🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (d) 3.425 — x₁ = ∛(6 × 3 + 20) = ∛38 = 3.3620 (unrounded, 3.36198...). x₂ = ∛(6 × 3.3620 + 20) = ∛40.172 = 3.425 (3 d.p.). Choosing 3.362 stops at x₁ instead of continuing to x₂. Choosing 2.722 leaves out the '+ 20' inside the root, working out ∛(6 × 3.3620) = ∛20.172 = 2.722. Choosing 0.556 subtracts 20 instead of adding it, working out ∛(6 × 3.3620 − 20) = ∛0.172 = 0.556.
2. (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
3. (c) 3/20 — Method: the second fraction is quoted for the perennials only, so it is a conditional probability and the two fractions multiply. Working: the probability that a plant is a perennial is 3/5, and given that it is a perennial the probability that it is in flower is 1/4. Multiplying gives 3 × 1 over 5 × 4, which is 3/20. Answer: the probability is 3/20. The distractors: 17/20 comes from adding the fractions, 12/20 plus 5/20, instead of multiplying, which would be right only for two outcomes that cannot both happen; 4/9 comes from adding the numerators and the denominators separately, the classic 3 + 1 over 5 + 4; 1/4 quotes the flowering fraction on its own, as though every plant in the garden centre were a perennial, so the 3/5 is never used.
4. (d) 25 ≤ t < 45 — Method: the height of a bar on a histogram is the frequency density, so work out frequency ÷ class width for every class and compare the four heights. Working: 20 ÷ 10 = 2 for the first class; 12 ÷ 15 = 0.8 for the second; 50 ÷ 20 = 2.5 for the third; 60 ÷ 30 = 2 for the fourth. Answer: the largest of 2, 0.8, 2.5 and 2 is 2.5, so the tallest bar is the one for 25 ≤ t < 45. The distractors: 45 ≤ t < 75 comes from picking the class with the greatest frequency, 60, and treating a frequency as a height — but that class is three times as wide, so its 60 competitors are spread thinly; 10 ≤ t < 25 comes from dividing the class width by the frequency, 15 ÷ 12, and picking the largest of those reversed values; 0 ≤ t < 10 comes from assuming the narrowest class must always give the tallest bar, which is only true when the frequencies are equal.
5. (b) 3.38 — r₁ = √(300 ÷ (π × 8)) = √11.9366 = 3.4550. r₂ = √(300 ÷ (π × 8.4550)) = √11.2947 = 3.3608. r₃ = √(300 ÷ (π × 8.3608)) = √11.4232 = 3.3798, which rounds to 3.38. Choosing 3.36 stops at r₂, one iteration too early. Choosing 4.82 leaves out the '+ 5' inside the bracket, dividing by π × rₙ instead of π × (rₙ + 5). Choosing 3.45 comes from using π ≈ 3 instead of the calculator's π key throughout.
6. (d) 1.2 pupils per mark — Method: on a histogram whose class intervals are not all the same width the height of a bar is not the frequency but the frequency density, found by dividing the frequency of the class by the width of that class, so that the area of the bar represents the frequency. Working: the class 40 ≤ m < 60 holds 24 pupils, and its width is 60 − 40 = 20 marks, so the frequency density is 24 ÷ 20 = 1.2. Answer: 1.2 pupils per mark. The distractors: 24 pupils per mark comes from plotting the frequency itself as the height, which is only correct when every class has the same width; 2.4 pupils per mark comes from dividing by 10, the width of the narrowest class, instead of by the width of this class; 0.2 pupils per mark comes from dividing by the 120 pupils in the year group, which gives the proportion of pupils in the class and not a frequency density.
7. (b) 1/3 — Method: the first sweet has already been taken and it was red, so work out the second probability from what is actually left in the bag. Working: one red sweet has gone, so 3 red sweets remain out of 9 sweets altogether, giving 3/9. Dividing the numerator and the denominator by 3 gives 1/3. Answer: the probability is 1/3. The distractors: 2/5 is 4/10, the probability for the first draw used again, which is only right if the first sweet is put back; 3/10 takes one off the red count but leaves the total at 10, updating half of the fraction; 4/9 takes one off the total but leaves the red count at 4, updating the other half of the fraction.
8. (d) x² − 7x − 3 = 0 — Method: an iteration settles where the next value equals the one before it, so both can be written as the same letter x; replace every xₙ by x, square both sides to clear the square root, and collect all the terms on one side. Working: x = √(7x + 3) gives x² = 7x + 3 on squaring both sides; subtracting 7x and 3 from both sides gives x² − 7x − 3 = 0. Answer: x² − 7x − 3 = 0. The distractors: x² + 7x − 3 = 0 moves the 7x across the equals sign without changing its sign; x² − 7x + 3 = 0 makes that same slip on the constant instead, leaving the 3 positive as it crosses; x² − 7x − 9 = 0 squares the expression term by term, squaring the 3 to give 9 as though squaring √(7x + 3) gave 7x + 9, which is the (a + b)² = a² + b² mistake dressed as a square root.
9. (d) (2/5)a + (3/5)b — Method: OP = OA + AP, and since AP : PB = 3 : 2 splits AB into 5 equal parts, AP is 3/5 of the whole of AB, with AB = b − a. Working: OP = a + 3/5(b − a) = a − (3/5)a + (3/5)b = (2/5)a + (3/5)b. Answer: OP = (2/5)a + (3/5)b. Using the ratio the wrong way round, as though it read AP : PB = 2 : 3, gives (3/5)a + (2/5)b; adding (3/5)b onto the whole of a without subtracting a inside the bracket first gives a + (3/5)b; and treating the ratio as 1 : 1 gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from A, matching the ORDER the ratio is stated in, and subtract before you scale.
10. (b) Q was faster on average and more consistent — Method: compare the medians for the average and the interquartile ranges for the spread, remembering that a shorter time is faster and a smaller interquartile range means more consistent. Working: the median for class Q is 35 seconds against 38 seconds for class P, so class Q was faster on average; the interquartile range for class P is 46 − 24 = 22 seconds and for class Q it is 44 − 30 = 14 seconds, so class Q's times are more tightly grouped. Answer: class Q was faster on average and more consistent. The distractors: calling Q slower comes from comparing the lower quartiles, 30 against 24, as though a quartile were the average; calling Q less consistent comes from using the gap between the median and the upper quartile as the spread, 44 − 35 = 9 against 46 − 38 = 8, instead of the full interquartile range; the statement that Q was both slower and less consistent comes from making both of those mistakes together.
11. (b) (12, −9) — Two enlargements one after the other combine into a single scale factor: 1.5 × 2 = 3. Multiplying a vector by a scalar means multiplying both the top number and the bottom number by it: top = 4 × 3 = 12, bottom = −3 × 3 = −9, giving (12, −9). A candidate who adds the scale factor to each number instead of multiplying gets (4 + 3, −3 + 3) = (7, 0). A candidate who multiplies the top number but leaves the bottom number unchanged gets (12, −3). A candidate who multiplies the bottom number but leaves the top number unchanged gets (4, −9). The correct column vector for the poster is (12, −9).
12. (b) (x + 2)/3 — Start with y = 3x − 2 and swap x and y: x = 3y − 2. Add 2 to both sides: x + 2 = 3y. Divide both sides by 3: y = (x + 2)/3, so f⁻¹(x) = (x + 2)/3. Writing x/3 + 2 comes from dividing only the 3y term by 3 and leaving the +2 outside the division — the 2 must be added before you divide, not after. Writing (x − 2)/3 comes from keeping the subtraction sign instead of flipping it to addition when the −2 is moved across the equals sign. Writing 3x + 2 comes from swapping x and y but never actually solving for y — just changing the sign of the constant term.
13. (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
14. (a) y = −x³ + 5x — Reflecting a graph in the y-axis replaces every x in the equation with −x: y = (−x)³ − 5(−x) = −x³ + 5x. Writing y = −x³ − 5x comes from substituting −x into the x³ term only and leaving the −5x term as it was. Writing y = x³ + 5x comes from substituting −x into the −5x term only and leaving the x³ term as it was. Writing y = x³ − 5x is the original equation with no reflection applied at all — every term needs the substitution, not just one of them.
15. (c) (10.5, −2) — Method: the vector from the first tile to the third tile is the pattern's vector doubled, since two translations happen between them. Working: doubling (4.5, −2.5) gives (9, −5); adding this to the starting corner (1.5, 3) gives x-coordinate 1.5 + 9 = 10.5 and y-coordinate 3 − 5 = −2. Answer: (10.5, −2). A candidate who only applies the vector once, translating to the second tile instead of the third, gets (6, 0.5). A candidate who adds 2.5 instead of subtracting it in the y-coordinate gets (10.5, 8). A candidate who doubles the x-part of the vector correctly but forgets to change the y-coordinate at all gets (10.5, 3).
16. (c) (2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 8(n² + 2n + 1) + 2, and n² + 2n + 1 is an integer, so the sum is always 2 more than a multiple of 8. — Expand each square carefully: (2n + 1)² = 4n² + 4n + 1 and (2n + 3)² = 4n² + 12n + 9, since the cross term is 2 × 2n × 3 = 12n. Adding gives 8n² + 16n + 10, and factorising out 8 from every term that can hold one gives 8(n² + 2n + 1) + 2; since n² + 2n + 1 is always an integer, the sum is always 2 more than a multiple of 8. The attempt reaching 8(n² + 2n) + 10 has the correct expansion but stops the factorisation one step early — it never pulls a further 8 out of the 10 (10 = 8 + 2), so 'always 10 more than a multiple of 8' should be reduced to 'always 2 more than a multiple of 8'. The attempt reaching 2(4n² + 8n + 5) also has the correct expansion, and the factorisation is true, but 'always even' only shows the sum is a multiple of 2 — being even is necessary but nowhere near sufficient to be a multiple of 8, and the argument never finds the extra factor of 4. The fourth attempt makes an expansion slip, using (2n + 3)² = 4n² + 9 instead of 4n² + 12n + 9 — dropping the 12n cross term entirely — so it works from the wrong expression 8n² + 4n + 10 throughout, and no amount of correct working afterwards can recover the right conclusion.
17. (a) 1.861 — x₁ = ∛(10 − 2²) = ∛6 = 1.817120593. x₂ = ∛(10 − 1.817120593²) = ∛6.698072751 = 1.885022855. x₃ = ∛(10 − 1.885022855²) = ∛6.446688837 = 1.861139399, which rounds to 1.861. Reporting x₂ instead of x₃ gives 1.885022855, which rounds to 1.885. Stopping after the first iteration and reporting x₁ instead of x₃ gives 1.817120593, which rounds to 1.817. A sign error inside the cube root, using xₙ₊₁ = ∛(10 + xₙ²) instead of ∛(10 − xₙ²), gives x₁ = ∛14 = 2.410142264, x₂ = ∛(10 + 2.410142264²) = 2.509763724, and x₃ = ∛(10 + 2.509763724²) = 2.535437381, which rounds to 2.535.
18. (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.