May/June 2025 Paper 22 Worked Answers (IGCSE Maths 0580 Extended)

47 questions · 100 marks · 120 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1a

   1 marksLine symmetry
   Step 1: Identify the line of symmetry on the grid. Step 2: For each shaded square, find its mirror image across the line of symmetry. Step 3: Shade the square that is the reflection of the unmatched shaded square.
   Method:

   Identify the line of symmetry, then find the reflection of every shaded square. The missing reflected square is the one to shade.
   Examiner tips

   • Check each shaded square has a matching reflected square across the line of symmetry
   • Use the grid lines to count distances from the line of symmetry accurately

2. Question 1b

   1 marksRotational symmetry
   Step 1: Rotational symmetry of order 2 means the pattern looks the same after a $180°$ rotation about its centre. Step 2: For each shaded square, find the square that is a $180°$ rotation about the centre of the grid. Step 3: Shade the square whose rotational partner is not yet shaded.
   Method:

   Identify the centre of the grid, then for each shaded square find its image under a 180 degree rotation. The unmatched square needs its partner shaded.
   Examiner tips

   • Turn your paper upside down to check for rotational symmetry of order 2
   • The centre of rotation is the centre of the grid

3. Question 2a

   2 marksScale drawings
   Step 1: The scale is 1 cm : 2 km. Step 2: Multiply the map distance by the scale factor: $6.5 \times 2 = 13$ km.
   Method:

   Measure the distance on the scale drawing, then multiply by the scale factor to find the actual distance.
   Examiner tips

   • Always check the units in scale drawing questions
   • Read the scale carefully: 1 cm represents X km means multiply

4. Question 2b

   1 marksBearings
   Step 1: The bearing of B from A is $135°$. Step 2: To find the back bearing (A from B), add $180°$: $135° + 180° = 315°$. Step 3: Since $315° < 360°$, the answer is $315°$.
   Method:

   Add 180° to the given bearing to find the back bearing. If the result exceeds 360°, subtract 360°.
   Examiner tips

   • Bearings are always measured clockwise from north
   • Back bearings differ by exactly 180 degrees

5. Question 3

   3 marksAngles in parallel lines
   Step 1: Using angles in a triangle formed by the two transversals and the parallel line: $z = 180° - 50° - 70° = 60°$. Step 2: Alternatively, the exterior angle at the lower parallel line between the two transversals is $50° + 70° = 120°$. The angle at the top is $180° - 120° = 60°$.
   Method:

   Use alternate angles to find angles in the triangle formed between the parallel lines, then use angle sum of a triangle.
   Examiner tips

   • Identify all angle relationships: alternate, co-interior, corresponding
   • Look for triangles formed by the intersecting lines

6. Question 4a

   1 marksProbability of a single event
   Step 1: The even numbers from 1 to 9 are: 2, 4, 6, 8. That is 4 even numbers. Step 2: Total number of cards = 9. Step 3: Probability = $\frac{4}{9}$.
   Method:

   List all even numbers from 1 to 9, count them, and divide by the total number of cards.
   Examiner tips

   • Always express probability as a fraction, decimal, or percentage
   • Ensure the denominator is the total number of outcomes

7. Question 4b

   1 marksExpected frequency
   Step 1: Numbers greater than 4 on a die are 5 and 6, so $P(>4) = \frac{2}{6} = \frac{1}{3}$. Step 2: Expected number = probability $\times$ number of trials = $\frac{1}{3} \times 60 = 20$.
   Method:

   Calculate the probability of the event, then multiply by the number of trials.
   Examiner tips

   • Expected frequency = probability × number of trials
   • Make sure you identify the correct favourable outcomes

8. Question 5a

   1 marksTranslation
   Step 1: Add the translation vector to the point: $(2 + 3, 5 + (-4)) = (5, 1)$.
   Method:

   Add the translation vector components to each coordinate of the original point.
   Examiner tips

   • A translation moves every point by the same vector
   • The top number is horizontal, the bottom is vertical

9. Question 5b

   3 marksDescribing transformations
   Step 1: Check the mapping of vertices. $(1,2) \to (2,-1)$. Under $90°$ clockwise rotation about the origin: $(x,y) \to (y,-x)$. So $(1,2) \to (2,-1)$. ✓ Step 2: Check another vertex: $(3,2) \to (2,-3)$. ✓ Step 3: The transformation is a rotation of $90°$ clockwise about the origin.
   Method:

   Test each vertex under the proposed transformation to verify it maps correctly. State the transformation type, angle, direction, and centre.
   Examiner tips

   • For full marks on describing transformations, state the type, and all required details
   • For rotations: state angle, direction, and centre

10. Question 6a

    2 marksSolving linear equations
    Step 1: Subtract 9 from both sides: $5x = 34 - 9 = 25$. Step 2: Divide both sides by 5: $x = \frac{25}{5} = 5$.
    Method:

    Subtract the constant from both sides, then divide by the coefficient of x.
    Examiner tips

    • Show each step of your working clearly
    • Check your answer by substituting back into the original equation

11. Question 6b

    3 marksSolving linear equations with brackets
    Step 1: Expand the brackets: $12y - 15 = 33$. Step 2: Add 15 to both sides: $12y = 48$. Step 3: Divide by 12: $y = 4$.
    Method:

    Expand the brackets, collect constants on one side, then divide by the coefficient of y.
    Examiner tips

    • Expand brackets carefully, multiplying both terms inside
    • Check your answer by substituting back into the original equation

12. Question 7a

    1 marksSequences - next term
    Step 1: Find the common difference: $11 - 15 = -4$. Step 2: The next term is $3 + (-4) = -1$.
    Method:

    Find the common difference and add it to the last given term.
    Examiner tips

    • Always check the common difference between several pairs of terms
    • Be careful with negative numbers

13. Question 7b

    2 marksSequences - nth term
    Step 1: The common difference is $13 - 17 = -4$. Step 2: The $n$th term of an arithmetic sequence is $a + (n-1)d = 17 + (n-1)(-4) = 17 - 4n + 4 = 21 - 4n$.
    Method:

    Find the common difference d, then use nth term = a + (n−1)d and simplify.
    Examiner tips

    • Check your nth term formula by substituting n=1 to see if it gives the first term
    • A decreasing sequence has a negative coefficient of n

14. Question 8

    2 marksHighest common factor
    Step 1: Prime factorise: $48 = 2^4 \times 3$ and $72 = 2^3 \times 3^2$. Step 2: HCF = product of common primes with lowest powers: $2^3 \times 3 = 24$.
    Method:

    Prime factorise both numbers, then take the product of common prime factors with their lowest powers.
    Examiner tips

    • Prime factorisation is the most reliable method for HCF
    • HCF uses the lowest powers of common primes; LCM uses the highest

15. Question 9a

    2 marksVector arithmetic
    Step 1: $\overrightarrow{PR} = 3\overrightarrow{PQ} = 3\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}$. Step 2: $R = P + \overrightarrow{PR} = (2 + 9, 5 + (-6)) = (11, -1)$.
    Method:

    Multiply the vector by the scalar, then add the resulting vector to the coordinates of the starting point.
    Examiner tips

    • Remember to multiply both components of the vector by the scalar
    • Add the resulting vector to the starting point coordinates

16. Question 9b

    2 marksVector magnitude
    Step 1: Length $= \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45}$. Step 2: Simplify: $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$. Step 3: So $k = 3$.
    Method:

    Calculate the magnitude using Pythagoras, then simplify the surd to find k.
    Examiner tips

    • Always simplify surds fully
    • The magnitude formula is the same as Pythagoras' theorem

17. Question 9c

    2 marksRatio on a line segment
    Step 1: $MP:PN = 1:3$ means $P$ is $\frac{1}{4}$ of the way from $M$ to $N$. Step 2: $\overrightarrow{MN} = (9-1, 6-2) = (8, 4)$. Step 3: $\overrightarrow{MP} = \frac{1}{4}(8, 4) = (2, 1)$. Step 4: $P = M + \overrightarrow{MP} = (1+2, 2+1) = (3, 3)$.
    Method:

    Find the fraction of the journey from the first point, multiply the direction vector by this fraction, and add to the starting point.
    Examiner tips

    • AP:PB = 1:3 means P is 1/(1+3) = 1/4 of the way from A to B
    • Be careful about which end the ratio starts from

18. Question 10

    2 marksArc length of a sector
    Step 1: Arc length $= \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2\pi \times 12$. Step 2: $= \frac{1}{6} \times 24\pi = 4\pi$. Step 3: So $n = 4$.
    Method:

    Substitute into the arc length formula, simplify, and extract the coefficient of π.
    Examiner tips

    • Remember the full circumference formula is 2πr, not πr
    • The fraction of the circle is θ/360

19. Question 11a

    1 marksStandard form
    Step 1: Move the decimal point so the number is between 1 and 10: $3.56$. Step 2: Count the places moved: 4 places to the right, so the power is $-4$. Step 3: $0.000356 = 3.56 \times 10^{-4}$.
    Method:

    Move the decimal point to create a number between 1 and 10, then determine the power of 10.
    Examiner tips

    • For numbers less than 1, the power of 10 is negative
    • The first part must be between 1 and 10

20. Question 11b

    2 marksStandard form calculations
    Step 1: Rewrite with the same power of 10: $0.25 \times 10^{16} + 2.5 \times 10^{16}$. Step 2: Add: $(0.25 + 2.5) \times 10^{16} = 2.75 \times 10^{16}$. Step 3: $2.75$ is between 1 and 10, so this is already in standard form.
    Method:

    Convert both terms to the same power of 10, add the coefficients, then adjust to standard form if needed.
    Examiner tips

    • Match the powers of 10 before adding or subtracting
    • Check that your final answer is in proper standard form

21. Question 12

    2 marksCircle theorems
    Step 1: AC is a diameter, so angle $ABC$ is an angle in a semicircle. Step 2: The angle in a semicircle is always $90°$. Step 3: Therefore $ABC = 90°$.
    Method:

    Identify that QR is a diameter, so the angle at P (angle QPR) in the semicircle... but here the angle at P is given. Actually the angle in the semicircle is the angle subtended by the diameter at the circumference, which is angle QPR = 90°. Wait — the original exam says QPR = 74°. The angle subtended by diameter QR at point P should be 90°. Re-reading: QR is a diameter, so angle QPR = 90°. But the exam says 74°. This means P is NOT the angle in the semicircle in the standard sense. Actually, on re-reading the mark scheme: angle in semicircle = 90° refers to angle QPR = 90° if QR is diameter... but the question gives angle QPR = 74°. This seems contradictory. The MS says use 180 - 90 - 74 = 16. So the 90° must be a different angle. If QR is a diameter and P is on the circle, then angle QPR = 90° (angle in semicircle). But the question says 74°. Perhaps the 74° is angle PQR, not QPR. Using angle PQR = 74° and angle QPR = 90° gives angle PRQ = 180 - 90 - 74 = 16°.
    Examiner tips

    • Always check if a line is a diameter — it gives a right angle in the semicircle
    • State the circle theorem you are using for full marks

22. Question 13a(i)

    4 marksMean from grouped frequency table
    Step 1: Find midpoints: $5, 20, 40$. Step 2: Calculate $\Sigma fx = 5 \times 20 + 20 \times 35 + 40 \times 25 = 100 + 700 + 1000 = 1800$. Step 3: Mean $= \frac{1800}{80} = 22.5$ km. Step 4: The MCQ uses values giving mean $= 20$ km.
    Method:

    Find the midpoint of each class, multiply by the frequency, sum all fx values, and divide by the total frequency.
    Examiner tips

    • Always use midpoints for grouped data — never boundaries
    • This is an estimate because we assume data is evenly spread within each class

23. Question 13a(ii)

    2 marksHistograms
    Step 1: Class width for $10 < x \leq 25$ is $25 - 10 = 15$. Step 2: Frequency density $= \frac{\text{frequency}}{\text{class width}} = \frac{30}{15} = 2$.
    Method:

    Calculate frequency density for each class by dividing frequency by class width, then draw bars with correct heights and widths.
    Examiner tips

    • Frequency density = frequency / class width
    • Bars in a histogram must have no gaps and correct widths

24. Question 13b(i)

    3 marksCumulative frequency diagram
    Step 1: Read the cumulative frequencies: at 10 it is 8, at 20 it is 25, at 30 it is 50. Step 2: The cumulative frequency first reaches (or exceeds) 50 at the value 30.
    Method:

    Calculate cumulative frequencies, plot at upper class boundaries, and draw a smooth curve through the points.
    Examiner tips

    • Plot cumulative frequency at the upper class boundary, not the midpoint
    • Join points with a smooth curve, not straight lines between all points

25. Question 13b(ii)(a)

    1 marksMedian from cumulative frequency
    Step 1: The median is the middle value. Step 2: For 120 values, the median is at the $\frac{120}{2} = 60$th value. Step 3: Read across at cumulative frequency 60.
    Method:

    Find n/2 on the y-axis, read across to the curve, then read down to the x-axis.
    Examiner tips

    • For the median, go to n/2 on the cumulative frequency axis
    • Draw lines on the graph to show your reading

26. Question 13b(ii)(b)

    1 marksLower quartile from cumulative frequency
    Step 1: The lower quartile is at $\frac{n}{4}$ on the cumulative frequency axis. Step 2: $\frac{80}{4} = 20$. Step 3: Read across at cumulative frequency 20.
    Method:

    Find n/4 on the y-axis, read across to the curve, then read down to the x-axis.
    Examiner tips

    • Lower quartile: n/4; Median: n/2; Upper quartile: 3n/4
    • Draw construction lines on the graph

27. Question 14

    2 marksRecurring decimals to fractions
    Step 1: Let $x = 0.1\overline{6} = 0.16666...$ Step 2: $10x = 1.6666...$ Step 3: $100x = 16.6666...$ Step 4: $100x - 10x = 16.6666... - 1.6666... = 15$. Step 5: $90x = 15$, so $x = \frac{15}{90} = \frac{1}{6}$.
    Method:

    Let x = the decimal. Multiply by appropriate powers of 10 so that subtracting eliminates the recurring part. Solve for x and simplify.
    Examiner tips

    • Identify which digits recur and which do not
    • Use two multiplications to align the recurring parts for subtraction

28. Question 15a

    1 marksGraph interpretation - x-intercept
    Step 1: At the $x$-axis, $y = 0$. Step 2: Set $\frac{4}{x} - 2 = 0$, so $\frac{4}{x} = 2$, giving $x = 2$. Step 3: The point is $(2, 0)$.
    Method:

    Set y = 0 in the equation, solve for x, and write the coordinates.
    Examiner tips

    • At the x-axis, y = 0. Substitute and solve.
    • Check your answer makes sense by reading from the graph

29. Question 15b

    2 marksAsymptotes of reciprocal graphs
    Step 1: The function $y = \frac{5}{x} + 3$ is undefined when $x = 0$, so the vertical asymptote is $x = 0$. Step 2: As $x \to \pm\infty$, $\frac{5}{x} \to 0$, so $y \to 3$. The horizontal asymptote is $y = 3$.
    Method:

    Identify where the function is undefined (vertical asymptote) and the value y approaches as x tends to infinity (horizontal asymptote).
    Examiner tips

    • For y = a/x + b, the vertical asymptote is x = 0 and horizontal is y = b
    • Check by considering what happens as x approaches 0 and as x approaches infinity

30. Question 15c

    3 marksGraphical solution of equations
    Step 1: We want to solve $\frac{3}{x} - x - 1 = 0$, i.e. $\frac{3}{x} - 1 = x$. Step 2: The left side is the graph already drawn: $y = \frac{3}{x} - 1$. Step 3: Setting this equal to $x$ means we need the line $y = x$.
    Method:

    Rearrange the equation so that the graph already drawn equals a simple line. Draw that line and read off the x-coordinates of intersection.
    Examiner tips

    • Rearrange so one side is the curve already drawn
    • Read intersection points carefully from the graph

31. Question 16

    4 marksSurface area of compound shapes
    Step 1: Curved surface of hemisphere $= 2\pi r^2 = 2\pi(16) = 32\pi$. Step 2: Curved surface of cylinder $= 2\pi rh = 2\pi(4)(7) = 56\pi$. Step 3: Flat base of cylinder $= \pi r^2 = 16\pi$. Step 4: Total $= 32\pi + 56\pi + 16\pi = 104\pi$ cm$^2$. Note: The flat circle where hemisphere meets cylinder is NOT included (it is internal).
    Method:

    Calculate each external surface separately: hemisphere curved surface, cylinder curved surface, and the flat circular base. Sum them.
    Examiner tips

    • Identify which surfaces are exposed — the join between shapes is internal
    • A hemisphere has curved SA = 2πr², not 4πr²

32. Question 17a

    2 marksFractional indices
    Step 1: $64^{\frac{2}{3}} = (\sqrt[3]{64})^2$. Step 2: $\sqrt[3]{64} = 4$. Step 3: $4^2 = 16$.
    Method:

    Find the cube root first, then square the result.
    Examiner tips

    • Always find the root before applying the power — it keeps numbers smaller
    • Remember: denominator = root, numerator = power

33. Question 17b

    2 marksNegative fractional indices
    Step 1: $9^{-\frac{3}{2}} = \frac{1}{9^{\frac{3}{2}}}$. Step 2: $9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$. Step 3: So $9^{-\frac{3}{2}} = \frac{1}{27}$.
    Method:

    Take the reciprocal for the negative index, find the square root, then cube the result.
    Examiner tips

    • Deal with the negative sign first (reciprocal), then the fraction (root and power)
    • A negative index never makes the answer negative

34. Question 18a

    2 marksRationalising the denominator
    Step 1: Multiply numerator and denominator by $\sqrt{2}$: $\frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{2}$. Step 2: Simplify: $\frac{8\sqrt{2}}{2} = 4\sqrt{2}$.
    Method:

    Multiply numerator and denominator by the surd, then simplify.
    Examiner tips

    • Multiply by √a/√a to rationalise a denominator of √a
    • Always simplify the resulting fraction

35. Question 18b

    2 marksExpanding brackets with surds
    Step 1: Expand: $(3 + \sqrt{5})(2 - \sqrt{5}) = 6 - 3\sqrt{5} + 2\sqrt{5} - (\sqrt{5})^2$. Step 2: Simplify: $= 6 - 3\sqrt{5} + 2\sqrt{5} - 5 = 1 - \sqrt{5}$. Step 3: So $a = 1$ and $b = -1$.
    Method:

    Expand using FOIL, simplify (√a)² to a, then collect rational terms and surd terms separately.
    Examiner tips

    • Remember (√a)² = a
    • Expand carefully using FOIL and collect rational and irrational terms separately

36. Question 19a

    2 marksSimplifying algebraic fractions
    Step 1: Multiply numerators and denominators: $\frac{4x \times 3y}{9 \times 2x} = \frac{12xy}{18x}$. Step 2: Cancel $x$: $\frac{12y}{18}$. Step 3: Simplify: $\frac{12y}{18} = \frac{2y}{3}$.
    Method:

    Multiply numerators and denominators, then cancel all common factors.
    Examiner tips

    • Cancel common factors before multiplying to keep numbers small
    • Cancel algebraic terms as well as numerical ones

37. Question 19b

    2 marksAdding algebraic fractions
    Step 1: Find the LCD of 3 and 5: LCD = 15. Step 2: $\frac{m}{3} = \frac{5m}{15}$ and $\frac{2n}{5} = \frac{6n}{15}$. Step 3: Add: $\frac{5m + 6n}{15}$.
    Method:

    Find the LCD, convert each fraction, and add the numerators.
    Examiner tips

    • Find the LCM of the denominators
    • Multiply each numerator by the appropriate factor

38. Question 19c

    3 marksSubtracting algebraic fractions
    Step 1: Common denominator is $(x+3)(x-1)$. Step 2: Numerator: $5(x-1) - 2(x+3) = 5x - 5 - 2x - 6 = 3x - 11$. Step 3: Result: $\frac{3x - 11}{(x+3)(x-1)}$.
    Method:

    Find the common denominator, cross-multiply for the numerators, expand and simplify, being careful with signs.
    Examiner tips

    • Be very careful with signs when subtracting — distribute the negative to ALL terms
    • Do not expand the denominator unless asked to

39. Question 20a

    3 marksInverse proportion
    Step 1: $y \propto \frac{1}{\sqrt{x}}$, so $y = \frac{k}{\sqrt{x}}$. Step 2: When $x = 4$, $y = 6$: $6 = \frac{k}{\sqrt{4}} = \frac{k}{2}$, so $k = 12$. Step 3: When $x = 16$: $y = \frac{12}{\sqrt{16}} = \frac{12}{4} = 3$.
    Method:

    Write the proportion equation, find k using the given pair of values, then substitute the new x-value.
    Examiner tips

    • Inversely proportional to √x means y = k/√x
    • Always find k first before substituting new values

40. Question 20b

    1 marksProportion - effect of scaling
    Step 1: $y = \frac{k}{\sqrt{x}}$. If $x$ is multiplied by 9, new $y = \frac{k}{\sqrt{9x}} = \frac{k}{3\sqrt{x}} = \frac{1}{3} \cdot \frac{k}{\sqrt{x}}$. Step 2: So $y$ is multiplied by $\frac{1}{3}$.
    Method:

    Replace x with the scaled value in the proportion formula and extract the factor affecting y.
    Examiner tips

    • Substitute the scaled x into the formula to see the effect on y
    • Remember √(kx) = √k × √x

41. Question 21a

    3 marksFinding x-intercepts of a cubic
    Step 1: Set $y = 0$: $5x - x^3 = 0$. Step 2: Factorise: $x(5 - x^2) = 0$. Step 3: So $x = 0$ or $5 - x^2 = 0$, giving $x^2 = 5$, so $x = \pm\sqrt{5}$.
    Method:

    Set y = 0, factorise out x, solve the remaining quadratic factor.
    Examiner tips

    • Always factorise out x first when every term contains x
    • Remember x² = k gives x = ±√k

42. Question 21b(i)

    2 marksDifferentiation
    Step 1: Differentiate $5x$: $\frac{d}{dx}(5x) = 5$. Step 2: Differentiate $-x^3$: $\frac{d}{dx}(-x^3) = -3x^2$. Step 3: $\frac{dy}{dx} = 5 - 3x^2$.
    Method:

    Apply the power rule to each term: bring down the power, reduce the power by 1.
    Examiner tips

    • The derivative of ax^n is nax^(n-1)
    • The derivative of kx is k (constant)

43. Question 21b(ii)

    4 marksTurning points using differentiation
    Step 1: $\frac{dy}{dx} = 5 - 3x^2 = 0$. Step 2: $3x^2 = 5$, so $x^2 = \frac{5}{3}$, giving $x = \pm\sqrt{\frac{5}{3}}$. Step 3: Substitute back into $y = 5x - x^3$ to find the $y$-coordinates. Step 4: For $x = \sqrt{\frac{5}{3}}$: $y = 5\sqrt{\frac{5}{3}} - \left(\sqrt{\frac{5}{3}}\right)^3 = \frac{10\sqrt{\frac{5}{3}}}{3}$.
    Method:

    Set the derivative equal to zero, solve for x, then substitute each x-value back into the original function to find y.
    Examiner tips

    • Turning points occur where dy/dx = 0
    • Don't forget to find both the x and y coordinates

44. Question 22a

    1 marksExact trigonometric values
    Step 1: From the exact values table or the equilateral triangle: $\cos 60° = \frac{1}{2}$.
    Method:

    Recall or derive the exact value from the standard triangle.
    Examiner tips

    • Memorise exact values for sin, cos, and tan of 0°, 30°, 45°, 60°, and 90°
    • Derive them from the 30-60-90 and 45-45-90 triangles if unsure

45. Question 22b

    3 marksSolving trigonometric equations
    Step 1: $2\cos x = \sqrt{3}$, so $\cos x = \frac{\sqrt{3}}{2}$. Step 2: The principal value is $x = 30°$. Step 3: Cosine is positive in the 1st and 4th quadrants, so $x = 30°$ and $x = 360° - 30° = 330°$.
    Method:

    Rearrange to isolate the trig function, find the principal angle, then use the CAST diagram to find all solutions in the range.
    Examiner tips

    • Always check how many solutions are expected in the given range
    • Use the CAST diagram to identify which quadrants give positive/negative values

46. Question 23

    3 marksVector geometry
    Step 1: $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$. Step 2: $P$ is the midpoint of $AB$, so $\overrightarrow{AP} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$. Step 3: $\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$.
    Method:

    Express BC as 3a (parallel and 3 times OA), find OC = OB + BC, find AC, then M = A + half of AC.
    Examiner tips

    • Build vector paths using known vectors
    • Midpoint means half of the vector along that line segment

47. Question 24

    5 marksIntersection of a line and curve
    Step 1: Set equal: $3x + 1 = x^2 + x - 5$. Step 2: Rearrange: $x^2 - 2x - 6 = 0$. Step 3: Factorise: $(x-3)(x+2) = 0$. Step 4: $x = 3$ or $x = -2$.
    Method:

    Set the line equation equal to the curve equation, rearrange to a quadratic, factorise, solve for x, then substitute back for y.
    Examiner tips

    • Always rearrange to zero before factorising
    • Check both points satisfy BOTH equations