Mathematics GCE Examination Study Guide

General Examination Guidelines and Instructions

• Examination Authority: Examinations Council of Zambia (ECZ).
• Assessment Level: General Certificate of Education (GCE) Ordinary Level.
• Subject: Mathematics Paper 1.
• Paper Code: $4024/1$.
• Candidate Instructions:
  • Write the centre number and examination number on every page of the question paper.
  • The paper contains $23$ questions in total.
  • All questions must be answered in the spaces provided within the question paper.
  • For any question, the working must be shown clearly in the space provided below that question.
  • Prohibited Materials: Calculators and mathematical tables are strictly forbidden for this specific paper.
  • Grading Caveat: Omission of essential working will result in the loss of marks.
  • Time Allocation: The duration of the examination is $2$ hours.
  • Materials Provided: Geometrical instruments and graph paper (where applicable for rough work).

Algebra: Evaluation, Expansion, and Equations

• Evaluating Indices:

  • Expression: $5^3 \times 5^{-2}$
  • Rule: $a^m \times a^n = a^{m+n}$
  • Calculation: $5^{3 + (-2)} = 5^1 = 5$
  • Final Answer: $5$

• Expansion and Simplification:

  • Expression: $(2a + 3)(3a - 4)$
  • Process:
    • $2a(3a) = 6a^2$
    • $2a(-4) = -8a$
    • $3(3a) = 9a$
    • $3(-4) = -12$
  • Simplified Result: $6a^2 + a - 12$

• Solving Quadratic Equations:

  • Equation: $4x^2 = 9x$
  • Rearrangement: $4x^2 - 9x = 0$
  • Factorization: $x(4x - 9) = 0$
  • Solutions: $x = 0$ or $4x = 9 \rightarrow x = \frac{9}{4}$

• Complete Factorization:

  • Expression: $2ab + xy - 2ay - bx$
  • Grouping: $(2ab - 2ay) + (xy - bx)$
  • Factorization: $2a(b - y) - x(b - y)$
  • Final Result: $(2a - x)(b - y)$

Commercial Arithmetic and Set Theory

• Annual Dividend Calculation:

  • Dividend Rate: $6\%$
  • Share Value: $K25.00$
  • Total Shares: $500$
  • Formula: $\text{Annual Dividend} = \text{Rate} \times \text{Share Value} \times \text{Number of Shares}$
  • Calculation: $0.06 \times 25 \times 500 = 1.5 \times 500 = K750.00$

• Set Theory and Shading:

  • Operation: Shade $(A \cup B)' \cap C$ on a Venn diagram.
  • Logic: Identify the union of sets $A$ and $B$, take the complement (everything outside the union), and find the intersection with set $C$.

• Listing Set Elements:

  • Universal Set $E = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
  • Set $A = \{Prime \text{ } numbers\} = \{2, 3, 5, 7\}$
  • Set $B = \{Multiples \text{ } of \text{ } 3\} = \{3, 6, 9\}$
  • Union $(A \cup B) = \{2, 3, 5, 6, 7, 9\}$
  • Complement $(A \cup B)' = \{1, 4, 8, 10\}$

Sequences and Earth Geometry

• Arithmetic Sequences:

  • Sequence: $5, 12, 19, \dots$
  • First term ($a$): $5$
  • Common difference ($d$): $12 - 5 = 7$
  • Sixth term ($n=6$): $a + (n-1)d = 5 + (5)(7) = 5 + 35 = 40$
  • nth term ($T_n$): $5 + (n-1)7 = 5 + 7n - 7 = 7n - 2$

• Nautical Speed and Distance:

  • Distance ($D$): $8100\text{ nautical miles}$
  • Time ($T$): $18\text{ hours}$
  • Formula: $\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}$
  • Calculation: $\frac{8100}{18} = 450\text{ knots}$

• Longitudinal Differences:

  • Town B is East of Town A.
  • Time at B: $11:00\text{ hours}$; Time at A: $07:00\text{ hours}$.
  • Time difference: $4\text{ hours}$.
  • Conversion: $1\text{ hour} = 15^\circ\text{ longitude}$.
  • Calculation: $4 \times 15^\circ = 60^\circ\text{ difference}$.

Matrices and Probability

• Matrix Operations:

  • Matrix $C = \begin{pmatrix} 7 & 3 \\ 1 & 4 \end{pmatrix}$
  • Transpose ($C^T$): $C^T = \begin{pmatrix} 7 & 1 \\ 3 & 4 \end{pmatrix}$
  • Matrix Multiplication: $\begin{pmatrix} 1 & 8 \end{pmatrix} \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} (1 \times 2 + 8 \times 0) & (1 \times 5 + 8 \times 3) \end{pmatrix} = \begin{pmatrix} 2 & 29 \end{pmatrix}$

• Probability Constants:

  • $P(\text{Waking up late}) = 0.3$
  • $P(\text{Waking up early}) = 1 - 0.3 = 0.7$

Mensuration and Measurement Error

• Sector Area:

  • Radius ($r$): $6\text{ cm}$
  • Area: $24.2\text{ cm}^2$
  • Formula: $\text{Area} = \frac{\theta}{360} \pi r^2$

• Wood Piece Measurement Limits:

  • Measured value: $25\text{ cm}$.
  • Assuming measurement to the nearest unit ($1\text{ cm}$):
  • Absolute Error: $0.5\text{ cm}$.
  • Upper Limit: $25 + 0.5 = 25.5\text{ cm}$.
  • Lower Limit: $25 - 0.5 = 24.5\text{ cm}$.
  • Percentage Error: $\frac{0.5}{25} \times 100\% = 2\%$

Functions and Coordinate Geometry

• Function Manipulations:

  • Functions: $f(x) = x - 4$, $h(x) = 2x + 3$
  • Inverse of $h(x)$: Let $y = 2x + 3 \rightarrow x = \frac{y - 3}{2}$. Therefore, $h^{-1}(x) = \frac{x - 3}{2}$.
  • Evaluate $h^{-1}(5)$: $\frac{5 - 3}{2} = 1$.
  • Composite Function $hf(x)$: $h(f(x)) = 2(x - 4) + 3 = 2x - 8 + 3 = 2x - 5$.

• Line Equation and Gradients:

  • Points: $A(-2, 5)$, $B(0, y)$.
  • Gradient ($m$): $-3$.
  • Gradient Formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$
  • Calculation: $-3 = \frac{y - 5}{0 - (-2)} = \frac{y - 5}{2} \rightarrow -6 = y - 5 \rightarrow y = -1$.
  • Coordinates of $B$: $(0, -1)$.

• Distance Between Points:

  • Points: $A(2, -3)$, $B(7, 9)$.
  • Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  • Calculation: $\sqrt{(7-2)^2 + (9-(-3))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ units}$.

Bearings and Trigonometry

• Bearing Calculations:

  • Given: Bearing of $P$ from $Q$ is $228^\circ$.
  • Bearing of $Q$ from $P$: Reverse bearing = $228^\circ - 180^\circ = 048^\circ$.
  • Bearing of $P$ from $R$: Calculated based on R being due South of Q and angle $QPR = 72^\circ$. Result: $300^\circ$.

• Trigonometric Lengths:

  • Given: $\tan(ABD) = -\frac{3}{4}$ in a geometric setup involving a straight line ABC and a right-angled triangle BCD.
  • Length determined: $BD = 5\text{ units}$.

Similarity and Geometry

• Geometric Similarity:

  • Object: Two cylinders A and B.
  • Ratio of heights ($h_1:h_2$): $m : 4$.
  • Surface areas: $Area_A = 144\text{ cm}^2$, $Area_B = 256\text{ cm}^2$.
  • Area Ratio: $(\frac{m}{4})^2 = \frac{144}{256}$
  • Calculation: $\frac{m^2}{16} = \frac{9}{16} \rightarrow m^2 = 9 \rightarrow m = 3$.

• Pyramid Symmetry:

  • Square-based right pyramid (axis AB).
  • Symmetry Property: $4$ planes of symmetry; rotational symmetry of order $4$.

• Circle Geometry:

  • Context: Circle with diameter $BD$, $AD = AC$, and angle $BDC = 22^\circ$.
  • Properties: Angles in same segment; angle in a semi-circle is $90^\circ$.
  • Results:
    • Angle $BAC = 22^\circ$.
    • Angle $CBD = 68^\circ$.
    • Angle $ACO = 34^\circ$.

Variation and Calculus

• Joint Variation:

  • Condition: $y \propto \frac{x}{z^2}$.
  • Equation: $y = k \frac{x}{z^2}$.
  • Find constant $k$: Given $y=8, x=4, z=1 \rightarrow 8 = k \frac{4}{1^2} \rightarrow k = 2$.
  • Find $y$ when $x=9, z=3$: $y = 2 \frac{9}{3^2} = 2 \frac{9}{9} = 2$.
  • Find $z$ when $y=3, x=54$: $3 = 2 \frac{54}{z^2} \rightarrow 3z^2 = 108 \rightarrow z^2 = 36 \rightarrow z = \pm 6$.

• Differentiation:

  • Function: $y = 4x^3 - 3x^2 + 5x$
  • Derivative: $\frac{dy}{dx} = 12x^2 - 6x + 5$

Graphical Analysis and Kinematics

• Graphing Inequalities:

  • Region R is defined by three boundaries.
  • Condition 1: Above $x-axis$ ($y \ge 0$).
  • Condition 2: Below/on the line passing through $(1, 1)$ and $(7, 7)$ ($y \le x$).
  • Condition 3: Defined by boundaries observed on given diagram.

• Quadratic Graphs:

  • Equation: $y = 3 - 2x - x^2$
  • Intercepts ($y=0$): $x^2 + 2x - 3 = 0 \rightarrow (x+3)(x-1) = 0 \rightarrow A(-3, 0), C(1, 0)$.
  • Turning Point: $x = \frac{-b}{2a} = \frac{-(-2)}{2(-1)} = -1$. When $x = -1, y = 3 - 2(-1) - (-1)^2 = 3 + 2 - 1 = 4$. Turning point is $(-1, 4)$.

• Vector Components:

  • Vectors: $OA = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, $OB = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$.
  • Vector $BA$: $OA - OB = \begin{pmatrix} 3 - 2 \\ 4 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (Transcript answer notes $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$; however, subtraction dictates the previous calculation).

• Speed-Time Graphs:

  • Initial Speed: $0\text{ m/s}$; Speed at $t=5\text{ s}$ is $20\text{ m/s}$.
  • Acceleration in first $5\text{ s}$: $\frac{20 - 0}{5} = 4\text{ m/s}^2$.
  • Total Distance ($550\text{ m}$): The area under the speed-time graph.
  • Calculating $t$: Area of trapezium = $550 = \frac{1}{2}(t + (t-5)) \times 20$. Result: $t = 35\text{ s}$.
  • Average Speed: $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{550}{35} \approx 15.71\text{ m/s}$.