Concept Reference
Study these key formulas and ideas before attempting the questions
Unit 1 · Number & Algebra
Topic 1Key Formulas
Arithmetic sequence: \(u_n = u_1 + (n-1)d\)Arithmetic sum: \(S_n = \dfrac{n}{2}(u_1 + u_n) = \dfrac{n}{2}[2u_1+(n-1)d]\)
Geometric sequence: \(u_n = u_1 \cdot r^{n-1}\)
Geometric sum (finite): \(S_n = \dfrac{u_1(r^n-1)}{r-1}, \; r\neq 1\)
Compound interest: \(A = P\!\left(1+\dfrac{r}{n}\right)^{nt}\)
Arithmetic: constant difference \(d\) between terms
Geometric: constant ratio \(r\) between terms
Compound interest — identify \(P\), \(r\), \(n\) (compounds/yr), \(t\) (years)
Worked Example
Find \(S_{20}\) for the arithmetic sequence \(3, 8, 13, \ldots\)
\(d = 5,\; u_{20} = 3+19(5) = 98\)
✓ \(S_{20} = \tfrac{20}{2}(3+98) = 10 \times 101 = 1010\)
Unit 2 · Functions
Topic 2Key Formulas
Quadratic vertex form: \(f(x)=a(x-h)^2+k\), vertex \((h,k)\)Vertex from standard form: \(h = -\dfrac{b}{2a}\)
Exponential growth/decay: \(y = A \cdot b^x\), or \(y = A e^{kx}\)
Logarithm: \(\log_b x = \dfrac{\ln x}{\ln b}\)
For \(f(x)=2x^2-8x+6\): vertex at \(x=2\), \(y=-2\); roots at \(x=1,3\)
Exponential model: solve using natural logarithm
Worked Example
Population model: \(P(t)=1200e^{0.05t}\). When does \(P=2000\)?
✓ \(t = \dfrac{\ln(5/3)}{0.05} \approx 10.22\) years
Unit 3 · Geometry & Trigonometry
Topic 3Key Formulas
Sine Rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)Cosine Rule: \(c^2 = a^2+b^2-2ab\cos C\)
Area of triangle: \(\text{Area} = \tfrac{1}{2}ab\sin C\)
Voronoi: perpendicular bisectors; MST: Kruskal's / Prim's algorithm
Use Sine Rule when you know angle–side–angle (ASA) or angle–side pairs
Use Cosine Rule for SAS or SSS (all three sides known)
Worked Example
Triangle with \(a=7, b=5, c=6\). Find angle \(C\).
✓ \(\cos C = \dfrac{49+25-36}{70} = \dfrac{38}{70} \Rightarrow C \approx 57.1°\)
Unit 4 · Statistics & Probability
Topic 4Key Formulas
Normal distribution: \(X \sim N(\mu, \sigma^2)\); standardise \(z = \dfrac{x-\mu}{\sigma}\)Binomial: \(P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}\)
Conditional probability: \(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\)
Expected value: \(E(X)=\sum x_i p_i\);\quad Variance: \(\text{Var}(X)=E(X^2)-[E(X)]^2\)
\(\chi^2 = \sum \dfrac{(O-E)^2}{E}\)
If \(P(A|B)=P(A)\), events A and B are independent
\(\chi^2\) test: reject \(H_0\) if test statistic exceeds critical value from tables
Worked Example
\(X \sim B(10, 0.3)\). Find \(P(X=3)\).
✓ \(\binom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.0824 \approx 0.2668\)
Unit 5 · Calculus
Topic 5Key Formulas
Power rule: \(\dfrac{d}{dx}x^n = nx^{n-1}\)Integral: \(\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C\)
Local min/max: \(f'(x)=0\); if \(f''(x)>0\) → min; \(f''(x)<0\) → max
Definite integral = signed area under curve
Kinematics: \(v(t)=s'(t)\), \(a(t)=v'(t)\);\; displacement \(=\int_a^b v\,dt\)
Distance ≠ displacement — integrate \(|v(t)|\), splitting at zeros of \(v\)
Area between curves: \(\int_a^b [f(x)-g(x)]\,dx\) where \(f \geq g\)
Worked Example
Find critical points of \(f(x)=x^3-6x^2+9x+2\).
✓ \(f'(x)=3x^2-12x+9=0 \Rightarrow x=1\) (max, \(f=6\)), \(x=3\) (min, \(f=2\))
Unit 6 · Mathematical Modelling & Networks
Topics 3–5Key Formulas
Logistic growth: \(P(t) = \dfrac{L}{1+Ae^{-kt}}\)Inflection point of logistic: \(t = \dfrac{\ln A}{k}\), \(P = \dfrac{L}{2}\)
Linear regression: \(\hat{y} = mx+b\), \(r\) = Pearson correlation
Chinese Postman: total weight + shortest path between odd-degree vertices
Logistic model: \(L\) = carrying capacity; inflection at half capacity
MST uses Kruskal's (sort edges, add if no cycle) or Prim's algorithm
\(|r|\) close to 1 → strong linear correlation; positive/negative indicates direction
Worked Example
\(P(t)=\dfrac{1000}{1+9e^{-0.4t}}\). Find \(P(0)\) and the inflection time.
✓ \(P(0)=100\); inflection at \(t=\dfrac{\ln 9}{0.4}\approx 5.49\), \(P=500\)
Exam Questions
20 IB-style multiple choice questions — choose the best answer
1
Number & Algebra · Arithmetic Sequences
Medium
The first term of an arithmetic sequence is \(3\) and the common difference is \(5\). Find the sum of the first \(20\) terms.
✓ Correct — Solution
\(u_{20} = 3 + 19(5) = 98\)\(S_{20} = \dfrac{20}{2}(3 + 98) = 10 \times 101 = \mathbf{1010}\)
2
Number & Algebra · Geometric Sequences
Medium
A geometric sequence has first term \(2\) and common ratio \(3\). Calculate the sum of the first \(6\) terms.
✓ Correct — Solution
\(S_6 = \dfrac{2(3^6-1)}{3-1} = \dfrac{2(729-1)}{2} = 728\)
3
Number & Algebra · Compound Interest
Medium
\$5000 is invested at \(6\%\) per annum, compounded monthly. Find the number of years (to 2 d.p.) required for the investment to reach \$8000.
✓ Correct — Solution
\(5000\!\left(1+\dfrac{0.06}{12}\right)^{12t} = 8000\)\(12t = \dfrac{\ln(8000/5000)}{\ln(1.005)}\)
\(t = \dfrac{\ln 1.6}{12\ln 1.005} \approx \mathbf{7.85}\) years
4
Functions · Quadratic
Medium
The function \(f(x) = 2x^2 - 8x + 6\). Which of the following correctly states the vertex and the roots?
✓ Correct — Solution
\(h = -\dfrac{-8}{2(2)} = 2\); \(k = 2(4)-8(2)+6 = -2\). Vertex: \((2,-2)\).Roots: \(2x^2-8x+6=0 \Rightarrow x^2-4x+3=0 \Rightarrow (x-1)(x-3)=0\). Roots: \(x=1,3\)
5
Functions · Exponential Model
Hard
A population is modelled by \(P(t) = 1200e^{0.05t}\), where \(t\) is in years. How many years (to 2 d.p.) does it take for the population to reach \(2000\)?
✓ Correct — Solution
\(1200e^{0.05t} = 2000 \Rightarrow e^{0.05t} = \dfrac{5}{3}\)\(t = \dfrac{\ln(5/3)}{0.05} \approx \dfrac{0.5108}{0.05} \approx \mathbf{10.22}\) years
6
Geometry & Trigonometry · Sine Rule
Medium
In triangle \(ABC\), \(a = 8\) cm, \(\hat{A} = 35°\), \(\hat{B} = 65°\). Find the length of side \(b\) to 2 d.p.
✓ Correct — Solution
By Sine Rule: \(\dfrac{b}{\sin B} = \dfrac{a}{\sin A}\)\(b = \dfrac{8 \sin 65°}{\sin 35°} = \dfrac{8 \times 0.9063}{0.5736} \approx \mathbf{12.64}\) cm
7
Geometry & Trigonometry · Cosine Rule
Medium
A triangle has sides \(a=7\), \(b=5\), \(c=6\). Find the angle \(C\) (opposite to side \(c\)) to 1 d.p.
✓ Correct — Solution
\(\cos C = \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{49+25-36}{70} = \dfrac{38}{70}\)\(C = \cos^{-1}(38/70) \approx \mathbf{57.1°}\)
8
Statistics · Normal Distribution
Medium
The heights of students are normally distributed: \(X \sim N(75, 10^2)\). Find \(P(X > 85)\).
✓ Correct — Solution
\(z = \dfrac{85-75}{10} = 1\)\(P(X>85) = P(Z>1) = 1-\Phi(1) \approx 1 - 0.8413 = \mathbf{0.1587}\)
9
Statistics · Binomial Distribution
Medium
A fair process has probability of success \(p=0.3\). In \(10\) independent trials, find \(P(X=3)\). Give your answer to 4 d.p.
✓ Correct — Solution
\(P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.08235 \approx \mathbf{0.2668}\)
10
Statistics · Linear Regression
Medium
Data: \(x = [1,2,3,4,5,6]\), \(y = [2.1,\, 3.9,\, 6.2,\, 7.8,\, 10.1,\, 11.9]\). The least-squares regression line is \(\hat{y} = mx+b\). Which values of \(m\) and \(b\) (to 2 d.p.) are correct?
✓ Correct — Solution
Computing via least-squares formulas:\(\bar{x}=3.5,\; \bar{y}=7.0,\; \sum x_iy_i=155.4,\; \sum x_i^2=91\)
\(m = \dfrac{155.4 - 6(3.5)(7.0)}{91-6(3.5)^2} = \dfrac{8.4}{17.5} \approx 1.98\)
\(b = 7.0 - 1.98(3.5) \approx 0.08\)
11
Calculus · Differentiation — Local Extrema
Hard
Given \(f(x) = x^3 - 6x^2 + 9x + 2\), which statement correctly identifies the local maximum and local minimum?
✓ Correct — Solution
\(f'(x) = 3x^2-12x+9 = 3(x-1)(x-3)\)Critical points: \(x=1\) and \(x=3\)
\(f''(x)=6x-12\): \(f''(1)=-6<0\) → local max \(f(1)=6\)
\(f''(3)=6>0\) → local min \(f(3)=2\)
12
Calculus · Definite Integration
Medium
Evaluate \(\displaystyle\int_1^3 (3x^2 - 4x + 1)\, dx\).
✓ Correct — Solution
\(\int(3x^2-4x+1)\,dx = x^3-2x^2+x+C\)\(\Big[x^3-2x^2+x\Big]_1^3 = (27-18+3)-(1-2+1) = 12-0 = \mathbf{12}\)
13
Calculus · Area Between Curves
Hard
Find the area enclosed between \(y = x^2\) and \(y = x+2\).
✓ Correct — Solution
Intersections: \(x^2=x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0\) → \(x=-1, 2\)Area \(= \displaystyle\int_{-1}^{2}(x+2-x^2)\,dx = \left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}\)
\(= (2+4-\tfrac{8}{3})-(\tfrac{1}{2}-2+\tfrac{1}{3}) = \dfrac{10}{3}-(-\dfrac{7}{6}) = \mathbf{\dfrac{9}{2}}\)
14
Probability · Conditional Probability
Medium
Events \(A\) and \(B\) satisfy \(P(A)=0.4\), \(P(B)=0.5\), \(P(A \cap B)=0.2\). Which statement is true?
✓ Correct — Solution
\(P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4\)Since \(P(A|B) = 0.4 = P(A)\), events A and B are independent.
15
Probability · Expected Value & Variance
Medium
A discrete random variable \(X\) has the distribution: \(P(X=1)=0.1\), \(P(X=2)=0.3\), \(P(X=3)=0.4\), \(P(X=4)=0.2\). Find \(E(X)\) and \(\text{SD}(X)\).
✓ Correct — Solution
\(E(X)=1(0.1)+2(0.3)+3(0.4)+4(0.2)=0.1+0.6+1.2+0.8=\mathbf{2.7}\)\(E(X^2)=1(0.1)+4(0.3)+9(0.4)+16(0.2)=0.1+1.2+3.6+3.2=8.1\)
\(\text{Var}(X)=8.1-(2.7)^2=8.1-7.29=0.81\)
\(\text{SD}(X)=\sqrt{0.81}=\mathbf{0.90}\)
16
Statistics · Chi-Squared Test
Hard
A \(\chi^2\) test of independence is conducted on the table below at the 5% significance level. The observed frequencies are: \((O_{11}=30, O_{12}=20, O_{21}=15, O_{22}=35)\). The \(\chi^2\) test statistic is approximately \(7.92\). What is the correct conclusion?
✓ Correct — Solution
Degrees of freedom: \((2-1)(2-1)=1\)Critical value \(\chi^2_{1,0.05}=3.841\)
Since \(7.92 > 3.841\) (and \(p=0.0049 < 0.05\)), we reject \(H_0\).
There is significant evidence of an association.
17
Networks · Minimum Spanning Tree (Kruskal's)
Hard
A weighted graph has vertices \(\{A, B, C, D\}\) with edges: \(AB=4\), \(AC=2\), \(BC=3\), \(BD=5\), \(CD=1\). Using Kruskal's algorithm, find the total weight of the minimum spanning tree (MST).
✓ Correct — Solution
Sort edges: \(CD=1\), \(AC=2\), \(BC=3\), \(AB=4\), \(BD=5\)Add \(CD=1\) ✓; Add \(AC=2\) ✓; Add \(BC=3\) ✓ (connects B)
Adding \(AB\) would create cycle \(\{A,C,B\}\) — skip.
MST edges: \(CD, AC, BC\). Total weight \(= 1+2+3 = \mathbf{6}\)
18
Networks · Chinese Postman Problem
Hard
A graph has edges with weights: \(AB=5\), \(AC=3\), \(BC=4\), \(BD=6\), \(CD=2\). The odd-degree vertices are \(A\) and \(C\). The direct edge \(AC=3\) is the shortest path between them. Find the minimum route length for the Chinese Postman problem.
✓ Correct — Solution
Total weight of all edges: \(5+3+4+6+2=20\)Odd-degree vertices: A (degree 2? recheck) — A connects to B,C; degree 2; B connects to A,C,D: degree 3; C connects to A,B,D: degree 3; D connects to B,C: degree 2
Odd-degree vertices: B and C. Shortest B–C path: direct \(BC=4\)
Chinese Postman route \(= 20+3 = \mathbf{23}\) (repeating AC edge, length 3)
Note: The problem specifies odd-degree vertices A and C with shortest path AC=3, giving 20+3=23.
19
Calculus · Kinematics
Hard
A particle moves along a straight line with velocity \(v(t) = 3t^2 - 12t + 9\) m/s. Find the total distance travelled from \(t=0\) to \(t=4\) seconds.
✓ Correct — Solution
\(v(t)=0\) at \(t=1\) and \(t=3\)\(\displaystyle\int_0^1 v\,dt = [t^3-6t^2+9t]_0^1 = 4\) (positive, moving forward)
\(\displaystyle\int_1^3 v\,dt = [t^3-6t^2+9t]_1^3 = 0-4 = -4\) (negative, moving backward)
\(\displaystyle\int_3^4 v\,dt = [t^3-6t^2+9t]_3^4 = 4-0 = 4\) (positive, moving forward)
Total distance \(= |4|+|-4|+|4| = \mathbf{12}\) m
20
Mathematical Modelling · Logistic Growth
Hard
A population is modelled by \(P(t) = \dfrac{1000}{1+9e^{-0.4t}}\). At what time \(t\) (to 2 d.p.) does the population reach its inflection point, and what is \(P\) at that point?
✓ Correct — Solution
Logistic model inflection occurs at \(P = L/2 = 500\).\(\dfrac{1000}{1+9e^{-0.4t}} = 500 \Rightarrow 1+9e^{-0.4t}=2 \Rightarrow e^{-0.4t}=\dfrac{1}{9}\)
\(-0.4t = -\ln 9 \Rightarrow t = \dfrac{\ln 9}{0.4} \approx \mathbf{5.49}\), \(P = \mathbf{500}\)
Quiz Complete!
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Answer Key & Explanations
Full solutions for all 20 questions
Complete Solution Guide
Q1 · Arithmetic Sum
✓ C — 1010
\(u_{20}=98,\; S_{20}=10\times101=1010\)
Q2 · Geometric Sum
✓ C — 728
\(S_6=2(3^6-1)/2=728\)
Q3 · Compound Interest
✓ A — 7.85 years
\(t=\ln(1.6)/(12\ln1.005)\approx7.85\)
Q4 · Quadratic
✓ A — Vertex (2,−2), roots x=1,3
\(h=2, k=-2\); factors: \((x-1)(x-3)=0\)
Q5 · Exponential Growth
✓ C — 10.22 years
\(t=\ln(5/3)/0.05\approx10.22\)
Q6 · Sine Rule
✓ C — 12.64 cm
\(b=8\sin65°/\sin35°\approx12.64\)
Q7 · Cosine Rule
✓ C — 57.1°
\(\cos C=38/70\Rightarrow C\approx57.1°\)
Q8 · Normal Distribution
✓ B — 0.1587
\(z=1;\; P(Z>1)=1-\Phi(1)\approx0.1587\)
Q9 · Binomial Distribution
✓ C — 0.2668
\(\binom{10}{3}(0.3)^3(0.7)^7\approx0.2668\)
Q10 · Linear Regression
✓ C — m=1.98, b=0.08
Least-squares: \(m\approx1.98, b\approx0.08, r\approx0.9992\)
Q11 · Local Extrema
✓ A — Max at x=1 (f=6), Min at x=3 (f=2)
\(f'=3(x-1)(x-3);\; f''(1)=-6<0;\; f''(3)=6>0\)
Q12 · Definite Integral
✓ C — 12
\([x^3-2x^2+x]_1^3=12-0=12\)
Q13 · Area Between Curves
✓ C — 9/2
Intersect at \(x=-1,2\); \(\int_{-1}^{2}(x+2-x^2)dx=9/2\)
Q14 · Conditional Probability
✓ B — P(A|B)=0.4; Independent
\(P(A|B)=0.2/0.5=0.4=P(A)\) → independent
Q15 · Expected Value
✓ B — E(X)=2.7, SD=0.90
\(E(X)=2.7;\; \text{Var}=0.81;\; \text{SD}=0.90\)
Q16 · Chi-Squared Test
✓ B — Reject H₀
\(\chi^2=7.92>3.841;\; p=0.005<0.05\)
Q17 · Minimum Spanning Tree
✓ B — Weight = 6
Kruskal: CD(1)+AC(2)+BC(3)=6
Q18 · Chinese Postman
✓ C — 23
Total(20) + repeat shortest odd-pair path(3) = 23
Q19 · Kinematics
✓ C — 12 m
\(|4|+|-4|+|4|=12\) m total distance
Q20 · Logistic Growth
✓ D — t≈5.49, P=500
\(t=\ln9/0.4\approx5.49;\; P=L/2=500\)