🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (c) (−35, −55) — Add the three stages component by component to find the drone's position relative to base: (30+(−10)+15, 40+20+(−5)) = (35, 55). The flight back to base is the negative of this vector, reversing both numbers: (−35, −55). (35, 55) is the vector from base to the drone's position — it forgets to reverse direction for the return flight. (−35, 55) only reverses the top number. (35, −55) only reverses the bottom number.
2. (c) A histogram, with frequency density up the vertical axis — Method: decide which diagram makes area stand for frequency, which is the property the question asks for. Working: on a histogram the vertical axis is frequency density, so the area of a bar is frequency density × class width, and that product is the frequency; this is exactly what is wanted, and it is what allows classes of unequal width to be shown fairly. Answer: a histogram, with frequency density up the vertical axis. The distractors: a bar chart plots frequency as the height, so with unequal widths a wide class would cover far more area than a narrow class holding the same number of batteries, and area would measure nothing; a cumulative frequency diagram plots running totals against upper class boundaries, so a point on it gives how many lie below a value rather than how many lie in a class; a pie chart shows each class as a share of the whole 300 and loses the class widths entirely, so no area on it is tied to a scale of hours.
3. (b) 143 — fg(4) means f(g(4)): work out g(4) first, then substitute the result into f. g(4) = 3 × 4 = 12, then f(12) = 12² − 1 = 144 − 1 = 143. Working out gf(4) instead swaps the order: f(4) = 4² − 1 = 15, then g(15) = 3 × 15 = 45 — that is the wrong composition. Treating f(x) as x − 1 (forgetting to square the input) gives f(12) = 12 − 1 = 11. Applying g twice instead of applying g then f gives g(g(4)) = g(12) = 3 × 12 = 36, which mixes up which function should be applied second.
4. (a) (6, 8) — Method: a point lies on the circle x² + y² = 100 exactly when the squares of its two coordinates add to 100, so square both coordinates of each point and add them. Working: for (6, 8), 6² + 8² = 36 + 64 = 100, which matches the right-hand side of the equation. Answer: (6, 8) lies on the circle. The distractors: (3, 4) is the 3, 4, 5 right-angled triangle recalled but never scaled up to a radius of 10, and 3² + 4² = 25, so it lies on the far smaller circle x² + y² = 25; (5, 5) has coordinates adding to 10, which compares the sum of the coordinates with the radius instead of the sum of their squares with r², and 5² + 5² = 50; (10, 10) takes each coordinate separately to equal the radius, and 10² + 10² = 200, which is twice too big.
5. (a) xₙ₊₁ = ∛(5xₙ + 3) — Starting from x³ − 5x − 3 = 0, add 5x and 3 to both sides to get x³ = 5x + 3, then take the cube root of both sides: x = ∛(5x + 3), giving the iterative formula xₙ₊₁ = ∛(5xₙ + 3). A sign error when moving the constant term across, treating x³ − 5x − 3 = 0 as x³ = 5x − 3, gives xₙ₊₁ = ∛(5xₙ − 3). Swapping the coefficient of x with the constant term gives xₙ₊₁ = ∛(3xₙ + 5), which does not come from x³ = 5x + 3 at all. Treating cubing as meaning multiply by 3 rather than raise to the power 3, and so undoing it by dividing by 3 instead of taking a cube root, gives xₙ₊₁ = (5xₙ + 3) ÷ 3.
6. (c) The class with times from 10 up to 20 — Method: to find the median class from a histogram, first turn each bar's frequency density into a frequency using density × class width, build up the cumulative frequency, and find the first class whose cumulative frequency reaches or passes n ÷ 2. Working: the four classes have widths 10, 10, 20 and 20, so their frequencies are 5 × 10 = 50, 2 × 10 = 20, 1.5 × 20 = 30 and 1 × 20 = 20, which add to the 120 visitors stated. The median sits at position 120 ÷ 2 = 60. The cumulative frequency is 50 after the first class and 50 + 20 = 70 after the second, so the 60th visitor is reached during the second class. Answer: the median lies in the class 10 ≤ t < 20. Watch which class each shortcut lands on: the tallest bar belongs to the first class, with the highest frequency density, 5 — but the tallest bar shows where visitors are packed most densely, not where the middle visitor falls, and picking it lands one class too early, at 0 ≤ t < 10; taking half of the total TIME span instead of half of the total NUMBER of visitors, 60 minutes ÷ 2 = 30 minutes, lands in the class 20 ≤ t < 40, confusing a value on the horizontal axis with a position in the data; and using the full 120 visitors as the target position, rather than 120 ÷ 2 = 60, reaches all the way to the last class, 40 ≤ t < 60, treating the whole data set's size as though it were the position of a single middle value.
7. (d) (6, 6) — y = f(x − 4) + 1 combines two translations: the −4 inside the brackets moves the graph 4 units to the RIGHT, and the +1 outside the brackets moves it 1 unit UP. Applying both to (2, 5): add 4 to the x-coordinate 2, and add 1 to the y-coordinate 5, giving (6, 6). Moving left instead of right, as the minus sign inside the bracket might suggest, gives (−2, 6) or (−2, 4); forgetting the horizontal shift altogether and only applying the vertical one gives (2, 6).
8. (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
9. (c) (1/2)c − a — Method: in parallelogram OABC, AB is equal and parallel to OC, so AB = c; M is the midpoint of AB, so AM = (1/2)c and OM = OA + AM = a + (1/2)c. MC runs from M to C, so MC = OC − OM. Working: MC = c − (a + (1/2)c) = (1/2)c − a. Answer: MC = (1/2)c − a. Subtracting in the wrong order gives a − (1/2)c, the same vector pointing the opposite way, from C to M rather than M to C; forgetting to halve the c-term gives c − a, which is AC, not MC; and adding instead of subtracting gives (1/2)c + a, which is OM itself. Always subtract the vector for the START of the journey, OM, from the vector for its END point, OC — and keep the fraction from the halving step.
10. (d) −11/60 — The radius from the origin to (11, 60) has gradient 60/11. The tangent is perpendicular to this radius, so its gradient is the negative reciprocal: −1 ÷ (60/11) = −11/60. Choosing 60/11 uses the radius's gradient unchanged, without applying perpendicularity. Choosing −60/11 negates the radius's gradient but forgets to take its reciprocal. Choosing 11/60 takes the reciprocal correctly but keeps the gradient positive instead of negative.
11. (d) It diverges, growing rapidly without limit. — x₁ = 2³ − 2 = 8 − 2 = 6. x₂ = 6³ − 2 = 216 − 2 = 214. x₃ = 214³ − 2 = 9800344 − 2 = 9800342. The values 6, 214, 9800342, … grow far larger at every step, so the sequence diverges rather than settling anywhere. Checking whether the sequence converges to a fixed value near 2 fails, since the terms grow enormously instead of levelling off. Checking for a repeating pair of values also fails, since 6, 214 and 9800342 are all different, with no sign of a return to 6. x₀ = 2 is a fixed point only if 2³ − 2 = 2, but 2³ − 2 = 6, not 2, so the sequence does not stay constant.
12. (c) 75 — Method: the number in a class is the area of its bar, frequency density × class width, so work out the frequency of each class that lies at or above 10 minutes and add them. Working: the class 10 ≤ t < 25 is 15 minutes wide with a frequency density of 3.2, giving 3.2 × 15 = 48 members; the class 25 ≤ t < 55 is 30 minutes wide with a frequency density of 0.9, giving 0.9 × 30 = 27 members; the total charged is 48 + 27 = 75. Answer: 75 members pay the extra charge. The distractors: 4.1 comes from adding the two frequency densities, 3.2 + 0.9, as though each height were a count; 93 comes from including the class 0 ≤ t < 10 as well, 1.8 × 10 = 18 added to 48 and 27, which charges every member; 27 comes from using only the class 25 ≤ t < 55 and forgetting that 10 ≤ t < 25 is also at or above 10 minutes.
13. (a) 1/3 — Method: since OABC is a parallelogram, B = OA + OC = a + c. M = (1/2)c, since M is the midpoint of OC. Since AN is twice NC, N is 2/3 of the way along AC from A, so N = a + 2/3(c − a) = (1/3)a + (2/3)c. Working: MN = N − M = (1/3)a + (1/6)c, and MB = B − M = a + (1/2)c. Comparing term by term, 1/3 × (a + (1/2)c) = (1/3)a + (1/6)c, which matches MN exactly. Answer: k = 1/3, so M, N and B lie on a straight line. Giving 2/3 instead is the scalar linking N to B (NB = (2/3)MB), not M to N; giving 1/6 is just MN's c-coefficient read off on its own, without comparing it to MB's c-coefficient at all; and giving 3 is the scalar the wrong way up — it is MB that equals 3 × MN, not the other way round, since MN = k × MB was what was asked for. Always match the direction of the scalar to the vectors exactly as the question states them.
14. (a) 4(x + 3) = 20 and 4x + 3 = 11 when x = 2, so the two expressions are not equivalent, because the bracket means the 3 must be added before multiplying by 4. — Substituting x = 2: 4(x + 3) = 4 × 5 = 20, and 4x + 3 = 8 + 3 = 11. The two values are different, and expanding 4(x + 3) algebraically gives 4x + 12, which can never equal 4x + 3 (that would require 12 = 3) — so the two expressions are never equivalent, for any value of x. The option claiming they become equal for a larger x is wrong: 4x + 12 = 4x + 3 has no solution at all. The option claiming they are equivalent because they share the terms 4x and 3 ignores that the bracket changes the constant term. The option that calculates 4(x + 3) as 11 ignores the bracket completely, applying the 4 only to the x term.
15. (d) Yes, because 3² + 4² = 25. — A point lies on the circle x² + y² = 25 exactly when substituting its coordinates makes the equation true. Squaring each coordinate separately and adding: 3² + 4² = 9 + 16 = 25, which matches the right-hand side, so (3, 4) does lie on the circle. Adding the coordinates without squaring them, 3 + 4 = 7, and then reasoning that 7 is less than 25 happens to reach the same verdict, but it is not testing the equation of the circle at all — the circle equation depends on x² + y², not x + y. Squaring the sum instead of summing the squares, (3 + 4)² = 49, not 25, wrongly rules the point out. Doubling each coordinate instead of squaring it, so that 4² is taken as 8, gives 9 + 8 = 17, not 25, which also wrongly rules the point out.
16. (c) 3.153 — x₁ = √(4 × 1 + 1) = √5 = 2.236067977. x₂ = √(4 × 2.236067977 + 1) = √9.944271908 = 3.153453965, which rounds to 3.153. Reporting x₁ instead of x₂ gives 2.236067977, which rounds to 2.236. A sign error inside the root, using xₙ₊₁ = √(4xₙ − 1) instead of √(4xₙ + 1), gives x₁ = √3 = 1.732050808 and x₂ = √(4 × 1.732050808 − 1) = √5.928203232 = 2.434790182, which rounds to 2.435. Applying the formula in the wrong order, working out √(4xₙ) + 1 at every step instead of √(4xₙ + 1), gives x₁ = √4 + 1 = 3 and x₂ = √(4 × 3) + 1 = 4.464101615, which rounds to 4.464.
17. (d) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
18. (a) 2.154 — Continuing xₙ₊₁ = ∛(3xₙ + 4) from x₁ = 2.154 gives x₂ = 2.187, x₃ = 2.195, settling towards the root near 2.196 — this is the formula that converges. Continuing xₙ₊₁ = (xₙ³ − 4) ÷ 3 from x₁ = 1.333 gives x₂ = −0.543 and then x₃ = −1.387, moving further from the root each time. Continuing xₙ₊₁ = 4 ÷ (xₙ² − 3) from x₁ = 4 gives x₂ = 0.308 and then x₃ = −1.377, swinging wildly rather than settling. Continuing xₙ₊₁ = xₙ³ − 2xₙ − 4 from x₁ = 0 gives x₂ = −4 and then x₃ = −60, running away from the root entirely.