🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (b) (6, 2) — The overall journey from house to park is the sum of the two vectors: top = 2 + 4 = 6, bottom = 5 + (−3) = 2, giving (6, 2). A candidate who subtracts the second vector from the first instead of adding gets (2 − 4, 5 − (−3)) = (−2, 8). A candidate who subtracts the other way round gets (4 − 2, −3 − 5) = (2, −8). A candidate who forgets the negative sign on the second vector's bottom number and adds 3 instead of −3 gets (6, 8). Because the journeys join end to end, the correct resultant vector is (6, 2).
2. (d) y = −f(x + 2) has a maximum turning point at (1, 5). — y = f(x + 2) translates the graph 2 units in the negative x-direction, so the turning point's x-coordinate moves from 3 to 1; reflecting in the x-axis then negates the y-coordinate, turning −5 into 5, and also turns a minimum into a maximum — giving a true maximum turning point at (1, 5). Translating the root at x = 1 in the wrong direction, adding 2 instead of subtracting it, gives x = 3, but the true image root is at 1 − 2 = −1, so that statement is false. Keeping the coordinates (1, 5) correct but forgetting that reflecting in the x-axis turns a minimum into a maximum gives a false "minimum turning point" statement. Getting the reflection's effect on the turning-point type right (a maximum) but translating the x-coordinate in the wrong direction, using 3 + 2 = 5 instead of 3 − 2 = 1, gives a false statement at (5, 5).
3. (a) (4, 1) — y = −f(x) reflects the graph of y = f(x) in the x-axis: every point (x, y) maps to (x, −y). Applying this to (4, −1): the x-coordinate stays 4, and the y-coordinate −1 becomes its negative, 1. Leaving the y-coordinate unchanged skips the reflection entirely, giving (4, −1); reflecting the x-coordinate instead, or reflecting both, mixes this up with a reflection in the y-axis or a rotation, giving (−4, −1) or (−4, 1).
4. (d) y = −2x + 10 — Method: a tangent is perpendicular to the radius drawn to the point where it touches, so work out the gradient of that radius, take its negative reciprocal for the tangent, then substitute into y − y₁ = m(x − x₁). Working: the radius joins (0, 0) to (4, 2), so its gradient is 2 ÷ 4 = 1/2; turning 1/2 upside down gives 2 and changing the sign gives −2. Substituting into y − 2 = −2(x − 4) gives y − 2 = −2x + 8, so y = −2x + 10. Answer: y = −2x + 10. The distractors: y = −0.5x + 4 changes the sign of the radius gradient but never turns it upside down, using −1/2 where −2 belongs; y = 2x − 6 turns the gradient upside down but leaves it positive, using 2 where −2 belongs; y = −2x − 10 has the correct gradient but substitutes the point with both signs reversed, writing y + 2 = −2(x + 4) instead of y − 2 = −2(x − 4).
5. (d) 12/17 — Method: the person picked is known to have passed, so the sample space is everyone who passed; divide the course takers who passed by that total. Working: 120 course takers and 50 others passed, so 170 people passed. The course takers who passed give 120/170, and dividing the numerator and the denominator by 10 gives 12/17. Answer: the probability is 12/17. The distractors: 4/5 is 120/150, the probability that someone passed given that they took the course, which reverses the condition and the event; 12/25 is 120/250, dividing by everyone who sat the test rather than by the 170 who passed; 17/25 is 170/250, the probability that a person picked from everyone sitting the test passed, which answers a different question altogether.
6. (c) 37.5 — The tangent at (18, 24) is 18x + 24y = 900 (using ax + by = r² with a = 18, b = 24, r² = 900). Setting x = 0 to find the y-intercept: 24y = 900, so y = 37.5. Choosing 900 skips the division by 24 and just repeats the constant. Choosing 50 divides the constant by the x-coefficient 18 instead of the y-coefficient 24. Choosing 1.25 uses the radius 30 instead of r² = 900 as the constant before dividing.
7. (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.
8. (a) 12.5 — Method: the tangent at A is perpendicular to the radius OA, so find the gradient of OA, take its negative reciprocal, write the equation of the tangent and find where it meets the y-axis; the length of OB is then the distance of that crossing from the origin. Working: OA runs from (0, 0) to (−6, 8), so its gradient is 8 ÷ (−6), which cancels to −4/3; the negative reciprocal of −4/3 is 3/4. Substituting into y − 8 = 3/4(x + 6) gives y = 0.75x + 4.5 + 8, so y = 0.75x + 12.5 and B is (0, 12.5). The length OB is therefore 12.5. Answer: 12.5. The distractors: 10 is the radius of the circle, quoted on the assumption that the tangent always meets an axis one radius from the centre, which is only true when the radius itself lies along that axis; 8 is the y-coordinate of A, quoted by treating the tangent as horizontal so that it keeps the height of A; 3.5 comes from turning the gradient of OA upside down without changing its sign, which gives y = −0.75x + 3.5.
9. (d) 45 — Method: a cumulative frequency is a running total — it counts everybody in every class up to and including the one that ends at the value given. Working: the classes that lie wholly below 30 minutes are 0 ≤ t < 10, 10 ≤ t < 20 and 20 ≤ t < 30, with frequencies 6, 14 and 25, so the running total is 6 + 14 = 20 and then 20 + 25 = 45. Answer: 45 people took less than 30 minutes. The distractors: 25 comes from quoting the frequency of the class 20 ≤ t < 30 on its own instead of the running total; 65 comes from accumulating one class too many and including 30 ≤ t < 40, which is 45 + 20; 35 comes from accumulating from the top downwards, 15 + 20, which counts the people who took 30 minutes or more rather than fewer.
10. (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
11. (a) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
12. (a) n² + n — n² + n = n(n + 1), the product of two consecutive integers. One of any two consecutive integers is always even, so their product is always even, whatever whole number n is. n² − n + 1 = n(n − 1) + 1 is always ODD, since n(n − 1) is even and adding 1 makes it odd — the opposite of what's asked. 2n + 1 is always odd by definition, not even. n² + 1 is not always even at all: it depends on whether n is odd or even, and testing n = 2 gives 5, which is odd.
13. (b) (5, 1) — First scale a by 2: 2a = (2×3, 2×(−2)) = (6, −4). Then add b component by component: (6+(−1), −4+5) = (5, 1). (2, 3) is a + b without doubling a first. (4, 6) doubles both a and b instead of only a. (7, −9) subtracts b from 2a instead of adding it.
14. (d) n and n + 1 are consecutive integers, so one of them must be even; this makes n(n + 1) even, so 4n(n + 1) is 4 × an even number, which is a multiple of 8. — The proof needs a reason why n(n + 1) is even, not just an assertion. n and n + 1 are consecutive integers, so exactly one of them is even; multiplying by that even number keeps n(n + 1) even, so 4n(n + 1) = 4 × (an even number), and 4 × an even number always has a further factor of 2 hidden inside it, making the whole product a multiple of 8. The option 'is a multiple of 4, and because n and n + 1 are consecutive integers, it must be a multiple of 8' asserts the multiple-of-8 conclusion directly from 'multiple of 4' and 'consecutive integers' without ever showing that n(n + 1) itself is even — the missing step is exactly what earns the mark. The option '4n is always a multiple of 4 ... which means it is a multiple of 8' mistakes 4n being a multiple of 4 for the whole product 4n(n + 1) being a multiple of 8; that extra factor of 2 only comes from n(n + 1) being even, not from 4n alone. The option that expands to 4n² + 4n and calls it 'clearly a multiple of 8' never checks for a factor of 2 beyond the 4 already there — the word 'clearly' is standing in for a missing argument.
15. (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.
16. (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.
17. (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.
18. (b) y = f(−x) — y = f(−x) reflects the graph of y = f(x) in the y-axis: every x-coordinate changes sign. The x-intercepts −3 and 7 become 3 and −7, matching the second graph's intercepts −7 and 3. A point already on the y-axis is unaffected, since −0 = 0, so the y-intercept (0, 21) stays exactly where it is — matching the second graph as well. y = −f(x) leaves the x-intercepts unchanged at −3 and 7, since f(x) = 0 exactly where −f(x) = 0, which does not match; it also sends the y-intercept to (0, −21), a second mismatch. y = −f(−x) does send the x-intercepts to the right places, −7 and 3, but it sends the y-intercept to (0, −21) instead of (0, 21), so it fails the second clue. y = f(x) − 4 moves every point down 4, sending the y-intercept to (0, 17) instead of (0, 21), so it fails the y-axis clue. Test each option against BOTH clues — the pair of x-intercepts and the point on the y-axis — because more than one option gets only one of the two right.