🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (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.
2. (d) P lies outside the circle — The squared distance from the origin to P is 8² + 9² = 64 + 81 = 145, which is greater than 100, so P lies outside the circle. Choosing 'P lies inside the circle' comes from adding the coordinates directly (8 + 9 = 17) instead of squaring and adding them, and comparing 17 with 100. Choosing 'P lies on the circle' comes from squaring 9 incorrectly as 36 (as if it were 6²) instead of 81, giving 64 + 36 = 100, equal to the right-hand side. Choosing 'Cannot be determined' wrongly assumes the equation of the line itself is needed, when the radius and the distance from the origin to P are already enough.
3. (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
4. (c) x + 18 — Expand −3(2x − 6) by multiplying both terms by −3: −3 × 2x = −6x and −3 × (−6) = 18, giving 7x − 6x + 18 = x + 18. Writing x − 18 comes from not flipping the sign of the −6 inside the bracket, so −3 × (−6) is treated as −18 instead of +18. Writing x + 6 comes from forgetting to multiply the −6 by 3, only carrying its sign. Writing 13x − 18 comes from treating the whole bracket as being added rather than subtracted, so 3(2x − 6) = 6x − 18 is added to 7x.
5. (d) 7/18 — Method: the pupil picked is known to study French, so the sample space shrinks to the 18 French students; divide the number who study both languages by 18. Working: 7 of the pupils study both French and German, and all 7 of them are among the 18 French students, so the probability is 7/18, which will not cancel. Answer: the probability is 7/18. The distractors: 7/30 divides by the whole class, keeping the restricted numerator but the full denominator; 1/2 is 7/14, which conditions on the German students instead, answering the probability that a German student also studies French; 7/25 uses 18 + 14 minus 7 = 25, the number who study at least one language, which is a larger group than the one the question restricts you to.
6. (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
7. (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.
8. (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.
9. (c) 42 — Method: find the target cumulative frequency, 90% of the total, locate the class it falls in from the plotted points, then interpolate: lower boundary, plus the extra distance needed into the class divided by the class's frequency, times its width. Working: 90% of 320 is 0.9 × 320 = 288. The plotted points show a cumulative frequency of 280 at d = 40 and 320 at d = 50, so the class 40 ≤ d < 50 has frequency 320 − 280 = 40 and width 50 − 40 = 10, and 288 falls inside it. The extra distance needed into the class is 288 − 280 = 8, and 8 ÷ 40 × 10 = 2, so the diameter is 40 + 2 = 42. Answer: the estimated diameter is 42 mm. Watch which point and which class the interpolation actually uses: reading off d = 40, the plotted point just below the target, instead of interpolating the extra 8 ball bearings into the next 10 mm, stops one step short of the true answer; finding the diameter below which only 10% lie instead of 90% gives a target of 0.1 × 320 = 32, which falls in the class 10 ≤ d < 20 — the extra distance into that class is 32 − 30 = 2, and 2 ÷ 60 × 10 = 0.3, so this route gives 10 + 0.3 = 10.3, the bottom decile rather than the top 90%; and interpolating within the class 30 ≤ d < 40 instead of 40 ≤ d < 50, as though 288 had not yet reached a cumulative frequency of 280, treats the extra distance as 288 − 190 = 98, and 98 ÷ 90 × 10 = 10.9, giving 30 + 10.9 = 40.9, one class too early.
10. (b) (x + 2)/3 — Start with y = 3x − 2 and swap x and y: x = 3y − 2. Add 2 to both sides: x + 2 = 3y. Divide both sides by 3: y = (x + 2)/3, so f⁻¹(x) = (x + 2)/3. Writing x/3 + 2 comes from dividing only the 3y term by 3 and leaving the +2 outside the division — the 2 must be added before you divide, not after. Writing (x − 2)/3 comes from keeping the subtraction sign instead of flipping it to addition when the −2 is moved across the equals sign. Writing 3x + 2 comes from swapping x and y but never actually solving for y — just changing the sign of the constant term.
11. (d) 60 — Method: on a histogram the frequency of a class is its frequency density × its class width, and the frequencies of all the classes add up to the total, so turn each labelled bar into a frequency and subtract their total from 250. Working: 10 ≤ age < 20 has width 20 − 10 = 10, so its frequency is 4.5 × 10 = 45; 20 ≤ age < 35 has width 35 − 20 = 15, so its frequency is 6 × 15 = 90; 50 ≤ age < 70 has width 70 − 50 = 20, so its frequency is 2.75 × 20 = 55. Those three come to 45 + 90 + 55 = 190, and the total is 250, so the missing frequency is 250 − 190 = 60. Answer: the class 35 ≤ age < 50 has 60 members. Watch what you do with the total and the three frequencies you have found: giving the total, 250, as the answer forgets that three bars have already accounted for some of the members; giving 190, the total of the other three classes, reports how many members are not in this class rather than how many are; and leaving one of the three out of the subtraction, for example 45 + 90 = 135 and 250 − 135 = 115, still owes the class at 50 ≤ age < 70 its 55 members.
12. (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
13. (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
14. (d) (−4, −3) — The combined translation is the sum of the two column vectors, added component by component: top numbers 3 + (−7) = −4, bottom numbers −5 + 2 = −3, giving (−4, −3). (10, −7) subtracts the second vector from the first instead of adding them. (−4, 3) gets the top number right but makes a sign error on the bottom, treating −5 + 2 as +3. (4, −3) gets the bottom number right but makes a sign error on the top, treating 3 + (−7) as +4.
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. (b) Translate +5 in x, then reflect in the x-axis. — Translating y = x² by 5 units in the positive x-direction gives y = (x − 5)². Reflecting this in the x-axis, which replaces y with −y, gives y = −(x − 5)², matching the target. Using a translation of −5 in x instead gives y = (x + 5)², and reflecting that in the x-axis gives y = −(x + 5)² — the sign inside the bracket is wrong. Reflecting in the y-axis first does nothing to y = x², since (−x)² = x², so translating afterwards only reaches y = (x − 5)² with no negative sign at all. Translating by 5 units in y instead of x gives y = x² + 5, and reflecting that in the x-axis gives y = −x² − 5, a different curve altogether — a vertical shift does not create the (x − 5)² term the target equation needs.
17. (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.
18. (c) 4.30 — Method: substitute the starting value into the right-hand side to get x₁, then feed each value back in, keeping the whole display and respecting the order of operations, which divides before it subtracts. Working: x₁ = 5 − 3 ÷ 2.5 = 5 − 1.2 = 3.8; x₂ = 5 − 3 ÷ 3.8 = 5 − 0.78947… = 4.21052…; x₃ = 5 − 3 ÷ 4.21052… = 5 − 0.7125 = 4.2875; x₄ = 5 − 3 ÷ 4.2875 = 5 − 0.69970… = 4.30029…, which is 4.30 correct to 3 significant figures. Answer: 4.30. The distractors: 4.29 is x₃ = 4.2875 rounded, reached by counting the starting value itself as the first iterate and so stopping one use of the formula early; 3.80 is x₁, the value after a single use of the formula; 2.50 comes from working out (5 − 3) ÷ xₙ instead of 5 − (3 ÷ xₙ), subtracting before dividing, which produces the sequence 0.8, 2.5, 0.8, 2.5 and lands on 2.5 at the fourth step.