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Lesson 16 min· Section 1 of 4

The straight-up lesson, built for the exam.

Functions

Formula sheet

Everything for this chapter, in one place.

Composition

Meaning(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)) — inner function first
Orderf(g(x))≠g(f(x))f(g(x)) \neq g(f(x)) in general
Domain of f(g(x))f(g(x))xx must be allowed in gg, and g(x)g(x) must be allowed in ff

Functional equations and what they force

f(x+y)=f(x)f(y)f(x+y) = f(x)f(y)f(x)=axf(x) = a^x, so f(n)=f(1)nf(n) = f(1)^n
f(xy)=f(x)f(y)f(xy) = f(x)f(y)multiplicative; always f(1)=1f(1) = 1 (or 0)
f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y)f(x)=cxf(x) = cx
f(x)+k f ⁣(1x)=g(x)f(x) + k\,f\!\left(\frac{1}{x}\right) = g(x)write it again with x→1xx \to \frac{1}{x} and solve the two together
f(x+2)=f(x)+f(x+1)f(x+2) = f(x) + f(x+1)a recurrence — step forward from two known values
f(xy)=f(x)f(y)+f(x)+f(y)f(xy) = f(x)f(y) + f(x) + f(y)set g=f+1g = f+1; then gg is multiplicative

Useful substitutions

x=y=1x = y = 1pins down f(1)f(1)
y=xy = xturns a two-variable law into one variable
x→1xx \to \frac{1}{x}pairs with the original to eliminate the unwanted term
A prime factorisationa multiplicative ff is determined by its values on primes

Symmetry

f(a+x)=f(a−x)f(a+x) = f(a-x) for all xxthe graph is symmetric about x=ax=a
So the roots pair upeach pair sums to 2a2a
With 2k2k rootstheir total is 2ak2ak

The larger or smaller of two curves

min⁡{u(x),v(x)}\min\{u(x), v(x)\}its maximum sits where the two curves cross
max⁡{u(x),v(x)}\max\{u(x), v(x)\}its minimum sits where they cross
Either waysolve u(x)=v(x)u(x) = v(x) and substitute

Range of a rational function

Methodset y=numdeny = \dfrac{\text{num}}{\text{den}}, cross-multiply into a quadratic in xx
Thenreal xx needs discriminant ≥0\ge 0 — that inequality is the range
Alternativesubstitute for a repeated block and track what it can be

Greatest integer, [x][x]

Meaningthe largest integer not exceeding xx
So[x]=n  ⟺  n≤x<n+1[x] = n \iff n \le x < n+1
Summing [ ⋅ ][\,\cdot\,] over a rangegroup the terms that share a value, then count each group
⌊n⌋=k\lfloor\sqrt{n}\rfloor = kfor the 2k+12k+1 values n=k2n = k^2 to k2+2kk^2+2k

Why this chapter matters

0.8
questions in a typical Quant section
15th
of 25 chapters

≈ 2 of 66 marks a section ride on it.

Asked in20 of 24 papers
20172025
Year by yearper section
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  7. 2023: 0.3 per section
  8. 2024: 1 per section
  9. 2025: 1 per section
Question type23 asked
MCQ 52%Type-in 48%

Type-in answers carry no negative marking.

From every Functions question in CAT papers since 2017.