🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (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.
2. (d) −2 — y = f(x) − 6 is f(x) shifted down by 6, so every y-value on the graph decreases by 6. At x = 0, f(0) = 4, so the new y-value is 4 − 6 = −2. Adding 6 instead of subtracting gives 10; writing down the shift itself, −6, or leaving the original value 4 unchanged both skip the translation altogether.
3. (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.
4. (c) Translate +1 in y; new range 0 ≤ y ≤ 2 — Adding 1 to sin x translates the graph 1 unit in the positive y-direction, and every y-value on the range increases by 1: −1 + 1 = 0 and 1 + 1 = 2, giving a new range of 0 ≤ y ≤ 2. A translation in the x-direction would not add anything to the y-values, so 'Translate +1 in x; range unchanged' correctly leaves the range at −1 ≤ y ≤ 1 for that (wrong) transformation, but the transformation itself is not what y = sin x + 1 shows. 'Translate +1 in y; new range −2 ≤ y ≤ 0' correctly spots the translation but subtracts 1 from each bound instead of adding it. 'Reflect in the x-axis' mistakes the transformation for a reflection rather than a translation; reflecting sin x in the x-axis does leave the range at −1 ≤ y ≤ 1 unchanged in size, but that is not the transformation y = sin x + 1 actually applies.
5. (a) 4.897 — Method: substitute the starting value into the right-hand side of the formula to get x₁, then feed each new value back in, keeping the whole calculator display every time and rounding only at the very end. Working: x₁ = √(5 × 2 + 2) = √12 = 3.46410…; x₂ = √(5 × 3.46410… + 2) = √19.32050… = 4.39551…; x₃ = √(5 × 4.39551… + 2) = √23.97755… = 4.89668…, which is 4.897 correct to 3 decimal places. Answer: 4.897. The distractors: 4.396 is x₂, written down by a candidate who counts the starting value x₀ as the first iterate and so stops one use of the formula early; 3.464 is x₁, the value after using the formula only once; 5.146 is x₄, one use of the formula too many — the mirror image of the first slip, made by a candidate who labels the first value worked out as x₀ rather than as x₁ and so runs the count a step long.
6. (d) 2 — Method: if f is 3 times e, then each part of f equals 3 times the matching part of e. Working: using the bottom numbers, 6 = 3 × k, so k = 2. Answer: k = 2. A candidate who multiplies instead of dividing, working out 6 × 3, gets 18. A candidate who uses the top numbers' ratio instead, 15 ÷ 5, and gives that ratio as k gets 3. A candidate who adds instead of using the multiple relationship, working out 6 + 3, gets 9.
7. (c) 70 — Method: estimate the median from the cumulative frequency table by interpolation: find its position, n ÷ 2, locate the class it falls in, then add the fraction of the way through that class (adjusted for the cumulative frequency reached before it) to the class's lower boundary. Working: there are 180 riders, so the median is at position 180 ÷ 2 = 90. Before the class 60 ≤ d < 80 the cumulative frequency is 60, and by the end of it, 120, so the 90th rider falls in this class; its frequency is 120 − 60 = 60 and its width is 80 − 60 = 20. The extra distance needed into the class is 90 − 60 = 30, and 30 ÷ 60 × 20 = 10, so the median is 60 + 10 = 70. Answer: the estimated median distance is 70 km. Watch which numbers the interpolation actually uses: reading off just the class's lower boundary, 60, ignores how far into the class the 90th rider falls; using the target position, 90, as the extra distance instead of subtracting the 60 riders already counted before the class gives 90 ÷ 60 × 20 = 30, so 60 + 30 = 90, overshooting by treating the whole position as if none of it had already been counted; and using the total number of riders, 180, instead of half of it as the target position lands in the very last class, giving an estimate of 130 km — further than any rider is known to have ridden by that point in the table.
8. (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.
9. (c) 10x − 15 — Expand the bracket first: 3(2x − 5) = 6x − 15. Then add the 4x: 6x − 15 + 4x = 10x − 15. The option 10x − 5 comes from forgetting to multiply the 5 inside the bracket by 3 (treating it as 6x − 5), then adding 4x. The option 10x + 15 comes from a sign error when expanding, treating 3 × (−5) as +15 instead of −15, then adding 4x. The option 6x − 15 comes from expanding the bracket correctly but forgetting to add the 4x term at all.
10. (a) A translation by vector (0, 3) — y = x² + 3 adds a constant outside the squaring, so it is a vertical translation of y = x² — every point moves the same distance parallel to the y-axis, with no change in x. The vector is (0, 3), matching the +3. A vector of (3, 0) confuses this with a horizontal shift; (0, −3) has the right axis but the wrong sign, since the graph moves up, not down; a stretch changes the shape of the curve, which adding a constant term never does.
11. (a) (30, 60) — Method: a cumulative frequency point is plotted at the upper boundary of its class, paired with the running total of all the frequencies up to and including that class. Working: the running totals are 7, then 7 + 19 = 26, then 26 + 34 = 60, then 60 + 40 = 100; the class 20 ≤ t < 30 has upper boundary 30, and the running total there is 60. Answer: the point for that class is plotted at 30 seconds against a cumulative frequency of 60. The distractors: (25, 60) comes from plotting at the class midpoint, which is what a frequency polygon uses and not what a cumulative frequency diagram uses; (30, 34) comes from plotting the class frequency, 34, rather than the running total; (20, 60) comes from plotting at the lower boundary of the class, which would claim that 60 calls took less than 20 seconds when only 26 did.
12. (a) 9/25 — Method: P(badminton | tennis) = n(tennis and badminton) ÷ n(tennis) — restrict to the tennis-players, then find what fraction of them also play badminton. Working: n(tennis and badminton) = 18, n(tennis) = 50, so P(badminton | tennis) = 18/50 = 9/25. Answer: 9/25. Watch out: dividing by 40 (the badminton total) finds P(tennis | badminton) instead of P(badminton | tennis) — the wrong direction. Dividing by 90 (all the members named in the question) ignores that you already know the member plays tennis. And dividing by 72 (50 + 40 − 18, the number who play at least one of the two sports) answers a question about the union, not the condition you were given.
13. (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.
14. (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
15. (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.
16. (b) 9x + 40y = 1681 — For a circle x² + y² = r² centred at the origin, the tangent at a point (a, b) on the circle has equation ax + by = r². Here (a, b) = (9, 40) and r² = 1681, so the tangent is 9x + 40y = 1681. Choosing 40x + 9y = 1681 swaps the coefficients, using the y-coordinate as the x-coefficient and the x-coordinate as the y-coefficient. Choosing 9x + 40y = 41 uses the radius 41 instead of r² = 1681 as the constant. Choosing 9x − 40y = 1681 has the correct coefficients and constant but the wrong sign on the y-term.
17. (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.
18. (a) False — 4(2x − 3) = 8x − 12, not 8x − 3. — Expand the bracket by multiplying both terms by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12, which is not 8x − 3 — the student is wrong. Saying 4(2x − 3) = 8x − 3 comes from multiplying only the 2x by 4 and copying the −3 across unchanged. Saying 4(2x − 3) = 2x − 12 comes from multiplying only the −3 by 4 and leaving 2x unmultiplied. Claiming it is true because both expressions are linear ignores that equivalence depends on the actual coefficients, not the type of expression.