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Mathematics I

BSMA1001 · 11 topics · 19 min read

0/11 revised

Week 1Most papers

Number systems & sets

Number systems

Chain: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ
IITM lectures take ℕ = {0, 1, 2, …}. Some questions start at 1, so read the definition given.
Rational: pq, p, q ∈ ℤ, q ≠ 0
Decimal expansion terminates or recurs.
Irrational: √2, √3, π, e
Non-terminating, non-recurring. √n is irrational unless n is a perfect square.
gcd × lcm: gcd(a, b) × lcm(a, b) = a × b
From prime factors: gcd = lowest powers of common primes · lcm = highest powers of all primes

Sets (|A| = n)

Power set: |P(A)| = 2ⁿ
Proper subsets: 2ⁿ − 1
Non-empty proper subsets: 2ⁿ − 2
Subsets of size k: C(n, k) = n!k!(n − k)!
Cartesian product: |A × B| = |A| · |B|
A × B ≠ B × A unless A = B or one of them is empty.
Difference: A \ B = A ∩ B^c, |A \ B| = |A| − |A ∩ B|
De Morgan: (A ∪ B)^c = A^c ∩ B^c · (A ∩ B)^c = A^c ∪ B^c

Number systems, nested

Every natural number is an integer, every integer is rational, and the rationals plus the irrationals make up ℝ.

∅ is a subset of every set, and every set is a subset of itself.
{∅} is not empty: it has one element.
Repeats and order don't matter: {1, 2, 2} = {2, 1}.
Set-builder: list it out before counting. {x ∈ ℤ | −2 < x ≤ 3} = {−1, 0, 1, 2, 3}.
Intervals: [a, b] includes both ends, (a, b) excludes them.

Traps to avoid

∈ is for elements, ⊆ is for sets: 1 ∈ {1, 2} but {1} ⊆ {1, 2}.
0 is an integer and a rational number (0/1). It is neither positive nor negative.
22/7 and 1.414 are rational; π and √2 are not.

Week 1Every paper

Venn diagrams & counting

Two sets: |A ∪ B| = |A| + |B| − |A ∩ B|
Three sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C|
Only A: |A| − |A ∩ B| − |A ∩ C| + |A ∩ B ∩ C|
A and B, not C: |A ∩ B| − |A ∩ B ∩ C|
Exactly two: (|A ∩ B| + |B ∩ C| + |A ∩ C|) − 3|A ∩ B ∩ C|
Exactly one: |A| + |B| + |C| − 2(|A ∩ B| + |B ∩ C| + |A ∩ C|) + 3|A ∩ B ∩ C|
Neither / none: Total − |A ∪ B ∪ C|
Largest overlap: |A ∩ B|_max = min(|A|, |B|)
Smallest overlap: |A ∩ B|_min = max(0, |A| + |B| − Total)

Shortcut: Fill the diagram from the centre outwards: the triple overlap first, then the 'exactly two' regions, then the 'only' regions, then 'none' outside. Every sub-question is then just adding regions.

The Jan 2024 PYQ, filled in

Fill order:① centre 5② exactly two③ only one④ none = 100 − 80

Start with the 5 in the centre. Take it off each pair (20 − 5, 15 − 5, 10 − 5), then fill the singles. The union is 80, so 20 bought none.

PYQ pattern (Sep 2023)

In a group of 60 people, 40 like cricket and 35 like football. (a) What is the minimum number who like both? (b) If 10 like neither, how many like only cricket?

Show solution

(a) Both ≥ 40 + 35 − 60 = 15.
(b) At least one = 60 − 10 = 50, so both = 40 + 35 − 50 = 25.
Only cricket = 40 − 25 = 15.

Answer: (a) 15 · (b) 15

PYQ pattern (Jan 2024)

Of 100 customers, 50 bought milk, 40 bread and 30 fruits. 20 bought milk and bread, 15 milk and fruits, 10 bread and fruits, and 5 bought all three. How many bought milk and fruits but not bread? How many bought none?

Show solution

Milk and fruits, not bread = 15 − 5 = 10.
Union = 50 + 40 + 30 − 20 − 15 − 10 + 5 = 80.
None = 100 − 80 = 20.

Answer: 10 and 20

Traps to avoid

'Both A and B' includes people who are also in C. 'Only A and B' does not.
'At least one' is the union. 'Neither' is total − union.
If the total isn't given, never assume 'none' is 0.

Week 1Every paper

Relations

Definitions (|A| = n)

Relation on A: R ⊆ A × A
Number of relations: 2^n²
Reflexive: (a, a) ∈ R for every a ∈ A
Number of reflexive relations: 2^{n² − n}
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R
Number of symmetric relations: 2^{n(n + 1)/2}
Antisymmetric: (a, b) ∈ R and (b, a) ∈ R ⇒ a = b
Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
Equivalence: reflexive + symmetric + transitive
Pairs not in R: |(A × A) \ R| = n² − |R|

The four properties as arrow diagrams

Reflexiveevery a → a

Symmetrica → b ⇒ b → a

Antisymmetricnever both ways (a ≠ b)

Transitive1 → 2 → 3 ⇒ 1 → 3

An arrow a → b means (a, b) ∈ R. A loop on a node is (a, a).

Standard relations

Relation	Refl.	Sym.	Anti.	Trans.
Identity {(a, a)}	✓	✓	✓	✓
Empty ∅ (A non-empty)	✗	✓	✓	✓
Universal A × A (\|A\| ≥ 2)	✓	✓	✗	✓
a ≤ b	✓	✗	✓	✓
a < b	✗	✗	✓	✓
a divides b (positive integers)	✓	✗	✓	✓
a − b is even (integers)	✓	✓	✗	✓
a ≠ b	✗	✓	✗	✗
Line ⊥ line	✗	✓	✗	✗

Identity {(a, a)}
Refl.:
✓
Sym.:
✓
Anti.:
✓
Trans.:
✓
Empty ∅ (A non-empty)
Refl.:
✗
Sym.:
✓
Anti.:
✓
Trans.:
✓
Universal A × A (|A| ≥ 2)
Refl.:
✓
Sym.:
✓
Anti.:
✗
Trans.:
✓
a ≤ b
Refl.:
✓
Sym.:
✗
Anti.:
✓
Trans.:
✓
a < b
Refl.:
✗
Sym.:
✗
Anti.:
✓
Trans.:
✓
a divides b (positive integers)
Refl.:
✓
Sym.:
✗
Anti.:
✓
Trans.:
✓
a − b is even (integers)
Refl.:
✓
Sym.:
✓
Anti.:
✗
Trans.:
✓
a ≠ b
Refl.:
✗
Sym.:
✓
Anti.:
✗
Trans.:
✗
Line ⊥ line
Refl.:
✗
Sym.:
✓
Anti.:
✗
Trans.:
✗

PYQ (May 2023)

The Cartesian product A × A has 9 elements. Two of them are (2, 0) and (0, 8). Find the sum of all the elements of A.

Show solution

|A|² = 9, so |A| = 3.
(2, 0) and (0, 8) put 2, 0 and 8 in A, so A = {0, 2, 8}.
Sum = 0 + 2 + 8.

Answer: 10

Worked example

S = {1, 2, 3, 4} and R = {(a, b) ∈ S × S : a divides b}. Find |R| and |(S × S) \ R|.

Show solution

a = 1: b = 1, 2, 3, 4 → 4 pairs.
a = 2: b = 2, 4 → 2 pairs. a = 3: (3, 3). a = 4: (4, 4).
|R| = 8, and |S × S| = 16, so 16 − 8 = 8 pairs are not in R.

Answer: |R| = 8 · |(S × S) \ R| = 8

Shortcut: Write R out as a list of pairs first (it's usually under 15 pairs), then test each property against that list. Reasoning from the rule alone is where marks are lost.

Traps to avoid

Transitivity trap: if (1, 2) and (2, 1) are in R, then (1, 1) and (2, 2) must be in R too.
Reflexive needs (a, a) for every element of A, including elements that never appear in R.
Symmetric and antisymmetric aren't opposites. The identity relation is both; {(1, 2), (2, 1), (1, 3)} is neither.
No chain (a, b), (b, c) at all? Then R is transitive (vacuously). ∅ is symmetric and transitive.

Week 1Most papers

Functions

|A| = m, |B| = n

Function f: A → B: every a ∈ A has exactly one image in B
Range: {f(a) : a ∈ A} ⊆ codomain B
One-one (injective): f(a₁) = f(a₂) ⇒ a₁ = a₂
Onto (surjective): range = codomain
Bijective: one-one + onto ⇒ f⁻¹ exists
Functions A → B: n^m
One-one functions (m ≤ n): n!(n − m)!
Bijections (m = n): n!
Composition: (g ∘ f)(x) = g(f(x))
Usually g ∘ f ≠ f ∘ g.

Mapping diagrams

One-one, not ontoone B element is never hit

Onto, not one-onetwo inputs share an image

Bijectivepairs up exactly: f⁻¹ exists

Not a functiontwo images, or none

Tests and domains

Check	Rule
Vertical line test	Every vertical line meets the graph at most once ⇒ it is a function
Horizontal line test	Every horizontal line meets the graph at most once ⇒ one-one
√g(x)	Domain: g(x) ≥ 0
1g(x)	Domain: g(x) ≠ 0
log g(x)	Domain: g(x) > 0

Standard functions, ℝ → ℝ

f(x)	One-one?	Onto?
ax + b (a ≠ 0)	✓	✓
x²	✗	✗ (range [0, ∞))
x³	✓	✓
\|x\|	✗	✗
e^x	✓	✗ (range (0, ∞))
x² on [0, ∞) → [0, ∞)	✓	✓

ax + b (a ≠ 0)
One-one?:
✓
Onto?:
✓
x²
One-one?:
✗
Onto?:
✗ (range [0, ∞))
x³
One-one?:
✓
Onto?:
✓
|x|
One-one?:
✗
Onto?:
✗
e^x
One-one?:
✓
Onto?:
✗ (range (0, ∞))
x² on [0, ∞) → [0, ∞)
One-one?:
✓
Onto?:
✓

Traps to avoid

Changing the domain or codomain changes the answer: x² is not one-one on ℝ, but is on [0, ∞).
If some element of A has no image, or two images, it is not a function at all.
Onto depends on the codomain the question states, not on the formula.

Week 2Most papers

Coordinate geometry

Distance: d = √(x₂ − x₁)² + (y₂ − y₁)²
Midpoint: (x₁ + x₂2, y₁ + y₂2)
Section, internal m : n: (mx₂ + nx₁m + n, my₂ + ny₁m + n)
Section, external m : n: (mx₂ − nx₁m − n, my₂ − ny₁m − n)
Centroid: (x₁ + x₂ + x₃3, y₁ + y₂ + y₃3)
Area of triangle: ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Collinear points: area = 0 (or equal slopes)
Distance from the axes: from x-axis: |y| · from y-axis: |x|

Reflections and quadrants

Reflect (x, y) in	Result
x-axis	(x, −y)
y-axis	(−x, y)
Origin	(−x, −y)
Line y = x	(y, x)
Quadrant signs	I (+, +) · II (−, +) · III (−, −) · IV (+, −)

Quadrants and reflections of P(3, 1)

PYQ pattern (Jan 2023)

Starting at the origin, Ajay walks 3 units right, 4 units up, then 6 units left. Where is he, and how far is he from home?

Show solution

(0, 0) → (3, 0) → (3, 4) → (−3, 4).
Distance = √(−3)² + 4² = √25.

Answer: (−3, 4) in quadrant II, 5 units from home

Traps to avoid

In the section formula, m multiplies the second point (x₂, y₂).
Area can't be negative: take the absolute value.

Week 2Every paper

Straight lines

Slope from 2 points: m = y₂ − y₁x₂ − x₁ = tan θ
Point–slope: y − y₁ = m(x − x₁)
Slope–intercept: y = mx + c
Intercept form: xa + yb = 1
x-intercept a, y-intercept b
General form: ax + by + c = 0 ⇒ slope −ab, x-int −ca, y-int −cb
Horizontal / vertical: y = k (slope 0) · x = k (slope undefined)
Parallel: m₁ = m₂
Line parallel to ax + by + c = 0: ax + by + k = 0
Perpendicular: m₁m₂ = −1
Line perpendicular to ax + by + c = 0: bx − ay + k = 0
Angle between lines: tan θ = |m₁ − m₂1 + m₁m₂|
Point to line: d = |ax₁ + by₁ + c|√a² + b²
Between parallel lines: d = |c₁ − c₂|√a² + b²
Foot of perpendicular: x − x₁a = y − y₁b = −ax₁ + by₁ + ca² + b²

Lines at a glance

Sign of the slopem > 0 · m < 0 · m = 0 · undefined

Parallelm₁ = m₂

Perpendicularm₁ · m₂ = −1

Point to lined = |ax₁ + by₁ + c| ÷ √(a² + b²)

PYQ (May 2024)

L₁ passes through P₁(3, −2) and has x-intercept 1. L₂ passes through P₂(−1, 5) and has y-intercept −1. If θ is the angle between them, find tan θ.

Show solution

L₁ passes through (1, 0) and (3, −2): m₁ = −2 − 03 − 1 = −1.
L₂ passes through (0, −1) and (−1, 5): m₂ = 5 − (−1)−1 − 0 = −6.
tan θ = |−1 − (−6)1 + (−1)(−6)| = 57.

Answer: tan θ = 5/7

Worked example

Where do 2x + y = 7 and x − y = −1 meet?

Show solution

Add the equations: 3x = 6, so x = 2.
Then y = 7 − 2(2) = 3.

Answer: (2, 3)

Traps to avoid

Slope of ax + by + c = 0 is −a/b. Not a/b, not −b/a.
Intercept form needs 1 on the right: 3x + 4y = 12 → x4 + y3 = 1.
1 + m₁m₂ = 0 means the lines are perpendicular (θ = 90°), so tan θ is undefined.
Before using the parallel-line distance, make the x and y coefficients identical: x + 2y − 3 = 0 becomes 2x + 4y − 6 = 0 to compare with 2x + 4y + 1 = 0.

Week 2Some papers

Straight-line fit

Error (residual): e_i = y_i − ŷ_i
actual − predicted
Sum of squared errors: SSE = Σ (y_i − ŷ_i)²
Better fit: the line with the smaller SSE
Least-squares line: m = nΣxy − ΣxΣynΣx² − (Σx)², c = y − mx

Errors are vertical gaps

actual ypredicted ŷ on the lineerror e (squared and summed = SSE)the line passes through (x̄, ȳ)

SSE adds up the squared red gaps. The least-squares line makes that total as small as possible.

Worked example

Points (1, 2), (2, 3), (3, 5). Which fits better: y = x + 1 or y = 1.5x + 0.5?

Show solution

y = x + 1 predicts 2, 3, 4 → errors 0, 0, 1 → SSE = 1.
y = 1.5x + 0.5 predicts 2, 3.5, 5 → errors 0, −0.5, 0 → SSE = 0.25.

Answer: y = 1.5x + 0.5 (SSE 0.25 < 1)

Shortcut: The least-squares line always passes through (x, y). Use it to knock out options without computing m.

Week 3Every paper

Quadratic functions & parabolas

Standard form: f(x) = ax² + bx + c, a ≠ 0
Vertex x: h = −b2a
Vertex y: k = f(h) = −D4a = c − b²4a
Axis of symmetry: x = −b2a
Vertex form: f(x) = a(x − h)² + k
Opens: a > 0: up, minimum = k · a < 0: down, maximum = k
Range: a > 0: [k, ∞) · a < 0: (−∞, k]
y-intercept: (0, c)
Slope at x (tangent): 2ax + b
Zero at the vertex.
Slope of secant x₁ → x₂: f(x₂) − f(x₁)x₂ − x₁ = a(x₁ + x₂) + b
Increasing / decreasing: a > 0: decreasing on (−∞, h), increasing on (h, ∞)
Reversed when a < 0.

Reading a parabola

a > 0: opens uprange [k, ∞)

a < 0: opens downrange (−∞, k]

vertex (h, k), h = −b/2a, k = −D/4aaxis of symmetryroots (x-intercepts)y-intercept c

PYQ (May 2024)

y = x² + 4x + 12. Find the vertex, the minimum value, the x-intercepts and the range.

Show solution

a = 1 > 0, so it opens up.
h = −4/2 = −2, k = 4 − 8 + 12 = 8 → vertex (−2, 8).
D = 16 − 48 = −32 < 0 → no x-intercepts. y-intercept = 12.

Answer: Vertex (−2, 8), minimum 8, no x-intercepts, range [8, ∞)

PYQ pattern (May 2023)

Profit P(x) = −2x² + 24x − 40, where x is tonnes of raw material. For what x is profit maximum, and when is profit zero?

Show solution

x = −242(−2) = 6, and P(6) = −72 + 144 − 40 = 32.
P(x) = 0 ⇒ x² − 12x + 20 = 0 ⇒ (x − 2)(x − 10) = 0.

Answer: Maximum profit 32 at x = 6; zero profit at x = 2 and x = 10

Traps to avoid

The max/min value is k (the y-coordinate). The input that gives it is h. Read which one is asked.
Vertex x is −b/(2a): keep both the minus sign and the 2.
D < 0 with a > 0 means the parabola is entirely above the x-axis, so f(x) > 0 for every x.

Week 3Every paper

Quadratic equations & roots

Roots: x = −b ± √D2a, D = b² − 4ac
Nature of roots: D > 0: two distinct real · D = 0: equal, x = −b2a · D < 0: no real roots
Rational roots: D is a perfect square (a, b, c rational)
Sum and product: α + β = −ba, αβ = ca
Equation from roots: x² − (α + β)x + αβ = 0
Sum of squares: α² + β² = (α + β)² − 2αβ
Difference: |α − β| = √D|a|
Reciprocals: 1α + 1β = α + βαβ
Cubes: α³ + β³ = (α + β)³ − 3αβ(α + β)
Signs of roots: ca < 0 ⇒ opposite signs · both positive ⇔ −ba > 0 and ca > 0
For real roots (D ≥ 0).
Completing the square: ax² + bx + c = a(x + b2a)² − D4a

The discriminant decides the picture

D > 0two real roots: crosses twice

D = 0equal roots: touches once

D < 0no real roots: never meets

PYQ pattern (Jan 2024)

Ram copied the constant term wrongly and got roots 3 and 4. Shyam copied the x-coefficient wrongly and got roots 1 and 10. Find the correct roots.

Show solution

Ram's mistake doesn't touch b, so the sum is right: α + β = 7.
Shyam's mistake doesn't touch c, so the product is right: αβ = 10.
x² − 7x + 10 = 0 ⇒ (x − 2)(x − 5) = 0.

Answer: 2 and 5

Worked example

For which k does x² − kx + 9 = 0 have equal roots?

Show solution

Equal roots ⇒ D = 0.
k² − 36 = 0.

Answer: k = ±6

Shortcut: Roots given and the equation asked? Don't expand anything. Check each option's −b/a and c/a against the sum and product.

Traps to avoid

Wrong constant ⇒ the sum is still right. Wrong x-coefficient ⇒ the product is still right.
x-intercepts of the graph are the real roots. D < 0 ⇒ the graph never meets the x-axis.

Week 4Most papers

Polynomials: algebra & division

Degree: highest power with a non-zero coefficient
Degree rules: deg(pq) = deg p + deg q · deg(p ± q) ≤ max(deg p, deg q)
Division algorithm: p(x) = d(x) · q(x) + r(x), deg r < deg d
Remainder theorem: p(x) ÷ (x − a) leaves remainder p(a)
÷ (ax − b) leaves p(b/a)
Factor theorem: (x − a) is a factor ⇔ p(a) = 0
Sum of coefficients: p(1)
Alternating sum: p(−1)
Rational root test: possible roots = ±factors of constantfactors of leading coefficient
Cubic ax³ + bx² + cx + d: α + β + γ = −ba, αβ + βγ + γα = ca, αβγ = −da

Identities

Square: (a ± b)² = a² ± 2ab + b²
Difference of squares: a² − b² = (a + b)(a − b)
Cube: (a + b)³ = a³ + b³ + 3ab(a + b)
Sum / difference of cubes: a³ ± b³ = (a ± b)(a² ∓ ab + b²)

Remainder and factor theorem

x³ − 2x² + 4x − 5 divided by (x − 2): p(2) = 3, which isn't 0, so the remainder is 3.

Worked example

Find the remainder when p(x) = x³ − 2x² + 4x − 5 is divided by (x − 2).

Show solution

Remainder = p(2) = 8 − 8 + 8 − 5.

Answer: 3

Worked example

(x − 1) is a factor of x³ + kx² − 4x + 2. Find k.

Show solution

p(1) = 0 ⇒ 1 + k − 4 + 2 = 0.

Answer: k = 1

Week 4Every paper

Graphs of polynomials

Real roots: degree n ⇒ at most n x-intercepts
Turning points: at most n − 1
Odd degree: at least 1 real root; the two ends point opposite ways
Even degree: both ends point the same way; may have no real root
Factor form: p(x) = a(x − r₁)^k₁(x − r₂)^k₂⋯, degree = k₁ + k₂ + ⋯
y-intercept: p(0)

End behaviour

Even, aₙ > 0↑ … ↑ both ends up

Even, aₙ < 0↓ … ↓ both ends down

Odd, aₙ > 0↓ … ↑ down-left, up-right

Odd, aₙ < 0↑ … ↓ up-left, down-right

End behaviour (leading term aₙxⁿ)

	aₙ > 0	aₙ < 0
n even	↑ … ↑ (both ends up)	↓ … ↓ (both ends down)
n odd	↓ … ↑ (down-left, up-right)	↑ … ↓ (up-left, down-right)

n even
aₙ > 0:
↑ … ↑ (both ends up)
aₙ < 0:
↓ … ↓ (both ends down)
n odd
aₙ > 0:
↓ … ↑ (down-left, up-right)
aₙ < 0:
↑ … ↓ (up-left, down-right)

Multiplicity at a root r

Factor	At x = r the graph
(x − r)¹	crosses straight through
(x − r)², any even power	touches the axis and bounces back (no sign change)
(x − r)³, odd power ≥ 3	crosses but flattens out (S-shape)

Reading a graph: a crossing root has odd multiplicity; a touching root has even multiplicity (usually 2).
Minimum degree from a graph = sum of the multiplicities you can see, and at least (turning points + 1).
Find the leading coefficient a by substituting the y-intercept or any labelled point.
p(x) changes sign only at odd-multiplicity roots. Use a sign chart for 'p(x) > 0 when?' questions.

Worked example

Describe the graph of p(x) = (x + 2)(x − 1)².

Show solution

Degree 3 with positive leading coefficient: starts down-left, ends up-right.
x = −2 (power 1): crosses. x = 1 (power 2): touches and bounces.
y-intercept p(0) = 2 × 1 = 2. Two turning points.
Sign: negative for x < −2, positive between −2 and 1, positive for x > 1.

Answer: Crosses at −2, bounces at 1, y-intercept 2

The worked example, drawn

p(x) = (x + 2)(x − 1)²Bottom strip: sign of p(x). It flips only at the odd-power root.

PYQ pattern (every paper has a graph)

A degree-4 polynomial crosses the x-axis at −1 and 3, touches it at 1, and p(0) = 6. Find p(x).

Show solution

p(x) = a(x + 1)(x − 3)(x − 1)²: degree 1 + 1 + 2 = 4 ✓.
p(0) = a(1)(−3)(1) = −3a = 6 ⇒ a = −2.
Leading coefficient negative ⇒ both ends point down.

Answer: p(x) = −2(x + 1)(x − 3)(x − 1)²

Traps to avoid

'Distinct roots' counts each root once. Counted with multiplicity, (x − 1)² is two roots.
Turning points aren't roots: x³ has 1 root and 0 turning points; (x + 2)(x − 1)² has 2 distinct roots and 2 turning points.
At an even-multiplicity root the graph does not change sign.

Exam morning

Last-minute checklist

The 10 things to read on the way to the exam centre.

1|A ∪ B ∪ C| = Σ|A| − Σ|A ∩ B| + |A ∩ B ∩ C|. Fill Venn diagrams from the centre out.
2Transitive trap: (a, b) and (b, a) in R ⇒ (a, a) and (b, b) must be in R.
3|P(A)| = 2ⁿ, relations on A: 2^n², functions A → B: |B|^|A|.
4Slope of ax + by + c = 0 is −a/b. Perpendicular ⇔ m₁m₂ = −1.
5tan θ = |(m₁ − m₂) / (1 + m₁m₂)|.
6Point to line: |ax₁ + by₁ + c| / √a² + b².
7Vertex at x = −b/2a, extreme value −D/4a.
8α + β = −b/a, αβ = c/a. Wrong constant keeps the sum; wrong x-coefficient keeps the product.
9Odd multiplicity crosses, even multiplicity touches.
10Degree n ⇒ at most n roots and at most n − 1 turning points.

Now test yourself

Turn these notes into marks

Sit a real paper in the exam portal, then come back to the Traps view for whatever you got wrong.

Previous-year papersThe real papers, official keys PracticeDrill PYQs by topic Full mockAll four courses, timed

Formulas and conventions follow the IIT Madras BS course material for weeks 1–4. “Every paper” and “Most papers” count how often a topic appears in the qualifier papers on BTechi (January 2023 to May 2024). BTechi is not affiliated with IIT Madras.

Mathematics I

BSMA1001 · 11 topics · 19 min read

0/11 revised

Week 1Most papers

Number systems & sets

Number systems

Chain: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ
IITM lectures take ℕ = {0, 1, 2, …}. Some questions start at 1, so read the definition given.
Rational: pq, p, q ∈ ℤ, q ≠ 0
Decimal expansion terminates or recurs.
Irrational: √2, √3, π, e
Non-terminating, non-recurring. √n is irrational unless n is a perfect square.
gcd × lcm: gcd(a, b) × lcm(a, b) = a × b
From prime factors: gcd = lowest powers of common primes · lcm = highest powers of all primes

Sets (|A| = n)

Power set: |P(A)| = 2ⁿ
Proper subsets: 2ⁿ − 1
Non-empty proper subsets: 2ⁿ − 2
Subsets of size k: C(n, k) = n!k!(n − k)!
Cartesian product: |A × B| = |A| · |B|
A × B ≠ B × A unless A = B or one of them is empty.
Difference: A \ B = A ∩ B^c, |A \ B| = |A| − |A ∩ B|
De Morgan: (A ∪ B)^c = A^c ∩ B^c · (A ∩ B)^c = A^c ∪ B^c

Number systems, nested

Every natural number is an integer, every integer is rational, and the rationals plus the irrationals make up ℝ.

∅ is a subset of every set, and every set is a subset of itself.
{∅} is not empty: it has one element.
Repeats and order don't matter: {1, 2, 2} = {2, 1}.
Set-builder: list it out before counting. {x ∈ ℤ | −2 < x ≤ 3} = {−1, 0, 1, 2, 3}.
Intervals: [a, b] includes both ends, (a, b) excludes them.

Traps to avoid

∈ is for elements, ⊆ is for sets: 1 ∈ {1, 2} but {1} ⊆ {1, 2}.
0 is an integer and a rational number (0/1). It is neither positive nor negative.
22/7 and 1.414 are rational; π and √2 are not.

Week 1Every paper

Venn diagrams & counting

Two sets: |A ∪ B| = |A| + |B| − |A ∩ B|
Three sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C|
Only A: |A| − |A ∩ B| − |A ∩ C| + |A ∩ B ∩ C|
A and B, not C: |A ∩ B| − |A ∩ B ∩ C|
Exactly two: (|A ∩ B| + |B ∩ C| + |A ∩ C|) − 3|A ∩ B ∩ C|
Exactly one: |A| + |B| + |C| − 2(|A ∩ B| + |B ∩ C| + |A ∩ C|) + 3|A ∩ B ∩ C|
Neither / none: Total − |A ∪ B ∪ C|
Largest overlap: |A ∩ B|_max = min(|A|, |B|)
Smallest overlap: |A ∩ B|_min = max(0, |A| + |B| − Total)

The Jan 2024 PYQ, filled in

Fill order:① centre 5② exactly two③ only one④ none = 100 − 80

Start with the 5 in the centre. Take it off each pair (20 − 5, 15 − 5, 10 − 5), then fill the singles. The union is 80, so 20 bought none.

PYQ pattern (Sep 2023)

In a group of 60 people, 40 like cricket and 35 like football. (a) What is the minimum number who like both? (b) If 10 like neither, how many like only cricket?

Show solution

(a) Both ≥ 40 + 35 − 60 = 15.
(b) At least one = 60 − 10 = 50, so both = 40 + 35 − 50 = 25.
Only cricket = 40 − 25 = 15.

Answer: (a) 15 · (b) 15

PYQ pattern (Jan 2024)

Show solution

Milk and fruits, not bread = 15 − 5 = 10.
Union = 50 + 40 + 30 − 20 − 15 − 10 + 5 = 80.
None = 100 − 80 = 20.

Answer: 10 and 20

Traps to avoid

'Both A and B' includes people who are also in C. 'Only A and B' does not.
'At least one' is the union. 'Neither' is total − union.
If the total isn't given, never assume 'none' is 0.

Week 1Every paper

Relations

Definitions (|A| = n)

Relation on A: R ⊆ A × A
Number of relations: 2^n²
Reflexive: (a, a) ∈ R for every a ∈ A
Number of reflexive relations: 2^{n² − n}
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R
Number of symmetric relations: 2^{n(n + 1)/2}
Antisymmetric: (a, b) ∈ R and (b, a) ∈ R ⇒ a = b
Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
Equivalence: reflexive + symmetric + transitive
Pairs not in R: |(A × A) \ R| = n² − |R|

The four properties as arrow diagrams

Reflexiveevery a → a

Symmetrica → b ⇒ b → a

Antisymmetricnever both ways (a ≠ b)

Transitive1 → 2 → 3 ⇒ 1 → 3

An arrow a → b means (a, b) ∈ R. A loop on a node is (a, a).

Standard relations

Relation	Refl.	Sym.	Anti.	Trans.
Identity {(a, a)}	✓	✓	✓	✓
Empty ∅ (A non-empty)	✗	✓	✓	✓
Universal A × A (\|A\| ≥ 2)	✓	✓	✗	✓
a ≤ b	✓	✗	✓	✓
a < b	✗	✗	✓	✓
a divides b (positive integers)	✓	✗	✓	✓
a − b is even (integers)	✓	✓	✗	✓
a ≠ b	✗	✓	✗	✗
Line ⊥ line	✗	✓	✗	✗

Identity {(a, a)}
Refl.:
✓
Sym.:
✓
Anti.:
✓
Trans.:
✓
Empty ∅ (A non-empty)
Refl.:
✗
Sym.:
✓
Anti.:
✓
Trans.:
✓
Universal A × A (|A| ≥ 2)
Refl.:
✓
Sym.:
✓
Anti.:
✗
Trans.:
✓
a ≤ b
Refl.:
✓
Sym.:
✗
Anti.:
✓
Trans.:
✓
a < b
Refl.:
✗
Sym.:
✗
Anti.:
✓
Trans.:
✓
a divides b (positive integers)
Refl.:
✓
Sym.:
✗
Anti.:
✓
Trans.:
✓
a − b is even (integers)
Refl.:
✓
Sym.:
✓
Anti.:
✗
Trans.:
✓
a ≠ b
Refl.:
✗
Sym.:
✓
Anti.:
✗
Trans.:
✗
Line ⊥ line
Refl.:
✗
Sym.:
✓
Anti.:
✗
Trans.:
✗

PYQ (May 2023)

The Cartesian product A × A has 9 elements. Two of them are (2, 0) and (0, 8). Find the sum of all the elements of A.

Show solution

|A|² = 9, so |A| = 3.
(2, 0) and (0, 8) put 2, 0 and 8 in A, so A = {0, 2, 8}.
Sum = 0 + 2 + 8.

Answer: 10

Worked example

S = {1, 2, 3, 4} and R = {(a, b) ∈ S × S : a divides b}. Find |R| and |(S × S) \ R|.

Show solution

a = 1: b = 1, 2, 3, 4 → 4 pairs.
a = 2: b = 2, 4 → 2 pairs. a = 3: (3, 3). a = 4: (4, 4).
|R| = 8, and |S × S| = 16, so 16 − 8 = 8 pairs are not in R.

Answer: |R| = 8 · |(S × S) \ R| = 8

Shortcut: Write R out as a list of pairs first (it's usually under 15 pairs), then test each property against that list. Reasoning from the rule alone is where marks are lost.

Traps to avoid

Transitivity trap: if (1, 2) and (2, 1) are in R, then (1, 1) and (2, 2) must be in R too.
Reflexive needs (a, a) for every element of A, including elements that never appear in R.
Symmetric and antisymmetric aren't opposites. The identity relation is both; {(1, 2), (2, 1), (1, 3)} is neither.
No chain (a, b), (b, c) at all? Then R is transitive (vacuously). ∅ is symmetric and transitive.

Week 1Most papers

Functions

|A| = m, |B| = n

Function f: A → B: every a ∈ A has exactly one image in B
Range: {f(a) : a ∈ A} ⊆ codomain B
One-one (injective): f(a₁) = f(a₂) ⇒ a₁ = a₂
Onto (surjective): range = codomain
Bijective: one-one + onto ⇒ f⁻¹ exists
Functions A → B: n^m
One-one functions (m ≤ n): n!(n − m)!
Bijections (m = n): n!
Composition: (g ∘ f)(x) = g(f(x))
Usually g ∘ f ≠ f ∘ g.

Mapping diagrams

One-one, not ontoone B element is never hit

Onto, not one-onetwo inputs share an image

Bijectivepairs up exactly: f⁻¹ exists

Not a functiontwo images, or none

Tests and domains

Check	Rule
Vertical line test	Every vertical line meets the graph at most once ⇒ it is a function
Horizontal line test	Every horizontal line meets the graph at most once ⇒ one-one
√g(x)	Domain: g(x) ≥ 0
1g(x)	Domain: g(x) ≠ 0
log g(x)	Domain: g(x) > 0

Standard functions, ℝ → ℝ

f(x)	One-one?	Onto?
ax + b (a ≠ 0)	✓	✓
x²	✗	✗ (range [0, ∞))
x³	✓	✓
\|x\|	✗	✗
e^x	✓	✗ (range (0, ∞))
x² on [0, ∞) → [0, ∞)	✓	✓

ax + b (a ≠ 0)
One-one?:
✓
Onto?:
✓
x²
One-one?:
✗
Onto?:
✗ (range [0, ∞))
x³
One-one?:
✓
Onto?:
✓
|x|
One-one?:
✗
Onto?:
✗
e^x
One-one?:
✓
Onto?:
✗ (range (0, ∞))
x² on [0, ∞) → [0, ∞)
One-one?:
✓
Onto?:
✓

Traps to avoid

Changing the domain or codomain changes the answer: x² is not one-one on ℝ, but is on [0, ∞).
If some element of A has no image, or two images, it is not a function at all.
Onto depends on the codomain the question states, not on the formula.

Week 2Most papers

Coordinate geometry

Distance: d = √(x₂ − x₁)² + (y₂ − y₁)²
Midpoint: (x₁ + x₂2, y₁ + y₂2)
Section, internal m : n: (mx₂ + nx₁m + n, my₂ + ny₁m + n)
Section, external m : n: (mx₂ − nx₁m − n, my₂ − ny₁m − n)
Centroid: (x₁ + x₂ + x₃3, y₁ + y₂ + y₃3)
Area of triangle: ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Collinear points: area = 0 (or equal slopes)
Distance from the axes: from x-axis: |y| · from y-axis: |x|

Reflections and quadrants

Reflect (x, y) in	Result
x-axis	(x, −y)
y-axis	(−x, y)
Origin	(−x, −y)
Line y = x	(y, x)
Quadrant signs	I (+, +) · II (−, +) · III (−, −) · IV (+, −)

Quadrants and reflections of P(3, 1)

PYQ pattern (Jan 2023)

Starting at the origin, Ajay walks 3 units right, 4 units up, then 6 units left. Where is he, and how far is he from home?

Show solution

(0, 0) → (3, 0) → (3, 4) → (−3, 4).
Distance = √(−3)² + 4² = √25.

Answer: (−3, 4) in quadrant II, 5 units from home

Traps to avoid

In the section formula, m multiplies the second point (x₂, y₂).
Area can't be negative: take the absolute value.

Week 2Every paper

Straight lines

Slope from 2 points: m = y₂ − y₁x₂ − x₁ = tan θ
Point–slope: y − y₁ = m(x − x₁)
Slope–intercept: y = mx + c
Intercept form: xa + yb = 1
x-intercept a, y-intercept b
General form: ax + by + c = 0 ⇒ slope −ab, x-int −ca, y-int −cb
Horizontal / vertical: y = k (slope 0) · x = k (slope undefined)
Parallel: m₁ = m₂
Line parallel to ax + by + c = 0: ax + by + k = 0
Perpendicular: m₁m₂ = −1
Line perpendicular to ax + by + c = 0: bx − ay + k = 0
Angle between lines: tan θ = |m₁ − m₂1 + m₁m₂|
Point to line: d = |ax₁ + by₁ + c|√a² + b²
Between parallel lines: d = |c₁ − c₂|√a² + b²
Foot of perpendicular: x − x₁a = y − y₁b = −ax₁ + by₁ + ca² + b²

Lines at a glance

Sign of the slopem > 0 · m < 0 · m = 0 · undefined

Parallelm₁ = m₂

Perpendicularm₁ · m₂ = −1

Point to lined = |ax₁ + by₁ + c| ÷ √(a² + b²)

PYQ (May 2024)

L₁ passes through P₁(3, −2) and has x-intercept 1. L₂ passes through P₂(−1, 5) and has y-intercept −1. If θ is the angle between them, find tan θ.

Show solution

L₁ passes through (1, 0) and (3, −2): m₁ = −2 − 03 − 1 = −1.
L₂ passes through (0, −1) and (−1, 5): m₂ = 5 − (−1)−1 − 0 = −6.
tan θ = |−1 − (−6)1 + (−1)(−6)| = 57.

Answer: tan θ = 5/7

Worked example

Where do 2x + y = 7 and x − y = −1 meet?

Show solution

Add the equations: 3x = 6, so x = 2.
Then y = 7 − 2(2) = 3.

Answer: (2, 3)

Traps to avoid

Slope of ax + by + c = 0 is −a/b. Not a/b, not −b/a.
Intercept form needs 1 on the right: 3x + 4y = 12 → x4 + y3 = 1.
1 + m₁m₂ = 0 means the lines are perpendicular (θ = 90°), so tan θ is undefined.
Before using the parallel-line distance, make the x and y coefficients identical: x + 2y − 3 = 0 becomes 2x + 4y − 6 = 0 to compare with 2x + 4y + 1 = 0.

Week 2Some papers

Straight-line fit

Error (residual): e_i = y_i − ŷ_i
actual − predicted
Sum of squared errors: SSE = Σ (y_i − ŷ_i)²
Better fit: the line with the smaller SSE
Least-squares line: m = nΣxy − ΣxΣynΣx² − (Σx)², c = y − mx

Errors are vertical gaps

actual ypredicted ŷ on the lineerror e (squared and summed = SSE)the line passes through (x̄, ȳ)

SSE adds up the squared red gaps. The least-squares line makes that total as small as possible.

Worked example

Points (1, 2), (2, 3), (3, 5). Which fits better: y = x + 1 or y = 1.5x + 0.5?

Show solution

y = x + 1 predicts 2, 3, 4 → errors 0, 0, 1 → SSE = 1.
y = 1.5x + 0.5 predicts 2, 3.5, 5 → errors 0, −0.5, 0 → SSE = 0.25.

Answer: y = 1.5x + 0.5 (SSE 0.25 < 1)

Shortcut: The least-squares line always passes through (x, y). Use it to knock out options without computing m.

Week 3Every paper

Quadratic functions & parabolas

Standard form: f(x) = ax² + bx + c, a ≠ 0
Vertex x: h = −b2a
Vertex y: k = f(h) = −D4a = c − b²4a
Axis of symmetry: x = −b2a
Vertex form: f(x) = a(x − h)² + k
Opens: a > 0: up, minimum = k · a < 0: down, maximum = k
Range: a > 0: [k, ∞) · a < 0: (−∞, k]
y-intercept: (0, c)
Slope at x (tangent): 2ax + b
Zero at the vertex.
Slope of secant x₁ → x₂: f(x₂) − f(x₁)x₂ − x₁ = a(x₁ + x₂) + b
Increasing / decreasing: a > 0: decreasing on (−∞, h), increasing on (h, ∞)
Reversed when a < 0.

Reading a parabola

a > 0: opens uprange [k, ∞)

a < 0: opens downrange (−∞, k]

vertex (h, k), h = −b/2a, k = −D/4aaxis of symmetryroots (x-intercepts)y-intercept c

PYQ (May 2024)

y = x² + 4x + 12. Find the vertex, the minimum value, the x-intercepts and the range.

Show solution

a = 1 > 0, so it opens up.
h = −4/2 = −2, k = 4 − 8 + 12 = 8 → vertex (−2, 8).
D = 16 − 48 = −32 < 0 → no x-intercepts. y-intercept = 12.

Answer: Vertex (−2, 8), minimum 8, no x-intercepts, range [8, ∞)

PYQ pattern (May 2023)

Profit P(x) = −2x² + 24x − 40, where x is tonnes of raw material. For what x is profit maximum, and when is profit zero?

Show solution

x = −242(−2) = 6, and P(6) = −72 + 144 − 40 = 32.
P(x) = 0 ⇒ x² − 12x + 20 = 0 ⇒ (x − 2)(x − 10) = 0.

Answer: Maximum profit 32 at x = 6; zero profit at x = 2 and x = 10

Traps to avoid

The max/min value is k (the y-coordinate). The input that gives it is h. Read which one is asked.
Vertex x is −b/(2a): keep both the minus sign and the 2.
D < 0 with a > 0 means the parabola is entirely above the x-axis, so f(x) > 0 for every x.

Week 3Every paper

Quadratic equations & roots

Roots: x = −b ± √D2a, D = b² − 4ac
Nature of roots: D > 0: two distinct real · D = 0: equal, x = −b2a · D < 0: no real roots
Rational roots: D is a perfect square (a, b, c rational)
Sum and product: α + β = −ba, αβ = ca
Equation from roots: x² − (α + β)x + αβ = 0
Sum of squares: α² + β² = (α + β)² − 2αβ
Difference: |α − β| = √D|a|
Reciprocals: 1α + 1β = α + βαβ
Cubes: α³ + β³ = (α + β)³ − 3αβ(α + β)
Signs of roots: ca < 0 ⇒ opposite signs · both positive ⇔ −ba > 0 and ca > 0
For real roots (D ≥ 0).
Completing the square: ax² + bx + c = a(x + b2a)² − D4a

The discriminant decides the picture

D > 0two real roots: crosses twice

D = 0equal roots: touches once

D < 0no real roots: never meets

PYQ pattern (Jan 2024)

Ram copied the constant term wrongly and got roots 3 and 4. Shyam copied the x-coefficient wrongly and got roots 1 and 10. Find the correct roots.

Show solution

Ram's mistake doesn't touch b, so the sum is right: α + β = 7.
Shyam's mistake doesn't touch c, so the product is right: αβ = 10.
x² − 7x + 10 = 0 ⇒ (x − 2)(x − 5) = 0.

Answer: 2 and 5

Worked example

For which k does x² − kx + 9 = 0 have equal roots?

Show solution

Equal roots ⇒ D = 0.
k² − 36 = 0.

Answer: k = ±6

Shortcut: Roots given and the equation asked? Don't expand anything. Check each option's −b/a and c/a against the sum and product.

Traps to avoid

Wrong constant ⇒ the sum is still right. Wrong x-coefficient ⇒ the product is still right.
x-intercepts of the graph are the real roots. D < 0 ⇒ the graph never meets the x-axis.

Week 4Most papers

Polynomials: algebra & division

Degree: highest power with a non-zero coefficient
Degree rules: deg(pq) = deg p + deg q · deg(p ± q) ≤ max(deg p, deg q)
Division algorithm: p(x) = d(x) · q(x) + r(x), deg r < deg d
Remainder theorem: p(x) ÷ (x − a) leaves remainder p(a)
÷ (ax − b) leaves p(b/a)
Factor theorem: (x − a) is a factor ⇔ p(a) = 0
Sum of coefficients: p(1)
Alternating sum: p(−1)
Rational root test: possible roots = ±factors of constantfactors of leading coefficient
Cubic ax³ + bx² + cx + d: α + β + γ = −ba, αβ + βγ + γα = ca, αβγ = −da

Identities

Square: (a ± b)² = a² ± 2ab + b²
Difference of squares: a² − b² = (a + b)(a − b)
Cube: (a + b)³ = a³ + b³ + 3ab(a + b)
Sum / difference of cubes: a³ ± b³ = (a ± b)(a² ∓ ab + b²)

Remainder and factor theorem

x³ − 2x² + 4x − 5 divided by (x − 2): p(2) = 3, which isn't 0, so the remainder is 3.

Worked example

Find the remainder when p(x) = x³ − 2x² + 4x − 5 is divided by (x − 2).

Show solution

Remainder = p(2) = 8 − 8 + 8 − 5.

Answer: 3

Worked example

(x − 1) is a factor of x³ + kx² − 4x + 2. Find k.

Show solution

p(1) = 0 ⇒ 1 + k − 4 + 2 = 0.

Answer: k = 1

Week 4Every paper

Graphs of polynomials

Real roots: degree n ⇒ at most n x-intercepts
Turning points: at most n − 1
Odd degree: at least 1 real root; the two ends point opposite ways
Even degree: both ends point the same way; may have no real root
Factor form: p(x) = a(x − r₁)^k₁(x − r₂)^k₂⋯, degree = k₁ + k₂ + ⋯
y-intercept: p(0)

End behaviour

Even, aₙ > 0↑ … ↑ both ends up

Even, aₙ < 0↓ … ↓ both ends down

Odd, aₙ > 0↓ … ↑ down-left, up-right

Odd, aₙ < 0↑ … ↓ up-left, down-right

End behaviour (leading term aₙxⁿ)

	aₙ > 0	aₙ < 0
n even	↑ … ↑ (both ends up)	↓ … ↓ (both ends down)
n odd	↓ … ↑ (down-left, up-right)	↑ … ↓ (up-left, down-right)

n even
aₙ > 0:
↑ … ↑ (both ends up)
aₙ < 0:
↓ … ↓ (both ends down)
n odd
aₙ > 0:
↓ … ↑ (down-left, up-right)
aₙ < 0:
↑ … ↓ (up-left, down-right)

Multiplicity at a root r

Factor	At x = r the graph
(x − r)¹	crosses straight through
(x − r)², any even power	touches the axis and bounces back (no sign change)
(x − r)³, odd power ≥ 3	crosses but flattens out (S-shape)

Reading a graph: a crossing root has odd multiplicity; a touching root has even multiplicity (usually 2).
Minimum degree from a graph = sum of the multiplicities you can see, and at least (turning points + 1).
Find the leading coefficient a by substituting the y-intercept or any labelled point.
p(x) changes sign only at odd-multiplicity roots. Use a sign chart for 'p(x) > 0 when?' questions.

Worked example

Describe the graph of p(x) = (x + 2)(x − 1)².

Show solution

Degree 3 with positive leading coefficient: starts down-left, ends up-right.
x = −2 (power 1): crosses. x = 1 (power 2): touches and bounces.
y-intercept p(0) = 2 × 1 = 2. Two turning points.
Sign: negative for x < −2, positive between −2 and 1, positive for x > 1.

Answer: Crosses at −2, bounces at 1, y-intercept 2

The worked example, drawn

p(x) = (x + 2)(x − 1)²Bottom strip: sign of p(x). It flips only at the odd-power root.

PYQ pattern (every paper has a graph)

A degree-4 polynomial crosses the x-axis at −1 and 3, touches it at 1, and p(0) = 6. Find p(x).

Show solution

p(x) = a(x + 1)(x − 3)(x − 1)²: degree 1 + 1 + 2 = 4 ✓.
p(0) = a(1)(−3)(1) = −3a = 6 ⇒ a = −2.
Leading coefficient negative ⇒ both ends point down.

Answer: p(x) = −2(x + 1)(x − 3)(x − 1)²

Traps to avoid

'Distinct roots' counts each root once. Counted with multiplicity, (x − 1)² is two roots.
Turning points aren't roots: x³ has 1 root and 0 turning points; (x + 2)(x − 1)² has 2 distinct roots and 2 turning points.
At an even-multiplicity root the graph does not change sign.

Exam morning

Last-minute checklist

The 10 things to read on the way to the exam centre.

1|A ∪ B ∪ C| = Σ|A| − Σ|A ∩ B| + |A ∩ B ∩ C|. Fill Venn diagrams from the centre out.
2Transitive trap: (a, b) and (b, a) in R ⇒ (a, a) and (b, b) must be in R.
3|P(A)| = 2ⁿ, relations on A: 2^n², functions A → B: |B|^|A|.
4Slope of ax + by + c = 0 is −a/b. Perpendicular ⇔ m₁m₂ = −1.
5tan θ = |(m₁ − m₂) / (1 + m₁m₂)|.
6Point to line: |ax₁ + by₁ + c| / √a² + b².
7Vertex at x = −b/2a, extreme value −D/4a.
8α + β = −b/a, αβ = c/a. Wrong constant keeps the sum; wrong x-coefficient keeps the product.
9Odd multiplicity crosses, even multiplicity touches.
10Degree n ⇒ at most n roots and at most n − 1 turning points.

Now test yourself

Turn these notes into marks

Sit a real paper in the exam portal, then come back to the Traps view for whatever you got wrong.

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