Every topic on the syllabus on one page. Highlighted items are the ones the mock test actually used. Part A is 40 one-mark MCQs; Part B is 4 of 5 fifteen-mark written questions (in the mock: 3 logic, 1 area between curves, 1 combinations + probability).
{1,2,3} · builder {x ∈ Z | −2 ≤ x < 3} = {−2,−1,0,1,2}A ∪ B in either · A ∩ B in both · A − B in A not B = A ∩ B′ · A′ = U − AA ⊕ B = (A−B) ∪ (B−A)(A∩B)′ = A′∪B′, (A∪B)′ = A′∩B′|A∪B| = |A|+|B|−|A∩B||A∪B∪C| = ΣA − Σ(pairs) + |A∩B∩C||U| − |A∪B|2ⁿ; proper subsets 2ⁿ − 1; power set size 2ⁿ|A×B| = |A|·|B|, pairs are ordered (a,b)A ∩ B = ∅. ∅ is a subset of every set.P(E) = favourable / total, 0 ≤ P ≤ 1P(A′) = 1 − P(A) · "at least one" = 1 − P(none)P(A∪B) = P(A)+P(B)−P(A∩B)P(A−B) = P(A) − P(A∩B)P(A|B) = P(A∩B) / P(B)P(A∩B) = P(A)·P(B)P(A∩B) = 0P(A∪B∪C) = ΣP − ΣP(pairs) + P(A∩B∩C)m×n = rows × columns. Add/subtract only same order.(m×n)(n×p) = m×p: entry = row of A · column of BAB ≠ BA in general · (AB)ᵀ = BᵀAᵀ · AI = IA = AAᵀ = A; skew: Aᵀ = −A[a b; c d]⁻¹ = (1/(ad−bc))·[d −b; −c a]A⁻¹ = adj(A)/|A|, adj = transpose of cofactor matrix|A| = 0 ⇔ no inverseAX = B ⇒ X = A⁻¹B|a b; c d| = ad − bc+ − +: a(ei−fh) − b(di−fg) + c(dh−eg)Cᵢⱼ = (−1)^(i+j)·Mᵢⱼ (Mᵢⱼ = minor, delete row i col j)|AB| = |A||B| · |Aᵀ| = |A| · |kA| = kⁿ|A| (n×n) · |A⁻¹| = 1/|A|x = Dₓ/D, y = D_y/D (replace that column with constants)½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)||v| = √(x²+y²+z²) · unit vector v/|v|a·b = a₁b₁+a₂b₂+a₃b₃ = |a||b|cos θa·b = 0 · parallel ⇔ a = k·ba×b = det of [i j k; a₁ a₂ a₃; b₁ b₂ b₃], magnitude = area of parallelogram( ) → ¬ → ∧ → ∨ → ⇒, ⇔ (⇔ evaluated last)P⇒Q is false only when T ⇒ F. False P ⇒ anything is T (vacuous)P⇔Q true when same · P⊕Q true when differentP⇒Q ≡ ¬P ∨ Q · ¬(P⇒Q) ≡ P ∧ ¬QQ⇒P · Inverse ¬P⇒¬Q · Contrapositive ¬Q⇒¬P (≡ original)2ⁿ (n distinct variables)P∨¬P) · Contradiction always F (P∧¬P) · Contingency mixed¬(P∧Q) ≡ ¬P∨¬Q, ¬(P∨Q) ≡ ¬P∧¬QP∧T≡P, P∨F≡P · Domination P∨T≡T, P∧F≡FP∨(P∧Q)≡P, P∧(P∨Q)≡P · Distributive P∧(Q∨R)≡(P∧Q)∨(P∧R)| Rule | From | Conclude |
|---|---|---|
| Modus Ponens | P⇒Q, P | Q |
| Modus Tollens | P⇒Q, ¬Q | ¬P |
| Hypothetical Syllogism | P⇒Q, Q⇒R | P⇒R |
| Disjunctive Syllogism | P∨Q, ¬P | Q |
| Resolution | P∨Q, ¬P∨R | Q∨R |
| Addition / Simplification | P / P∧Q | P∨Q / P |
| Conjunction | P, Q | P∧Q |
Fallacies: affirming the consequent (P⇒Q, Q ∴ P ✗), denying the antecedent (P⇒Q, ¬P ∴ ¬Q ✗). Contradiction proof: assume ¬conclusion, derive X ∧ ¬X.
∀x(S(x)⇒L(x)) · "Some S are L": ∃x(S(x)∧L(x))∀x(S(x)⇒¬H(x)) ≡ ¬∃x(S(x)∧H(x))∀x(D(x)⇒P(x)) (the thing after "only" goes on the right)¬∃x∀y Q ≡ ∀x∃y ¬Q¬∀x(P⇒Q) ≡ ∃x(P ∧ ¬Q)∀x P(x) get P(c) for any named c·, OR +, NOT A′. NAND/NOR are universal gates00, 01, 11, 10 (Gray code); edges wrap aroundⁿPᵣ = n!/(n−r)! order matters · ⁿCᵣ = n!/(r!(n−r)!) order doesn'tⁿCᵣ = ⁿCₙ₋ᵣ · 0! = 1 · ⁷C₃ = 35n!/(p!q!…)(n−1)! · necklace/bracelet (n−1)!/2(k+1)PᵣⁿC₂; lines from n points ⁿC₂; triangles ⁿC₃; diagonals n(n−3)/2(m+n)Cₘ(a+b)ⁿ = Σ ⁿCₖ aⁿ⁻ᵏ bᵏ, n+1 termsTₖ₊₁ = ⁿCₖ aⁿ⁻ᵏ bᵏ; for a coefficient, set power of x = target, solve k2ⁿⁿC₀ − ⁿC₁ + … = 0aₙ = a + (n−1)d · Sₙ = n/2·[2a+(n−1)d] = n/2·(a+l)n = (l−a)/d + 1 · AM = (a+b)/2 · 3 terms: a−d, a, a+daₙ = a·rⁿ⁻¹ · Sₙ = a(rⁿ−1)/(r−1)S∞ = a/(1−r), needs |r| < 1√(ab) · AM ≥ GM · 3 terms: a/r, a, ar0.ababab… = ab/99; 0.363636… = 36/99 = 4/11h + 2·hr/(1−r)an²+bn+c, with 2a = 2nd differencen(n+1)/2 · Pentagonal n(3n−1)/2 · Hexagonal n(2n−1)n²+(n−1)² · squares in n×n grid n(n+1)(2n+1)/6[(n+1)C₂]²log(ab)=log a+log b, log(a/b)=log a−log b, log aⁿ = n log a, e^(ln x) = xd/dx xⁿ = n xⁿ⁻¹ · constant → 0 · d/dx (kx) = kd/dx eˣ = eˣ · d/dx ln x = 1/x(f(b)−f(a))/(b−a) (secant slope)y − f(a) = f′(a)(x − a)∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1) · ∫1/x dx = ln|x| + C · ∫eˣ dx = eˣ + C∫ₐᵇ f dx = F(b) − F(a) = net signed area∫(3f+2g) = 3∫f + 2∫g · ∫ₐᵇ = −∫ᵇₐ · ∫ₐᶜ = ∫ₐᵇ + ∫ᵇᶜ∫ₐᵇ (top − bottom) dx, limits = intersection x'sdu = u′ dx, change bounds for definite∫₀³ 2x dx = 9 · ∫₀^(ln2) eˣ dx = 1μ = Σx/N · median = middle of sorted (avg of 2 middles if even) · mode = most frequentσ² = Σ(x−μ)²/N = Σx²/N − μ²σ = √variance, so variance = σ²(n₁m₁ + n₂m₂)/(n₁+n₂)(a+1)(b+1)(c+1); odd count ⇔ perfect squareGCD×LCM = a×b3x + 4 = 25x − 5 (order flips)