8️⃣ Revision and Final-Exam Preparation

Published

September 17, 2025

Abstract

Consolidate the key ideas of the course — functions, systems of equations, and financial mathematics — and practise on an exam-style problem drawn from the final examination.

Learning objectives
  • Revise the essential results of Modules 1–7.
  • Connect the concepts across topics (sequences ↔︎ interest, functions ↔︎ optimization).
  • Practise on an exam-style problem under exam conditions.
Warning

This revision session consolidates some of what you have learned; it does not replace the modules themselves. Work through every module before the exam.

Part 1 — Key Results to Remember

Numbers and algebra (Modules 1–2)
  • Order of operations (PEMDAS); \(|a|\) is a distance, so \(|a| \geq 0\).
  • Powers: \(a^{n}a^{m} = a^{n+m}\), \((a^{n})^{m} = a^{nm}\), \(a^{-n} = 1/a^{n}\).
  • Notable identities: \((a \pm b)^{2} = a^{2} \pm 2ab + b^{2}\), \((a-b)(a+b) = a^{2} - b^{2}\).
  • First-degree equation: \(ax + b = 0 \Rightarrow x = -b/a\) (if \(a \neq 0\)).
Curves, functions, optimization (Modules 3, 6, 7)
  • Line: \(y = mx + p\); system solution = intersection point.
  • Quadratic \(ax^{2}+bx+c=0\): \(\Delta = b^{2}-4ac\); \(x = \dfrac{-b \pm \sqrt{\Delta}}{2a}\).
  • Derivative: \((x^{n})' = n x^{n-1}\); extremum of a quadratic at \(x = -\dfrac{b}{2a}\) (max if \(a<0\), min if \(a>0\)).
Financial mathematics (Modules 4–5)
  • Arithmetic sequence: \(u_n = u_0 + nr\); sum \(= (\text{terms}) \times \dfrac{\text{first} + \text{last}}{2}\).
  • Geometric sequence: \(u_n = u_0 q^{n}\); sum \(= u_0 \dfrac{1 - q^{\text{terms}}}{1 - q}\) (for \(q \neq 1\)).
  • Simple interest → arithmetic: \(C_n = C_0(1 + in)\).
  • Compound interest → geometric: \(C_n = C_0(1 + i)^{n}\).

Part 2 — Exam-Style Problem (financing an investment)

To finance an investment, a firm needs \(1{,}000{,}000\) €. Two banks make a loan offer:

  • Bank A — simple interest: \(6\%\) per year;
  • Bank B — compound interest: \(5\%\) per year.

1. Type of sequence. Simple interest → arithmetic sequence; compound interest → geometric sequence.

2. First term, ratio/difference, general term.

  • Bank A: \(u_0 = 1{,}000{,}000\), common difference \(r = 0.06 \times 1{,}000{,}000 = 60{,}000\), general term \(u_n = 1{,}000{,}000 + 60{,}000\,n\).
  • Bank B: \(v_0 = 1{,}000{,}000\), common ratio \(q = 1.05\), general term \(v_n = 1{,}000{,}000 \times 1.05^{\,n}\).

3. Total to repay over 10 years — Bank A. \(u_{10} = 1{,}000{,}000 + 60{,}000 \times 10 = 1{,}600{,}000\) €.

4. Total to repay over 10 years — Bank B. \(v_{10} = 1{,}000{,}000 \times 1.05^{10} \approx 1{,}628{,}895\) €.

5. Is the lower annual rate the cheaper bank? No. Over \(10\) years Bank A (the higher \(6\%\) simple rate) is cheaper than Bank B (\(5\%\) compound), because \(1{,}600{,}000 < 1{,}628{,}895\). The regime matters as much as the rate.

6. Bank A at a monthly simple rate of \(0.5\%\). Simple rates convert proportionally: \(0.005 \times 12 = 6\%\) per year — exactly Bank A’s original offer.

7. Bank B at a monthly compound rate of \(0.5\%\). Compound rates do not convert proportionally: the equivalent annual rate is \((1 + 0.005)^{12} - 1 \approx 0.0617 = 6.17\%\), i.e. higher than \(6\%\).

Exam takeaways
  • Always identify the regime (simple vs compound) before computing.
  • Simple rates scale proportionally across periods; compound rates do not — use \((1 + i_m)^{12} - 1\).
  • A higher headline rate is not always more expensive: compare the amounts actually repaid over the whole horizon.

Part 3 — Extra Revision Exercises

  1. Solve the system \(\begin{cases} y = 2x + 1 \\ y = -x + 7 \end{cases}\). \(2x + 1 = -x + 7 \Rightarrow x = 2\), \(y = 5\). Solution \((2, 5)\).
  2. Factor \(9x^{2} - 25\) → \((3x - 5)(3x + 5)\).
  3. Find the maximum of \(R(x) = -x^{2} + 12x\). \(R'(x) = -2x + 12 = 0 \Rightarrow x = 6\); maximum \(R(6) = 36\).
  4. A capital of \(10{,}000\) € is invested at \(4\%\) compound for \(8\) years. Acquired capital? \(10{,}000 \times 1.04^{8} \approx 13{,}685.69\) €.