🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (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.
2. (b) 2x + 3 — fg(x) means f(g(x)): apply g first, then apply f to the result. g(x) = 2x, so f(g(x)) = f(2x) = 2x + 3. Writing 2x + 6 comes from working out gf(x) instead — g(f(x)) = g(x + 3) = 2(x + 3) = 2x + 6 — which applies the functions in the wrong order. Writing 3x + 3 comes from adding f(x) and g(x) together, (x + 3) + 2x = 3x + 3, instead of composing them. Writing 2x² + 6x comes from multiplying f(x) and g(x) together, (x + 3)(2x) = 2x² + 6x, instead of substituting one into the other.
3. (d) g is one-to-one: f(3) = f(−3), removed by x ≥ 0 — A function has an inverse only if it is one-to-one: every output must come from exactly one input. f(3) = 9 and f(−3) = 9, so two different inputs give the same output, and there is no way to send 9 back to a single input — f is not one-to-one over all real x. Restricting the domain to x ≥ 0 removes one of the two inputs behind every such pair, so g is one-to-one and does have an inverse. 'g's outputs are positive; f's could be negative' is wrong because f(x) = x² also only gives outputs of 0 or more — the outputs of f and g are identical sets; it is the INPUTS that differ, not the outputs. 'Restricting any domain always creates an inverse' is wrong because a restriction only helps if it actually removes the repeated outputs: restricting f(x) = x² to x ≥ −3 still leaves f(1) = f(−1) = 1, so that restricted function is still not one-to-one and still has no inverse. 'Squares can never be reversed, under any conditions' is wrong because √9 = 3 does reverse 3² = 9 once you know the input was non-negative — a square root just cannot tell you WHICH of two inputs you started from unless the domain has already ruled one of them out.
4. (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).
5. (a) −3/4 — Method: the tangent at a point on a circle is perpendicular to the radius drawn to that point, so find the gradient of the radius and then take its negative reciprocal. Working: the radius joins (0, 0) to (6, 8), so its gradient is 8 ÷ 6, which cancels to 4/3. Turning 4/3 upside down gives 3/4, and changing the sign gives −3/4. Answer: the gradient of the tangent is −3/4. The distractors: 4/3 is the gradient of the radius itself, quoted without taking the perpendicular at all; −4/3 changes the sign but leaves the fraction the same way up, so the two gradients do not multiply to give −1; 3/4 turns the fraction upside down but keeps it positive, which is the other half of the same rule left undone.
6. (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.
7. (a) 16 — A and B share the same y-coordinate, so the distance between them is the horizontal difference between their x-coordinates: 8 − (−8) = 16, giving a chord of length 16. Choosing 8 gives only the x-coordinate of one point, not the full distance between the two points. Choosing 34 is the diameter of the circle (2 × 17), not the length of this particular chord. Choosing 17 is the radius — the distance from the centre to A or to B, not from A to B.
8. (b) 2/11 — Method: restrict the 36 equally likely outcomes to those where at least one die shows a 5, then find what fraction of THOSE give a total of 8. Working: outcomes with at least one 5: (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) — 11 outcomes. Among these, the total is 8 for (5, 3) and (3, 5) — 2 outcomes. P(total 8 | at least one 5) = 2/11. Answer: 2/11. Watch out: 5/36 is P(total 8) over the full 36 outcomes — it ignores that you already know one die shows a 5. Treating the condition as 'the first die shows a 5' instead of 'at least one die shows a 5' restricts you to only 6 outcomes and misses the (3, 5) case, giving 1/6. And counting only (5, 3) but not its reverse (3, 5) inside the correct 11-outcome list gives 1/11 instead of 2/11.
9. (b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
10. (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.
11. (c) 0.04 — Method: 'made by machine B and faulty' is the second branch of a tree followed after the first, so multiply the probability of machine B by the probability of a fault given machine B. Working: machine B makes 0.4 of the bolts, and 0.1 of those bolts are faulty, so the probability is 0.4 × 0.1 = 0.04. Answer: the probability is 0.04. The distractors: 0.5 comes from adding 0.4 and 0.1 instead of multiplying, treating two stages of one journey as two separate outcomes; 0.1 gives the fault rate for machine B on its own, as though every bolt in the factory came from machine B, so the 40% share is never used; 0.07 is 0.6 × 0.05 added to 0.4 × 0.1, the probability that a bolt is faulty whichever machine made it, which answers a question about all the production rather than about machine B.
12. (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.
13. (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
14. (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.
15. (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
16. (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.
17. (b) 8/35 — Method: the pupil picked is known to play at least one of the two sports, so first count how many pupils that is, then divide the number who play both by it. Working: 25 play football and 18 play tennis, but the 8 who play both have been counted in each figure, so the number who play at least one sport is 25 + 18 minus 8, which is 35. The pupils who play both give 8/35, which will not cancel. Answer: the probability is 8/35. The distractors: 2/15 is 8/60, dividing by the whole year group instead of by the 35 pupils who play at least one sport; 8/43 uses 25 + 18 as the denominator, forgetting that the 8 pupils who play both have been counted twice; 8/25 conditions on the footballers alone, answering the probability that a footballer also plays tennis rather than using every pupil who plays a sport.
18. (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.