MHT-CET Maths · Sets, Relations and Functions

Sets, Relations and Types of Functions — One-One, Onto and the Greatest-Integer Equation

The definitions the chapter runs on: counting with sets and Cartesian products, whether a function is one-one and onto, and reading an equation in [x] as an interval.

All Sets, Relations and Functions notes MHT-CET Maths strategy

Why this matters

10 PYQs, none HARD — the vocabulary page, and the one with the most repeated stem in the chapter: [x]² − 5[x] + 6 = 0 has been set in three sittings with the same four options every time. The rest are a subsets-of-A×B count, a double-counting argument, two trig sets that turn out equal, a one-one/onto verdict on a linear-fractional function, and an identity in f(x + 1) − f(x). None needs more than the definition, applied once.

Concept 1 of 4

Sets and Cartesian Products: Counting Subsets and Double Counting

Intuition

A \times B

has

n(A)\,n(B)

ordered pairs, and a set with

m

elements has

2^m

subsets. 'At least

3

elements' is

2^m

minus the subsets with

0

1

2

elements.

Definition

$n(A \times B) = n(A)\cdot n(B)$ ; subsets of an $m$ -set: $2^m$ ; subsets with exactly $k$ elements: ${}^mC_k$ .
$n(A) = 4$ , $n(B) = 2$ : $A \times B$ has $8$ elements, $2^8 = 256$ subsets; with at least $3$ elements: $256 - (1 + 8 + 28) = 219$ .
Double counting: $300$ students each read $5$ papers, each paper read by $60$ students. Count (student, paper) pairs both ways: $300 \times 5 = 60 \times N$ , so $N = 25$ .
Equality of sets defined by conditions: $P = \{\theta : \sin\theta - \cos\theta = \sqrt2\cos\theta\}$ is $\tan\theta = \sqrt2 + 1$ ; $Q = \{\theta : \sin\theta + \cos\theta = \sqrt2\sin\theta\}$ is $\cot\theta = \sqrt2 - 1$ , i.e. $\tan\theta = \dfrac{1}{\sqrt2 - 1} = \sqrt2 + 1$ . Same condition, so $P = Q$ .
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ for two-set survey stems.

Counting with sets

n(A \times B) = n(A)\,n(B) \qquad \#\text{subsets} = 2^m \qquad n(A \cup B) = n(A) + n(B) - n(A \cap B)

Worked example

n(A) = 3

and

n(B) = 2

, how many subsets of

A \times B

have at least

2

elements?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Sets, Relations and FunctionsMODERATE

n(A)=4, n(B)=2

. Then the number of subsets of the set

A\times B

each having at least 3 elements are

[Q113 · 15th May Shift 1 · 2023]

Subtracting only the empty set

'At least

3

elements' removes the subsets of size

0

1

AND

2

1 + 8 + 28 = 37

. Removing only

1

gives

255

; removing

1 + 8

gives

247

Drill 2 more on sets and cartesian products: counting subsets and double counting

Concept 2 of 4

One-One and Onto: Test Injectivity by f(x₁) = f(x₂), Surjectivity by Solving for x

Intuition

One-one: different inputs never share an output — assume

f(x_1) = f(x_2)

and show

x_1 = x_2

. Onto: every value in the codomain is hit — solve

y = f(x)

for

x

and see which

y

fail.

Definition

$f(x) = \dfrac{2x + 3}{3x + 4}$ : cross-multiplying $f(x_1) = f(x_2)$ gives $x_1 = x_2$ , so one-one. Solving $y = f(x)$ : $x = \dfrac{3 - 4y}{3y - 2}$ , defined for all $y \ne \dfrac23$ — onto $\mathbb{R} - \{\tfrac23\}$ , not onto $\mathbb{R}$ .
A linear-fractional $\dfrac{ax + b}{cx + d}$ ( $ad \ne bc$ ) is always one-one on its domain, and misses exactly $y = \dfrac{a}{c}$ .
A strictly monotone function is one-one: $f(x) = x^3 + x - 1$ has $f'(x) = 3x^2 + 1 > 0$ , so it is increasing and has EXACTLY ONE real root.
Even functions and quadratics on $\mathbb{R}$ are not one-one ( $f(-x) = f(x)$ ); restricting the domain, as in $f(x) = (x+1)^2 - 1$ for $x \ge -1$ , makes them one-one and invertible.

One-one and onto

f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \ (\text{one-one}) \qquad \forall y\ \exists x: f(x) = y \ (\text{onto}) \qquad f' > 0 \Rightarrow \text{one-one}

Worked example

f : \mathbb{R} - \{1\} \to \mathbb{R}

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

, one-one? Is it onto?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Sets, Relations and FunctionsMODERATE

The function defined by

f(x) =\frac{2x+ 3}{3x+ 4},x\neq -\frac{4}{3}

[Q128 · 20 April Shift I · 2025]

Calling a linear-fractional function onto ℝ

It always misses

y = \dfrac{a}{c}

. Option (C) 'onto for

y \ne \frac23

and one-one' is the honest statement; 'only onto' and 'neither' are the traps.

Drill 1 more on one-one and onto: test injectivity by f(x₁) = f(x₂), surjectivity by solving for x

Concept 3 of 4

Equations in [x]: Solve for the Integer, Then Widen to the Interval

Intuition

[x] = k

means

k \le x < k + 1

— a half-open interval, closed on the left. Solve the equation for

[x]

as an ordinary unknown, then translate each integer solution into its interval and take the union.

Definition

$[x]^2 - 5[x] + 6 = 0 \Rightarrow ([x] - 2)([x] - 3) = 0 \Rightarrow [x] = 2$ or $3$ . $[x] = 2 \Rightarrow x \in [2, 3)$ ; $[x] = 3 \Rightarrow x \in [3, 4)$ . Union $[2, 4)$ .
$2[2x - 5] - 1 = 7 \Rightarrow [2x - 5] = 4 \Rightarrow 4 \le 2x - 5 < 5 \Rightarrow \dfrac92 \le x < 5$ .
Non-integer solutions for $[x]$ are discarded: $[x] = 2.5$ has no $x$ .
The left end is always included and the right end always excluded; every option list offers all four bracket combinations.

Greatest integer

[x] = k \iff k \le x < k + 1

Worked example

Solve

[x]^2 - 3[x] - 4 = 0

for

x

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Sets, Relations and FunctionsMODERATE

\lbrack x\rbrack^{2}- 5\lbrack x\rbrack + 6 = 0

, where

\lbrack \cdot \rbrack

denotes the greatest integer function, then

[Q114 · 26 April Shift I · 2025]

Closing the right end

[x] = 3

stops strictly before

4

: at

x = 4

[x] = 4

and

16 - 20 + 6 \ne 0

. The answer is

[2, 4)

, never

[2, 4]

Drill 3 more on equations in [x]: solve for the integer, then widen to the interval

Concept 4 of 4

Identities Like f(x + 1) − f(x) = 8x + 3: Compare Coefficients

Intuition

An identity holds for EVERY

x

, so after substituting the polynomial and simplifying, the coefficients of each power of

x

must match on both sides.

Definition

$f(x) = bx^2 + cx + d$ : $f(x + 1) - f(x) = b(2x + 1) + c = 2bx + (b + c)$ . Equal to $8x + 3$ : $2b = 8$ , $b + c = 3$ , so $b = 4$ , $c = -1$ ; $d$ is free.
A difference $f(x + 1) - f(x)$ lowers the degree by one — a quadratic's difference is linear — so the given right-hand side tells you the degree of $f$ .
The same move solves 'find $f$ given $f(g(x))$ ' on the composite page: match the shape, then the coefficients.

Comparing coefficients

f(x) = bx^2 + cx + d \Rightarrow f(x+1) - f(x) = 2bx + (b + c)

Worked example

f(x) = ax^2 + bx + 1

and

f(x + 1) - f(x) = 6x + 5

for all

x

, find

a

and

b

Practice this conceptself-check

From the bank · past-year question

Example 4Sets, Relations and FunctionsEASY

The values of

b

and

c

for which the identity

f(x + 1) - f(x) = 8x + 3

is satisfied, where

f(x) = bx^{2}+ cx + d

, are

[Q113 · 22 April Shift II · 2025]

Substituting one value of x

Putting

x = 0

gives one equation in two unknowns. An identity is matched coefficient by coefficient; that yields as many equations as there are unknowns.

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 (4)

Sets and Cartesian Products: Counting Subsets and Double Counting
Counting with sets
$n(A \times B) = n(A)\,n(B) \qquad \#\text{subsets} = 2^m \qquad n(A \cup B) = n(A) + n(B) - n(A \cap B)$
One-One and Onto: Test Injectivity by f(x₁) = f(x₂), Surjectivity by Solving for x
One-one and onto
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \ (\text{one-one}) \qquad \forall y\ \exists x: f(x) = y \ (\text{onto}) \qquad f' > 0 \Rightarrow \text{one-one}$
Equations in [x]: Solve for the Integer, Then Widen to the Interval
Greatest integer
$[x] = k \iff k \le x < k + 1$
Identities Like f(x + 1) − f(x) = 8x + 3: Compare Coefficients
Comparing coefficients
$f(x) = bx^2 + cx + d \Rightarrow f(x+1) - f(x) = 2bx + (b + c)$

Watch out for (4)

Subtracting only the empty set
→ Sets and Cartesian Products: Counting Subsets and Double Counting
Calling a linear-fractional function onto ℝ
→ One-One and Onto: Test Injectivity by f(x₁) = f(x₂), Surjectivity by Solving for x
Closing the right end
→ Equations in [x]: Solve for the Integer, Then Widen to the Interval
Substituting one value of x
→ Identities Like f(x + 1) − f(x) = 8x + 3: Compare Coefficients

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