🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (a) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
2. (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.
3. (d) (3/4)a + (1/4)c — Method: OX = OA + AX, and since AX is a third of XC, AX is 1/4 of the whole of AC, with AC = c − a. Working: OX = a + 1/4(c − a) = a − (1/4)a + (1/4)c = (3/4)a + (1/4)c. Answer: OX = (3/4)a + (1/4)c. Measuring 1/4 of AC from C's end instead of A's swaps the fractions round, giving (1/4)a + (3/4)c; adding (1/4)c onto the whole of a without subtracting a inside the bracket first gives a + (1/4)c; and treating the ratio as though AX and XC were equal gives the midpoint, (1/2)a + (1/2)c. Convert the ratio to a fraction of AC measured from A, subtract before you scale, and then add the result to OA.
4. (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.
5. (c) −4 — f(2) = 7, and y = f(x) + a passing through (2, 3) means f(2) + a = 3, so 7 + a = 3, giving a = 3 − 7 = −4. Writing a = 4 comes from subtracting the wrong way round, 7 − 3 instead of 3 − 7. Writing a = 10 comes from adding the two y-values instead of subtracting one from the other. Writing a = 3 comes from taking the new y-coordinate as the value of a directly, without accounting for the original y-value of 7 at all.
6. (a) 0.45, different from 0.4 for all the households — Method: work out the probability inside the restricted group of garden owners, then work out the probability across the whole survey, and compare the two. Working: 54 of the 120 households with a garden own a dog, so the conditional probability is 54 divided by 120, which is 0.45. Across the whole survey 80 of the 200 households own a dog, which is 0.4. Since 0.45 is not 0.4, having a garden changes the chance of owning a dog and the two events are not independent. Answer: 0.45, different from 0.4 for all the households. The distractors: 0.27 is 54/200, dividing the households with both by the whole survey instead of by the 120 with a garden; 0.675 is 54/80, the probability that a household has a garden given that it owns a dog, which is the condition and the event the wrong way round; 0.4 is 80/200, the probability of owning a dog with the garden information never used, which is why that route also reports no difference.
7. (d) 4, 3, 2, 6 — Method: frequency density = frequency ÷ class width for each class in turn; do not assume the classes are all the same width. Working: the four classes have widths 10 − 0 = 10, 30 − 10 = 20, 45 − 30 = 15 and 50 − 45 = 5. Dividing each frequency by its own width gives 40 ÷ 10 = 4, 60 ÷ 20 = 3, 30 ÷ 15 = 2 and 30 ÷ 5 = 6. Answer: the frequency densities, in order, are 4, 3, 2 and 6. Watch the width of each class separately: treating the last class as if it were also 10 units wide, like the first, gives 30 ÷ 10 = 3 instead of 30 ÷ 5 = 6 — the classes here are deliberately unequal, so no width can be borrowed from another class; dividing the width by the frequency instead of the frequency by the width for the third class gives 15 ÷ 30 = 0.5 in place of 2, the formula the wrong way round; and reading the frequency column straight off the table, 40, 60, 30, 30, skips the division by width altogether and reports how many fish are in each class rather than how densely packed each bar is.
8. (b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
9. (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.
10. (b) xₙ₊₁ = (xₙ² + 3) ÷ 5 — Starting from x² = 5x − 3, add 3 to both sides: x² + 3 = 5x. Divide both sides by 5: x = (x² + 3) ÷ 5. Writing this as an iteration gives xₙ₊₁ = (xₙ² + 3) ÷ 5. xₙ₊₁ = (xₙ² − 3) ÷ 5 comes from a sign error when moving the −3 across the equals sign — it should become +3, not stay as −3. xₙ₊₁ = 5(xₙ² + 3) comes from multiplying by 5 instead of dividing by 5 when isolating x. xₙ₊₁ = (xₙ + 3) ÷ 5 comes from dropping the index on x², using xₙ instead of xₙ².
11. (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.
12. (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.
13. (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.
14. (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.
15. (d) 2/3 — Method: the pupil picked is known to be a girl, so the sample space is the 45 girls and not all 80 pupils; divide the number of girls who walk by the number of girls. Working: 30 of the 45 girls walk to school, which gives 30/45. Dividing the numerator and the denominator by 15 gives 2/3. Answer: the probability is 2/3. The distractors: 3/8 is 30/80, dividing the girls who walk by every pupil in the group, which is the commonest slip on a conditional probability because it keeps the restricted numerator but the whole denominator; 1/3 is 15/45, counting the 15 girls who do not walk to school and so answering the opposite event inside the correct group; 11/20 is 44/80, adding the 30 girls and the 14 boys who walk and dividing by the whole group, which throws away the information that the pupil picked is a girl.
16. (a) xₙ₊₁ = ∛(5xₙ + 3) — Starting from x³ − 5x − 3 = 0, add 5x and 3 to both sides to get x³ = 5x + 3, then take the cube root of both sides: x = ∛(5x + 3), giving the iterative formula xₙ₊₁ = ∛(5xₙ + 3). A sign error when moving the constant term across, treating x³ − 5x − 3 = 0 as x³ = 5x − 3, gives xₙ₊₁ = ∛(5xₙ − 3). Swapping the coefficient of x with the constant term gives xₙ₊₁ = ∛(3xₙ + 5), which does not come from x³ = 5x + 3 at all. Treating cubing as meaning multiply by 3 rather than raise to the power 3, and so undoing it by dividing by 3 instead of taking a cube root, gives xₙ₊₁ = (5xₙ + 3) ÷ 3.
17. (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
18. (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.