🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
2. (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.
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. (d) 25 — Method: the median is estimated at position n ÷ 2 in the cumulative frequency table, then interpolated across the class it falls in: lower boundary, plus the fraction of the way through the class, times the class width. Working: there are 80 sacks, so the median sits at position 80 ÷ 2 = 40. Before the class 20 ≤ m < 30 the cumulative frequency is 22, and by the end of it, it is 58, so this class holds the 40th sack; its frequency is 58 − 22 = 36 and its width is 30 − 20 = 10. The extra distance needed into the class is 40 − 22 = 18, and 18 ÷ 36 × 10 = 5, so the median is 20 + 5 = 25. Answer: the estimated median mass is 25 kg. Watch which numbers the interpolation uses: reading off just the lower boundary of the median class, 20, ignores how far into that class the 40th sack actually falls; treating n ÷ 2 = 40 itself as the median mass mistakes a position in the list for a mass in kilograms; and using the target position, 40, as the extra distance into the class instead of subtracting the sacks already counted changes the calculation to 20 + 40 ÷ 36 × 10. That comes to 20 + 11.1 = 31.1, overshooting the class because it never subtracts the 22 sacks already counted before it.
5. (c) 15/23 — Method: find P(rough and delayed) and the overall P(delayed) using the tree, then divide. Working: P(rough and delayed) = 0.2 × 0.75 = 0.15. P(calm and delayed) = 0.8 × 0.1 = 0.08. P(delayed) = 0.15 + 0.08 = 0.23. P(rough | delayed) = 0.15 ÷ 0.23 = 15/23. Answer: 15/23. Watch out: leaving the answer as 0.15 (3/20) gives P(rough and delayed) itself, without dividing by the overall probability that a crossing is delayed. Giving 0.75 (3/4) is the probability you were told to start with — that a crossing is delayed GIVEN the sea is rough — which is the reverse of what's being asked. And 0.2 (1/5) is just the original probability that the sea is rough, before you take the fact that the crossing was delayed into account.
6. (d) x₂ closer: f(x₂) = 0.0225, nearer to 0 — f(2.6) = 2.6² − 7 = 6.76 − 7 = −0.24, and f(2.65) = 2.65² − 7 = 7.0225 − 7 = 0.0225. The closer a value of x is to the root, the closer f(x) is to zero — regardless of sign. Since |0.0225| = 0.0225 is much smaller than |−0.24| = 0.24, x₂ = 2.65 is closer to the root. 'x₁ closer: −0.24 is the smaller value' comes from comparing the SIGNED values of f(x) rather than their distances from zero — −0.24 is indeed less than 0.0225 as a number, but that does not mean x₁ is closer to the root. 'x₁ closer: f(x₁) negative ⇒ nearer root' invents a rule that a negative f(x) means x is closer to the root; the sign of f(x) only tells you which side of the root x is on, not how close it is. 'x₂ is the exact root, since f(x₂) ≈ 0' misreads f(x₂) = 0.0225 as zero; the true root is √7 ≈ 2.6458, so f(2.65) is close to zero but not equal to it, and x₂ is an approximation, not the exact root.
7. (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.
8. (c) 0.5 per gram — Method: turn the two known bars into frequencies using area, subtract from the total to find how many letters are left, then divide that frequency by the width of the last class to get its height. Working: the first bar covers 50 g at a frequency density of 1.2, giving 1.2 × 50 = 60 letters, and the second covers 50 g at 1.8, giving 1.8 × 50 = 90 letters; together that is 60 + 90 = 150 letters, so 200 − 150 = 50 letters remain; the class 100 ≤ m < 200 is 100 g wide, so its frequency density is 50 ÷ 100 = 0.5 per gram. Answer: 0.5 per gram. The distractors: 0.25 per gram comes from dividing the remaining 50 letters by the upper class boundary, 200, instead of by the class width of 100; 2 per gram comes from dividing the class width by the frequency, 100 ÷ 50, reversing the formula; 1.4 per gram comes from subtracting only the first bar's 60 letters, leaving 140, and then dividing by 100.
9. (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.
10. (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
11. (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.
12. (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
13. (c) −0.5 — fg(x) = f(g(x)) = f(x²) = x² + 2. gf(x) = g(f(x)) = g(x + 2) = (x + 2)² = x² + 4x + 4. Setting fg(x) = gf(x): x² + 2 = x² + 4x + 4. Subtract x² from both sides: 2 = 4x + 4. Subtract 4 from both sides: −2 = 4x, so x = −0.5. Writing 1.5 comes from adding the 4 instead of subtracting it: 4x = 2 + 4 = 6, giving x = 1.5. Writing 'no solution' comes from expanding (x + 2)² as x² + 4 using (a + b)² = a² + b², losing the middle term — the equation then reads x² + 2 = x² + 4, which has no solution, but the expansion itself is wrong. Writing 0 comes from treating gf(x) as g(x) + f(x) instead of g(f(x)): x² + (x + 2) = x² + 2 gives x = 0, but that adds the two functions rather than composing them.
14. (c) 27 — Method: a frequency is the area of the part of the bar being asked about, so frequency = frequency density × the width of that part. Working: the part asked about runs from 20 to 35, so its width is 35 − 20 = 15 metres; the frequency density there is 1.8 per metre, so the estimate is 1.8 × 15 = 27. Answer: about 27 throws. The distractors: 72 comes from taking the whole bar, 1.8 × 40, and so counting every throw from 20 up to 60; 1.8 comes from reading the height of the bar as a frequency, when a height is a density and only an area is a count; 63 comes from using the upper value 35 as the width, 1.8 × 35, instead of the width 35 − 20.
15. (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.
16. (b) (2, 6) — y = f(x − 3) translates y = f(x) horizontally by 3 units to the RIGHT — inside the brackets, subtracting moves the graph in the positive x-direction. Turning point (−1, 6) → (−1 + 3, 6) = (2, 6). The common slip is to move LEFT instead, since the sign inside the bracket is negative — that gives (−4, 6). Changing the y-coordinate instead of the x-coordinate, as in (−1, 3) or (−1, 9), treats this as a vertical shift, which y = f(x − 3) is not.
17. (a) y = −2x + 3 — Method: find the step in the outputs for each step of 1 in the input — falling outputs mean a negative multiplier — then read off the output when the input is 0, because that is the number added on. Working: the outputs 5, 3, 1, −1 fall by 2 each time x rises by 1, so x is multiplied by −2; the output at x = 0 is 3, so 3 is added. Answer: y = −2x + 3, checked at x = 2 by −2 × 2 + 3 = −1. The distractors: y = 2x + 3 comes from taking the size of the step, 2, as the multiplier and ignoring the fact that the outputs are falling; y = −2x − 3 comes from using the correct multiplier but writing the number added on as −3 instead of the output 3 listed at x = 0; y = −x + 4 comes from taking the multiplier as −1, its size read from the step of 1 in the inputs instead of the step of 2 in the outputs and its sign from the fact that the outputs fall, and then fitting the number added on to the pair x = −1, y = 5.
18. (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.