🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (c) 7/19 — Method: 'at least two black' covers two cases — all three black, and exactly two black. Work out the probability of each along a tree, add them, then use P(all three black | at least two black) = P(all three black) ÷ P(at least two black). Working: P(all three black) = 9/13 × 8/12 × 7/11 = 504/1716 = 42/143. For exactly two black, one order is black, black, white = 9/13 × 8/12 × 4/11 = 288/1716; the white sock could be drawn first, second or third, so there are 3 such orders, giving 3 × 288/1716 = 864/1716 = 72/143. P(at least two black) = 42/143 + 72/143 = 114/143. P(all three black | at least two black) = (42/143) ÷ (114/143) = 42/114 = 7/19. Answer: 7/19. Watch out: stopping at 42/143 gives the unconditioned probability that all three are black — it ignores that you already know at least two of them are. Dividing by the 'exactly two black' probability on its own gives 7/12, and forgets that the all-black outcomes are themselves part of the 'at least two black' group, so they must be inside the denominator, not left out of it. And 7/11 answers a different, easier question — the probability the THIRD sock is black given the FIRST TWO specifically are black — not 'at least two of the three, in any order, are black'.
2. (c) a + (1/2)c — Method: OABC is a parallelogram, so OB = OA + AB, and since AB is equal and parallel to OC, AB = c; this gives OB = a + c. M is the midpoint of AB, so AM = (1/2)AB = (1/2)c. Working: OM = OA + AM = a + (1/2)c. Answer: OM = a + (1/2)c. Adding the whole of AB instead of half of it gives a + c, which is OB, not OM; halving the whole diagonal OB instead of just AB gives (1/2)a + (1/2)c, the midpoint of OB rather than of AB; and flipping the sign on the c-term gives a − (1/2)c, which points back the wrong way along AB. Halve only the side you are told to halve, and check the sign before you commit to an answer.
3. (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
4. (d) −3 — Substitute y = 11 into y = 5 − 2x, giving 11 = 5 − 2x. Subtracting 5 from both sides gives 6 = −2x, so x = 6 ÷ (−2) = −3. A candidate who mishandles the negative sign when rearranging, treating the equation as 6 = 2x, gets x = 3. A candidate who correctly finds −2x = 6 but forgets to divide by 2 at all gets x = 6. A candidate who adds 5 and 11 instead of subtracting, getting 2x = 16, gets x = 8.
5. (a) 8 — f(x − 1) means every input to f is reduced by 1 before it is looked up. At x = 2, the input to f becomes 2 − 1 = 1, so f(x − 1) at x = 2 is f(1) = 8. Reading f(2) = 4 directly from the table, without applying the shift, gives 4. Shifting in the wrong direction, using x + 1 = 3 instead of x − 1 = 1, gives f(3) = 1. Confusing f(x − 1) with f(x) − 1 — taking f(2) = 4 and then subtracting 1 — gives 3.
6. (b) 3.38 — r₁ = √(300 ÷ (π × 8)) = √11.9366 = 3.4550. r₂ = √(300 ÷ (π × 8.4550)) = √11.2947 = 3.3608. r₃ = √(300 ÷ (π × 8.3608)) = √11.4232 = 3.3798, which rounds to 3.38. Choosing 3.36 stops at r₂, one iteration too early. Choosing 4.82 leaves out the '+ 5' inside the bracket, dividing by π × rₙ instead of π × (rₙ + 5). Choosing 3.45 comes from using π ≈ 3 instead of the calculator's π key throughout.
7. (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.
8. (a) 2.55 — x₁ = 20 ÷ (2 + 5) = 20 ÷ 7 = 2.857142857. x₂ = 20 ÷ (2.857142857 + 5) = 20 ÷ 7.857142857 = 2.545454545, which rounds to 2.55. Reporting x₁ instead of x₂ gives 2.857142857, which rounds to 2.86. Dropping the +5 in the denominator, using xₙ₊₁ = 20 ÷ xₙ, gives x₁ = 20 ÷ 2 = 10 and x₂ = 20 ÷ 10 = 2, which is 2.00. A sign error in the denominator, using xₙ₊₁ = 20 ÷ (xₙ − 5), gives x₁ = 20 ÷ (2 − 5) = −6.666666667 and x₂ = 20 ÷ (−6.666666667 − 5) = −1.714285714, which rounds to −1.71.
9. (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
10. (d) It diverges, growing rapidly without limit. — x₁ = 2³ − 2 = 8 − 2 = 6. x₂ = 6³ − 2 = 216 − 2 = 214. x₃ = 214³ − 2 = 9800344 − 2 = 9800342. The values 6, 214, 9800342, … grow far larger at every step, so the sequence diverges rather than settling anywhere. Checking whether the sequence converges to a fixed value near 2 fails, since the terms grow enormously instead of levelling off. Checking for a repeating pair of values also fails, since 6, 214 and 9800342 are all different, with no sign of a return to 6. x₀ = 2 is a fixed point only if 2³ − 2 = 2, but 2³ − 2 = 6, not 2, so the sequence does not stay constant.
11. (b) 143 — fg(4) means f(g(4)): work out g(4) first, then substitute the result into f. g(4) = 3 × 4 = 12, then f(12) = 12² − 1 = 144 − 1 = 143. Working out gf(4) instead swaps the order: f(4) = 4² − 1 = 15, then g(15) = 3 × 15 = 45 — that is the wrong composition. Treating f(x) as x − 1 (forgetting to square the input) gives f(12) = 12 − 1 = 11. Applying g twice instead of applying g then f gives g(g(4)) = g(12) = 3 × 12 = 36, which mixes up which function should be applied second.
12. (c) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
13. (d) line 2 — Line 1 correctly represents three consecutive integers using n. Line 2 adds them: n + (n + 1) + (n + 2). Collecting terms: the n-terms give 3n, and the constants give 1 + 2 = 3, so the correct sum is 3n + 3, not 3n + 2 as Line 2 states — this is the first error, an arithmetic slip in collecting the constant terms. Lines 3 and 4 both follow correctly from Line 2's incorrect result, but that result itself is wrong: the true sum, 3n + 3 = 3(n + 1), is a multiple of 3 for every whole number n. Check the working of each line against what came before it, in order, rather than judging whether the final conclusion feels right — an error that flips the conclusion can sit several lines before the line that states it.
14. (b) 5 ≤ m < 10 — Method: with 60 values the median is the 60 ÷ 2 = 30th value in order, so build a running total until it first reaches 30. Working: the running totals are 22 after the first class, 22 + 20 = 42 after the second, 51 after the third, 56 after the fourth and 60 after the fifth; the 30th parcel is past 22 but not past 42, so it lies in the second class. Answer: the median lies in the class 5 ≤ m < 10. The distractors: 0 ≤ m < 5 comes from giving the class with the greatest frequency, 22, which is the modal class and not the median class; 10 ≤ m < 20 comes from choosing the middle class in the list of five instead of counting to the middle value; 20 ≤ m < 30 comes from halving the range of the data, 50 ÷ 2 = 25, and giving the class that contains 25 kg rather than the class that contains the 30th parcel.
15. (d) (6, 6) — y = f(x − 4) + 1 combines two translations: the −4 inside the brackets moves the graph 4 units to the RIGHT, and the +1 outside the brackets moves it 1 unit UP. Applying both to (2, 5): add 4 to the x-coordinate 2, and add 1 to the y-coordinate 5, giving (6, 6). Moving left instead of right, as the minus sign inside the bracket might suggest, gives (−2, 6) or (−2, 4); forgetting the horizontal shift altogether and only applying the vertical one gives (2, 6).
16. (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).
17. (a) (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1) — Two consecutive odd numbers can be written as 2n + 1 and 2n + 3, for a whole number n. Adding them: (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), which is a multiple of 4 for every whole number n, proving the general result. Using 2n + 1 twice does not represent two different numbers, so it proves nothing about a sum of two numbers; writing n + (n + 2) drops the +1 that makes the numbers odd in the first place, and only shows a multiple of 2; and check every constant term is added correctly — 1 + 3 is 4, not 3.
18. (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.