🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (b) (3, 10) — Method: multiply every part of p by 2, then add the matching parts of q. Working: 2p = (4, 6); adding q gives top 4 + (−1) = 3 and bottom 6 + 4 = 10. Answer: 2p + q = (3, 10). A candidate who forgets to double p first, working out p + q instead, gets (1, 7). A candidate who doubles q instead of p, working out p + 2q, gets (0, 11). A candidate who subtracts q instead of adding it, working out 2p − q, gets (5, 2).
2. (b) (6, −8) — Method: a scalar multiple of m has the same ratio between its top and bottom numbers as m does. Working: m = (3, −4); multiplying both parts by 2 gives 2 × 3 = 6 and 2 × (−4) = −8, so (6, −8) is a scalar multiple of m. Answer: (6, −8). The vector (6, −4) needs a multiplier of 2 for the top number but only 1 for the bottom number, so it is not a multiple. The vector (−6, −8) needs a multiplier of −2 for the top number but 2 for the bottom number, so it is not a multiple. The vector (9, −8) needs a multiplier of 3 for the top number but 2 for the bottom number, so it is not a multiple.
3. (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
4. (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
5. (d) 5√2 — For a circle x² + y² = r², the 50 on the right-hand side is r², not r, so the radius is √50. Writing 50 as 25 × 2, the largest square factor times what remains, gives √50 = √25 × √2 = 5√2. Forgetting to square-root 50 at all and giving the value of r² instead gives 50. Halving 50 instead of taking its square root gives 25. Using 25 as the number left outside the square root sign, instead of as the number under it, gives the wrongly simplified 25√2.
6. (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.
7. (b) (1/2)b − (1/2)a — Method: MN runs from M to N, so MN = ON − OM, with OM = (1/2)a and ON = (1/2)b. Working: MN = (1/2)b − (1/2)a. Answer: MN = (1/2)b − (1/2)a. Subtracting the other way round gives (1/2)a − (1/2)b, the reverse vector from N to M; subtracting the wrong way round AND forgetting to halve gives a − b, which is BA, not MN; and adding the two halved vectors instead of subtracting them gives (1/2)a + (1/2)b, which is the position vector of the midpoint of AB. Always subtract the START point's vector from the END point's vector, and halve OA and OB before you combine them, not after.
8. (c) Week 16, £37,000 — y = f(x − 6) − 8000 combines a horizontal translation of 6 units RIGHT (subtracting 6 inside the brackets) with a vertical translation of £8000 DOWN (subtracting 8000 outside). Applying both to the maximum (10, 45000): 10 + 6 = 16, so the new maximum is in week 16. And 45000 − 8000 = 37000, so the maximum weekly profit is £37,000.
9. (d) 2/3 — Method: since A is the midpoint of OE, OE = 2a, so E = 2a. Since FB is twice AF, F is 1/3 of the way along AB from A, so F = a + 1/3(b − a) = (2/3)a + (1/3)b. G is the midpoint of OB, so G = (1/2)b. Working: EF = F − E = (2/3)a + (1/3)b − 2a = −(4/3)a + (1/3)b, and EG = G − E = −2a + (1/2)b. Comparing term by term, 2/3 × (−2a + (1/2)b) = −(4/3)a + (1/3)b, which matches EF exactly — the same scalar works on both the a-term and the b-term, so the two vectors are parallel, and since they share the point E the three points are collinear. Answer: k = 2/3, so E, F and G lie on a straight line. Giving 1/3 instead is the scalar linking F to G (FG = (1/3)EG), not E to F; giving 3/2 is the reciprocal — it is EG that equals 3/2 × EF, not the other way round, since EF = k × EG was what was asked for; and giving 4/3 is EF's a-coefficient read off raw, without ever dividing it by EG's a-coefficient to form the comparison. Always match the direction of the scalar to the vectors exactly as the question states them.
10. (b) 4.39 — x₁ = ∛(150 − 3 × 4²) = ∛(150 − 48) = ∛102 = 4.672 (unrounded). x₂ = ∛(150 − 3 × 4.672²) = ∛(150 − 65.49) = ∛84.51 = 4.39 (2 d.p.). Choosing 4.67 stops after only one iteration, giving x₁ instead of x₂. Choosing 84.51 finds the value inside the cube root for x₂ but never takes the cube root. Choosing 6.32 comes from adding 3xₙ² instead of subtracting it inside the root, which does not match the given formula.
11. (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.
12. (b) (0, 6) and (0, −6) — Method: every point on the y-axis has x-coordinate 0, so substitute x = 0 into the equation of the circle and solve for y, remembering that a square root has a negative value as well as a positive one. Working: putting x = 0 into x² + y² = 36 leaves y² = 36, so y = 6 or y = −6, and the two crossings are (0, 6) and (0, −6). Answer: (0, 6) and (0, −6). The distractors: (0, 36) and (0, −36) use 36 itself as the distance from the centre, which reads r² as r; (6, 0) and (−6, 0) are the right distance from the centre but are the crossings of the x-axis, found by setting y = 0 instead of x = 0; (0, 18) and (0, −18) halve 36, treating the right-hand side of the equation as a diameter.
13. (c) 1/56 — Method: for draws without replacement, multiply a chain of three fractions where both the numerator (reds remaining) and the denominator (counters remaining) fall by one after each draw. Working: P(all three red) = 3/8 × 2/7 × 1/6 = 6/336 = 1/56. Answer: 1/56. Watch out: using 3/8 for all three draws (27/512) treats the counters as if they were replaced each time. Reducing only the numerator each draw (3/8 × 2/8 × 1/8) forgets that the total number of counters left in the bag also falls. And reducing only the denominator while keeping the numerator at 3 each time (3/8 × 3/7 × 3/6) forgets that a red counter has actually left the bag.
14. (a) Outside: OQ² = 221 > r² = 169 — OQ² = 10² + 11² = 100 + 121 = 221. Comparing this with r² = 169: since 221 > 169, OQ > r, so Q lies outside the circle — this is the correct verdict AND the correct working. 'Outside: OQ = 21 (10 + 11) > r = 13' reaches the same Outside verdict, but by invalid working: it adds the coordinates instead of squaring them (10 + 11 = 21, rather than 10² + 11² = 221), so the stated 'OQ' of 21 is not a distance at all — the verdict happens to match, but the method is wrong. 'Inside: OQ ≈ 14.87 < r² = 169' correctly finds the distance OQ = √221 ≈ 14.87, but then compares that DISTANCE with r² = 169 instead of with r = 13 — comparing two different kinds of quantity gives a meaningless, and here wrong, verdict. 'Inside: OQ² = 221 < (2r)² = 676' confuses the radius with the diameter: it compares OQ² with the diameter squared, (2 × 13)² = 676, instead of with r² = 169.
15. (b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
16. (b) 6 — Reflecting y = f(x) in the x-axis turns the minimum point (2, −3) into a maximum point at (2, 3), since −f(x) negates every y-value: −(−3) = 3. Adding a then gives 3 + a = 9, so a = 9 − 3 = 6. Forgetting the reflection and using the original y-value of −3 gives −3 + a = 9, so a = 12 — this ignores that −f(x) changes the sign of the y-coordinate before a is added. Writing a = −12 comes from subtracting in the wrong order, working out 9 − (−3) as −3 − 9 instead. Writing a = −6 comes from taking the negative of the correct answer, as if the final value of a needed to be reflected too, on top of the turning point.
17. (d) 25 — Method: the median is estimated at position n ÷ 2 in the cumulative frequency table, then interpolated across the class it falls in: lower boundary, plus the fraction of the way through the class, times the class width. Working: there are 80 sacks, so the median sits at position 80 ÷ 2 = 40. Before the class 20 ≤ m < 30 the cumulative frequency is 22, and by the end of it, it is 58, so this class holds the 40th sack; its frequency is 58 − 22 = 36 and its width is 30 − 20 = 10. The extra distance needed into the class is 40 − 22 = 18, and 18 ÷ 36 × 10 = 5, so the median is 20 + 5 = 25. Answer: the estimated median mass is 25 kg. Watch which numbers the interpolation uses: reading off just the lower boundary of the median class, 20, ignores how far into that class the 40th sack actually falls; treating n ÷ 2 = 40 itself as the median mass mistakes a position in the list for a mass in kilograms; and using the target position, 40, as the extra distance into the class instead of subtracting the sacks already counted changes the calculation to 20 + 40 ÷ 36 × 10. That comes to 20 + 11.1 = 31.1, overshooting the class because it never subtracts the 22 sacks already counted before it.
18. (c) Correct: 3y = 4 − x gives f⁻¹(x) = (4 − x)/3 — Swap x and y: x = 4 − 3y. Add 3y to both sides and subtract x from both sides: 3y = 4 − x. Divide by 3: y = (4 − x)/3, which is exactly what Ben wrote — his rearrangement is correct. Check with a value: f(1) = 4 − 3 = 1, and Ben's formula gives (4 − 1)/3 = 1, which matches. 'Correct, but only because f is its own inverse' gives the right verdict for a false reason — f(f(x)) = 4 − 3(4 − 3x) = 9x − 8, which is not x, so f is not self-inverse; Ben's rearrangement is correct for the ordinary algebraic reason above, not because of any special property of f. 'Wrong: sign kept, giving (−4 − x)/3' comes from not carrying the swap through consistently — testing x = 1 gives (−4 − 1)/3 = −5/3, which does not equal 1, so it is wrong. 'Wrong: correct inverse is (x − 4)/3' comes from writing 3y = x − 4 instead of 3y = 4 − x, a sign slip when isolating y — testing x = 1 gives (1 − 4)/3 = −1, which again does not equal 1.