Sets, Relations and Functions — MHT-CET Maths Formulas and Traps

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

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)

Common traps

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

.

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.

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]

.

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.

Domain and Range — Where a Formula Is Defined and What It Produces

Learn this subtopic in the notes

Domain: Write One Condition Per Piece, Then Intersect

Standard conditions

\sqrt{u}: u \ge 0 \quad \frac{1}{\sqrt{u}}: u > 0 \quad \log u: u > 0 \quad \sin^{-1}u: -1 \le u \le 1 \quad \frac{1}{u}: u \ne 0

Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x|

Clearing a positive denominator

\frac{p(x)}{q(x)} \le 1,\ q(x) > 0 \iff p(x) \le q(x)

Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x

Discriminant method

y = f(x) \ \Rightarrow\ a(y)x^2 + b(y)x + c(y) = 0 \ \Rightarrow\ b(y)^2 - 4a(y)c(y) \ge 0

Common traps

Closing the bracket at a root in the denominator

\sqrt{9 - x^2}

underneath means

9 - x^2 > 0

, strictly.

[2, 3]

is option (C) on both sittings; the answer is

[2, 3)

.

Solving x² − x − 4 ≥ 0 as x ≥ the smaller root

x^2 - x - 4 \ge 0

holds OUTSIDE the roots. With

x \ge 0

imposed by the

|x|

split, only

x \ge \dfrac{1 + \sqrt{17}}{2}

survives;

\dfrac{\sqrt{17} - 1}{2}

is the wrong sign's root, option (B).

Guessing the bracket from the shape

\dfrac{x^2 + x + 2}{x^2 + x + 1}

attains its maximum

\dfrac73

but never its infimum

1

, so the range is

\left(1, \dfrac73\right]

; every other bracket pairing is on the list. Test each endpoint.

Composite Functions — f∘g, Iteration and Functional Identities

Learn this subtopic in the notes

Evaluate a Composite at a Point: Innermost First

Composition

(f \circ g)(x) = f(g(x)) \qquad \text{evaluate inside} \to \text{outside}

Composite as a Formula: Substitute in Two Steps, Then Simplify

Two-step composition

(g \circ g \circ f)(x) = g\big(g(u)\big)\Big|_{u = f(x)}

Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x

Shape matching

f(g(x)) = \Phi\big(g(x)\big) \ \Rightarrow\ f = \Phi

Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin

The two identities

f(x) = \log\frac{1 - x}{1 + x}:\quad f\!\left(\frac{2x}{1 + x^2}\right) = 2f(x),\qquad f\!\left(\frac{3x + x^3}{1 + 3x^2}\right) = 3f(x)

Common traps

Reading f(g(g(f(x)))) as (f∘g)² or as f²g²

It is a four-step chain, evaluated one function at a time from the inside. Option (A)

4

is

f(g(1))

-style short-cutting; the chain gives

5

.

Matching α² = 1 alone

\alpha = 1

satisfies

\alpha^2 = 1

but not

\alpha + 1 = 0

;

f(f(x)) = x

needs BOTH coefficient equations, so

\alpha = -1

, option (D).

Solving for f(x) when only f(2) is asked

Finding

f

as a formula from

(g \circ f)(x) = 4x^2 - 10x + 5

needs a square root of a quadratic. One substitution,

x = 2

, gives a quadratic in the number

f(2)

.

Reading the sign of the ratio backwards

\dfrac{1 - u}{1 + u}

with

u = \dfrac{2x}{1 + x^2}

is

\left(\dfrac{1 - x}{1 + x}\right)^2

, positive power. Option (B)

-2f(x)

is the inverted ratio.

Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

Learn this subtopic in the notes

Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut

Linear-fractional inverse

f(x) = \frac{ax + b}{cx + d} \ \Rightarrow\ f^{-1}(x) = \frac{dx - b}{-cx + a}

Self-Inverse: f(f(x)) = x Fixes the Parameter

Involution test

f(f(x)) = x \ \iff\ f = f^{-1} \qquad \frac{ax + b}{cx + d} \text{ self-inverse} \iff a + d = 0

Inverse With a Square Root: Choose the Branch From the Domain

Order of inverses

(f \circ g)^{-1} = g^{-1} \circ f^{-1}

Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x

Fixed points

f \text{ increasing}:\quad f(x) = f^{-1}(x) \iff f(x) = x

Common traps

Taking the reciprocal

f^{-1}

is not

\dfrac1f

.

\dfrac{3x - 4}{2x - 3}

-type options are the reciprocal; the inverse swaps and negates the coefficients instead.

Solving f(x) = x instead of f(f(x)) = x

Fixed points of

f

are not the same as

f

being an involution. The parameter comes from the identity

f(f(x)) = x

holding for EVERY

x

.

Keeping the '−' branch

\dfrac{x - \sqrt{x^2 - 4}}{2}

is option (C), and it is at most

1

for

x \ge 2

— it lands OUTSIDE the domain

[1, \infty)

except at the endpoint. The domain chooses the sign.

Including the complex roots

Option (C) lists two non-real numbers alongside

0

and

-1

. The set is real-valued and domain-restricted; the fixed-point equation

x^2 + x = 0

has exactly the two real solutions.

Sets, Relations and Functions formulas

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

Sets and Cartesian Products: Counting Subsets and Double Counting

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

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

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

Common traps

Domain and Range — Where a Formula Is Defined and What It Produces

Domain: Write One Condition Per Piece, Then Intersect

Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x|

Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x

Common traps

Composite Functions — f∘g, Iteration and Functional Identities

Evaluate a Composite at a Point: Innermost First

Composite as a Formula: Substitute in Two Steps, Then Simplify

Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x

Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin

Common traps

Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut

Self-Inverse: f(f(x)) = x Fixes the Parameter

Inverse With a Square Root: Choose the Branch From the Domain

Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x

Common traps

More MHT-CET Maths formula sheets