🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (c) 50 — Method: the edge of the track is a circle centred on the origin, so it has equation x² + y² = r²; use the floodlight whose coordinates are given to find r², then substitute x = 0 to find where that circle meets the positive y-axis. Working: 30² + 40² = 900 + 1600 = 2500, so the edge of the track is x² + y² = 2500. Substituting x = 0 gives k² = 2500, so k = √2500 = 50, taking the positive root because k is positive. Answer: k = 50. The distractors: 2500 is r² used as though it were the radius, which would place the second floodlight fifty times too far from the centre; 70 adds the two coordinates, 30 + 40, instead of adding their squares; 40 takes the y-coordinate of the first floodlight to be the radius, which would only be true if that floodlight stood on the y-axis itself.
2. (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.
3. (d) y = x² − 10x + 21 — A translation by the vector (3, 0) moves the graph 3 units in the positive x-direction, which means replacing every x in the equation with (x − 3). Substitute into x² − 4x: (x − 3)² − 4(x − 3). Expand (x − 3)² to x² − 6x + 9, and expand −4(x − 3) to −4x + 12. Collecting like terms, x² − 6x + 9 − 4x + 12 = x² − 10x + 21, so the image is y = x² − 10x + 21. Substituting (x + 3) instead of (x − 3) — translating in the wrong direction — gives y = x² + 2x − 3. Adding 3 straight onto the original equation, treating the translation as vertical instead of horizontal, gives y = x² − 4x + 3. Expanding (x − 3)² as x² − 3x + 9, using −3x instead of −6x for the middle term, and then combining with −4(x − 3) gives y = x² − 7x + 21.
4. (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.
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) 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.
7. (a) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
8. (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.
9. (a) −8 — −f(x) means take the output value from the table and change its sign, without changing which x-value is looked up. From the table, f(1) = 8, so −f(1) = −8. Reading f(1) = 8 from the table but forgetting to apply the negative sign gives 8. Misreading the row and using f(0) = 5 instead of f(1) = 8, then negating it, gives −5. Confusing −f(x) with f(x) − 1 — taking f(1) = 8 and subtracting 1 instead of negating — gives 7.
10. (a) 1.6 — Method: on a histogram the height of a bar is the frequency density, and frequency density = frequency ÷ class width. Working: the class 15 ≤ t < 30 runs from 15 to 30, so its width is 30 − 15 = 15 minutes; the frequency is 24, so the frequency density is 24 ÷ 15 = 1.6. Answer: 1.6 people per minute. The distractors: 360 comes from multiplying the frequency by the class width, 24 × 15, which uses the area rule backwards — area gives the frequency, so the frequency must be divided by the width to give the height; 0.625 comes from dividing the class width by the frequency, 15 ÷ 24, reversing the formula; 0.8 comes from dividing by the upper class boundary, 24 ÷ 30, instead of by the width of the class.
11. (c) A reflection in the y-axis — Replacing x with −x reflects the graph in the y-axis: each point (x, y) maps to (−x, y). Reflecting the OUTPUT instead, y = −f(x), gives a reflection in the x-axis — that is a different function. Combining both reflections gives a rotation of 180° about the origin, and reflecting in the line y = x swaps the x- and y-values, which is what produces the inverse function, not f(−x). Check first which side of f the minus sign sits on.
12. (d) 44 — Method: a cumulative frequency counts everything below a value, so the frequency of a class is the running total at the top of the class minus the running total at the bottom of it. Working: the running total below 20 kg is 96 and the running total below 10 kg is 52, so the number of boxes in the class 10 ≤ m < 20 is 96 − 52 = 44. Answer: 44 boxes. The distractors: 96 comes from quoting the running total at 20 kg itself, which counts every box below 20 kg rather than only those in this class; 34 comes from subtracting the wrong pair, 52 − 18, which gives the class 5 ≤ m < 10 instead; 54 comes from subtracting from the grand total, 150 − 96, which gives the boxes of 20 kg or more.
13. (c) y = (5/12)x − 169/12 — The gradient of the radius to (5, −12) is (−12 − 0) ÷ (5 − 0) = −12/5. A tangent is perpendicular to the radius at that point, so its gradient is the negative reciprocal, 5/12. Using y − y₁ = m(x − x₁) with (5, −12): y + 12 = (5/12)(x − 5), which gives y = (5/12)x − 169/12. y = −(12/5)x comes from using the radius's own gradient, −12/5, instead of turning it into the perpendicular gradient, and building the line through the origin (as the radius itself does). y = −(5/12)x − 119/12 comes from taking the reciprocal of −12/5 correctly as a size but keeping the wrong sign, using −5/12 instead of 5/12. y = (5/12)x − 25/12 comes from using the correct gradient 5/12 but building the line through (5, 0) instead of (5, −12) — dropping the point's y-coordinate.
14. (a) y = 2ˣ + 3 — A translation of 3 units in the positive y-direction shifts the whole graph up, which means adding to the output: y = f(x) + k with k = 3, so the image is y = 2ˣ + 3. Adding the 3 inside the power instead of outside it, which translates the graph horizontally instead of vertically, gives y = 2ˣ⁺³. Using a negative 3, which moves the graph down instead of up, gives y = 2ˣ − 3. Mistaking 2ˣ for the linear expression 2x and adding 3 inside brackets gives y = 2(x + 3).
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. (a) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
17. (c) 145 g — Method: find the position of the median from the total frequency, locate the class that contains it, then use linear interpolation inside that class, assuming the apples in it are spread evenly. Working: the median is the 100 ÷ 2 = 50th apple; the running totals are 10, then 10 + 30 = 40, then 40 + 40 = 80, so the 50th apple lies in the class 140 ≤ m < 160; it is the 50 − 40 = 10th of the 40 apples in that class, and the class is 20 g wide, so the median is 140 + (10 ÷ 40) × 20 = 140 + 5 = 145. Answer: an estimated median of 145 g. The distractors: 150 g comes from giving the midpoint of the class that contains the median instead of interpolating inside it; 140 g comes from stopping at the lower boundary of that class, which locates the class but not the value; 155 g comes from measuring the 5 g step down from the upper boundary, 160 − 5, instead of up from the lower boundary.
18. (b) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.