🔺 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. (b) 3 — At the limit, L = √(2L + 3). Squaring both sides: L² = 2L + 3, so L² − 2L − 3 = 0, which factorises as (L − 3)(L + 1) = 0, giving L = 3 or L = −1. Since the sequence of iterates stays positive throughout, the limit is L = 3. Taking the other, negative root without rejecting it gives −1. Treating the equation L = 2L + 3 as already linear, forgetting to square both sides first, gives −L = 3, so L = −3. A sign error when factorising, writing (L + 3)(L − 1) = 0 instead of (L − 3)(L + 1) = 0, gives L = 1.
3. (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.
4. (d) 2x² − 4x + 3 — fg(x) means f(g(x)): substitute g(x) into f in place of x. g(x) = x − 1, so fg(x) = f(x − 1) = 2(x − 1)² + 1. Expanding (x − 1)² = x² − 2x + 1, so fg(x) = 2(x² − 2x + 1) + 1 = 2x² − 4x + 2 + 1 = 2x² − 4x + 3. Writing 2x² comes from working out gf(x) instead — g(f(x)) = f(x) − 1 = (2x² + 1) − 1 = 2x², which applies the functions in the wrong order. Writing 2x² − 1 comes from expanding (x − 1)² as x² − 1, dropping the middle term, so f(x − 1) becomes 2(x² − 1) + 1 = 2x² − 2 + 1 = 2x² − 1. Writing 2x² − 4x + 2 comes from expanding correctly but forgetting the final + 1 from f, stopping at 2(x² − 2x + 1) = 2x² − 4x + 2.
5. (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.
6. (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.
7. (c) 5/18 — Method: knowing the total is even cuts the 36 equally likely outcomes down to the even ones, so count those first and then count how many of them give 8. Working: the even totals occur as 2 once, 4 three times, 6 five times, 8 five times, 10 three times and 12 once, which is 18 outcomes. The total is 8 for 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2, which is 5 outcomes. The probability is 5/18, which will not cancel. Answer: the probability is 5/18. The distractors: 5/36 keeps the right count of ways to make 8 but divides by all 36 outcomes, ignoring the fact that the odd totals have already been ruled out; 1/6 treats the six even totals 2, 4, 6, 8, 10 and 12 as equally likely and picks one of them, which they are not; 1/11 treats the eleven possible totals from 2 to 12 as equally likely and uses neither the counting nor the condition.
8. (a) (9, −5) — Method: multiply every part of q by 2, then subtract the matching part from p. Working: 2q = (−4, 6); p − 2q gives top 5 − (−4) = 9 and bottom 1 − 6 = −5. Answer: p − 2q = (9, −5). A candidate who forgets to double q first, working out p − q instead, gets (7, −2). A candidate who doubles p instead of q, working out 2p − q, gets (12, −1). A candidate who adds 2q instead of subtracting it gets (1, 7).
9. (d) 2 — Method: if f is 3 times e, then each part of f equals 3 times the matching part of e. Working: using the bottom numbers, 6 = 3 × k, so k = 2. Answer: k = 2. A candidate who multiplies instead of dividing, working out 6 × 3, gets 18. A candidate who uses the top numbers' ratio instead, 15 ÷ 5, and gives that ratio as k gets 3. A candidate who adds instead of using the multiple relationship, working out 6 + 3, gets 9.
10. (c) 0.04 — Method: 'made by machine B and faulty' is the second branch of a tree followed after the first, so multiply the probability of machine B by the probability of a fault given machine B. Working: machine B makes 0.4 of the bolts, and 0.1 of those bolts are faulty, so the probability is 0.4 × 0.1 = 0.04. Answer: the probability is 0.04. The distractors: 0.5 comes from adding 0.4 and 0.1 instead of multiplying, treating two stages of one journey as two separate outcomes; 0.1 gives the fault rate for machine B on its own, as though every bolt in the factory came from machine B, so the 40% share is never used; 0.07 is 0.6 × 0.05 added to 0.4 × 0.1, the probability that a bolt is faulty whichever machine made it, which answers a question about all the production rather than about machine B.
11. (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
12. (d) x = 4 and x = −4 — Substituting y = 3 gives x² + 9 = 25, which simplifies to x² = 16, so x = 4 or x = −4. Choosing 'x = 3 and x = −3' uses the given value y = 3 as if it were the x-coordinate. Choosing 'x = 4' alone finds the positive square root of 16 but forgets the negative root. Choosing 'x = 5 and x = −5' skips subtracting 3² = 9 from 25 and takes the square root of 25 directly.
13. (a) y = −x³ + 5x — Reflecting a graph in the y-axis replaces every x in the equation with −x: y = (−x)³ − 5(−x) = −x³ + 5x. Writing y = −x³ − 5x comes from substituting −x into the x³ term only and leaving the −5x term as it was. Writing y = x³ + 5x comes from substituting −x into the −5x term only and leaving the x³ term as it was. Writing y = x³ − 5x is the original equation with no reflection applied at all — every term needs the substitution, not just one of them.
14. (d) 3 — Method: the height of a bar is its frequency density, frequency ÷ class width, so work out both heights and divide one by the other. Working: the class 0 ≤ t < 4 is 4 minutes wide and holds 30 visits, so its frequency density is 30 ÷ 4 = 7.5 per minute; the class 4 ≤ t < 20 is 16 minutes wide and holds 40 visits, so its frequency density is 40 ÷ 16 = 2.5 per minute; dividing the heights, 7.5 ÷ 2.5 = 3. Answer: the first bar is 3 times as tall. The distractors: 0.75 comes from comparing the frequencies, 30 ÷ 40, as though the frequencies were the heights, which is the mistake the unequal widths are there to expose; 4 comes from comparing the class widths, 16 ÷ 4, instead of the heights; 5 comes from subtracting the two frequency densities, 7.5 − 2.5, which answers how much taller rather than how many times taller.
15. (a) 31.5 — Method: to estimate the mean from a histogram, first turn each bar into a frequency (frequency density × class width), then use mean = Σ(frequency × midpoint) ÷ Σfrequency, with the midpoint standing in for every value in that class. Working: the four classes have widths 20, 10, 20 and 20, so their frequencies are 1 × 20 = 20, 3 × 10 = 30, 2 × 20 = 40 and 0.5 × 20 = 10, which do add to the 100 vehicles stated. Their midpoints are 10, 25, 40 and 60, so Σfx = 20 × 10 + 30 × 25 + 40 × 40 + 10 × 60 = 200 + 750 + 1600 + 600 = 3150, and the mean is 3150 ÷ 100 = 31.5. Answer: the estimated mean speed is 31.5 mph. Watch which numbers you treat as the frequencies and which as the values: using the frequency densities themselves as the frequencies, without multiplying by the class widths first, gives 1 × 10 + 3 × 25 + 2 × 40 + 0.5 × 60 = 195 spread over 1 + 3 + 2 + 0.5 = 6.5, and 195 ÷ 6.5 = 30, a mean built from the wrong 'frequencies' altogether; averaging the four midpoints on their own, (10 + 25 + 40 + 60) ÷ 4 = 33.75, ignores how many vehicles are actually in each class; and using each class's lower boundary in place of its midpoint, 20 × 0 + 30 × 20 + 40 × 30 + 10 × 50 = 2300 and 2300 ÷ 100 = 23, systematically underestimates every class by roughly half its width.
16. (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).
17. (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.
18. (d) 2/3 — Method: since A is the midpoint of OE, OE = 2a, so E = 2a. Since FB is twice AF, F is 1/3 of the way along AB from A, so F = a + 1/3(b − a) = (2/3)a + (1/3)b. G is the midpoint of OB, so G = (1/2)b. Working: EF = F − E = (2/3)a + (1/3)b − 2a = −(4/3)a + (1/3)b, and EG = G − E = −2a + (1/2)b. Comparing term by term, 2/3 × (−2a + (1/2)b) = −(4/3)a + (1/3)b, which matches EF exactly — the same scalar works on both the a-term and the b-term, so the two vectors are parallel, and since they share the point E the three points are collinear. Answer: k = 2/3, so E, F and G lie on a straight line. Giving 1/3 instead is the scalar linking F to G (FG = (1/3)EG), not E to F; giving 3/2 is the reciprocal — it is EG that equals 3/2 × EF, not the other way round, since EF = k × EG was what was asked for; and giving 4/3 is EF's a-coefficient read off raw, without ever dividing it by EG's a-coefficient to form the comparison. Always match the direction of the scalar to the vectors exactly as the question states them.