out

Common mistakes

[−2.3] = −2[x] rounds down: [−2.3] = −3.

{−2.3} = 0.3{−2.3} = −2.3 − (−3) = 0.7.

Solving for x instead of [x]Solve for [x], then turn each value into an interval.

Practice

Try these

1. Find [−3.7], {−3.7} and [π] + [−π].

−4; 0.3; −1.

2. Solve [x] = 2.5.

No solution: [x] is always an integer.

3. Solve 2[x] = x + {x}.

x = 0 or x = 3/2.

4. Domain of 1/√([x] − 1).

[2, ∞).

5. Minimum value of max{x, 1 − x}.

1/2, at x = 1/2.