🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (b) x² + y² = 169 — Method: a circle centred on the origin has equation x² + y² = r², and every point on it satisfies that equation, so substituting the coordinates of a point that lies on the circle gives r² directly. Working: substituting x = 5 and y = 12 gives 5² + 12² = 25 + 144 = 169, so r² = 169 and the circle is x² + y² = 169. Answer: x² + y² = 169. The distractors: x² + y² = 13 uses the radius, √169 = 13, where r² belongs, which is the confusion between r and r² made in the other direction; x² + y² = 17 adds the two coordinates, 5 + 12, instead of adding their squares; x² + y² = 119 subtracts the squares, 144 − 25, treating 12 as the hypotenuse of the right-angled triangle rather than as one of the shorter sides.
2. (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.
3. (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
4. (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.
5. (a) 842.0 — The radius to (20, 21) has gradient 21/20, so the tangent's gradient is −20/21. The tangent line is y − 21 = −20/21(x − 20), i.e. y = −20/21x + 841/21. Setting y = 0 gives the x-intercept x = 841/20 = 42.05; setting x = 0 gives the y-intercept y = 841/21 ≈ 40.048. The area of triangle OPQ is 1/2 × 42.05 × 40.048 ≈ 842.0. 1684.0 comes from multiplying the two intercepts without the 1/2 that a triangle's area needs — twice the correct area. 580.7 comes from using the circle's radius, 29, as a side of the triangle instead of the x-intercept, 42.05: 1/2 × 29 × 40.048 ≈ 580.7. 2.0 comes from a sign error in the tangent's gradient — using 20/21 instead of −20/21 — which gives a different line, with intercepts x ≈ −2.05 and y ≈ 1.952, and area 1/2 × 2.05 × 1.952 ≈ 2.0.
6. (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.
7. (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
8. (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).
9. (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.
10. (d) Reflect in the y-axis, +2 in x; roots 3, −2 — f(2 − x) is zero exactly when 2 − x equals one of f's roots: 2 − x = −1 or 2 − x = 4. Solving each correctly (x = 2 − (−1) = 3, and x = 2 − 4 = −2) gives the new roots x = 3 and x = −2. This is the same as reflecting y = f(x) in the y-axis to get f(−x), then translating 2 units in the positive x-direction to get f(−(x − 2)) = f(2 − x). Solving 2 − x = k as x = k − 2 instead of x = 2 − k is a sign slip when rearranging, and gives x = −3 and x = 2. Translating +2 in x with no reflection at all uses f(x − 2), whose roots are the original roots plus 2: x = 1 and x = 6 — this misses the reflection completely. Assuming 'no overall change' wrongly treats a reflection-and-translation pair as always cancelling out, when here the roots genuinely move, from x = −1 and x = 4 to x = 3 and x = −2.
11. (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.
12. (b) y = −3x/4 + 25 — Method: a straight line that touches a circle at one point is a tangent there, so it is perpendicular to the radius drawn to that point; find the gradient of the radius, take its negative reciprocal, then substitute the point of contact into y − y₁ = m(x − x₁). Working: the radius from (0, 0) to (12, 16) has gradient 16 ÷ 12, which cancels to 4/3, so the tangent has gradient −3/4. Substituting gives y − 16 = −3/4(x − 12), so y − 16 = −3x/4 + 9 and y = −3x/4 + 25. Answer: y = −3x/4 + 25. The distractors: y = 3x/4 + 7 turns the gradient of the radius upside down but leaves it positive, so the perpendicular step is only half done; y = −4x/3 + 32 changes the sign of the radius gradient without turning it upside down, which is the other half left undone; y = −3x/4 − 25 uses the correct gradient but substitutes the point of contact with both signs reversed, writing y + 16 = −3/4(x + 12).
13. (b) y = f(−x) — y = f(−x) reflects the graph of y = f(x) in the y-axis: every x-coordinate changes sign. The x-intercepts −3 and 7 become 3 and −7, matching the second graph's intercepts −7 and 3. A point already on the y-axis is unaffected, since −0 = 0, so the y-intercept (0, 21) stays exactly where it is — matching the second graph as well. y = −f(x) leaves the x-intercepts unchanged at −3 and 7, since f(x) = 0 exactly where −f(x) = 0, which does not match; it also sends the y-intercept to (0, −21), a second mismatch. y = −f(−x) does send the x-intercepts to the right places, −7 and 3, but it sends the y-intercept to (0, −21) instead of (0, 21), so it fails the second clue. y = f(x) − 4 moves every point down 4, sending the y-intercept to (0, 17) instead of (0, 21), so it fails the y-axis clue. Test each option against BOTH clues — the pair of x-intercepts and the point on the y-axis — because more than one option gets only one of the two right.
14. (b) 1/3 — Method: the first sweet has already been taken and it was red, so work out the second probability from what is actually left in the bag. Working: one red sweet has gone, so 3 red sweets remain out of 9 sweets altogether, giving 3/9. Dividing the numerator and the denominator by 3 gives 1/3. Answer: the probability is 1/3. The distractors: 2/5 is 4/10, the probability for the first draw used again, which is only right if the first sweet is put back; 3/10 takes one off the red count but leaves the total at 10, updating half of the fraction; 4/9 takes one off the total but leaves the red count at 4, updating the other half of the fraction.
15. (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.
16. (d) (1/2)a + (1/2)c — Method: X is the midpoint of AC, so OX = OA + (1/2)AC, with AC = c − a. Working: OX = a + 1/2(c − a) = a − (1/2)a + (1/2)c = (1/2)a + (1/2)c. Answer: OX = (1/2)a + (1/2)c. Since OB = a + c, this is exactly half of OB, so OX = (1/2)OB, meaning X lies on OB at its midpoint too — the two diagonals bisect each other. Forgetting to halve AC at all gives a + c, which is OB itself, not its midpoint; halving only the c-term gives (1/2)a + c; and a sign error on the c-term gives (1/2)a − (1/2)c. Halve the whole of AC, both terms together, and add it to OA rather than to a alone.
17. (c) y = (5/12)x − 169/12 — The gradient of the radius to (5, −12) is (−12 − 0) ÷ (5 − 0) = −12/5. A tangent is perpendicular to the radius at that point, so its gradient is the negative reciprocal, 5/12. Using y − y₁ = m(x − x₁) with (5, −12): y + 12 = (5/12)(x − 5), which gives y = (5/12)x − 169/12. y = −(12/5)x comes from using the radius's own gradient, −12/5, instead of turning it into the perpendicular gradient, and building the line through the origin (as the radius itself does). y = −(5/12)x − 119/12 comes from taking the reciprocal of −12/5 correctly as a size but keeping the wrong sign, using −5/12 instead of 5/12. y = (5/12)x − 25/12 comes from using the correct gradient 5/12 but building the line through (5, 0) instead of (5, −12) — dropping the point's y-coordinate.
18. (d) x² − 7x − 3 = 0 — Method: an iteration settles where the next value equals the one before it, so both can be written as the same letter x; replace every xₙ by x, square both sides to clear the square root, and collect all the terms on one side. Working: x = √(7x + 3) gives x² = 7x + 3 on squaring both sides; subtracting 7x and 3 from both sides gives x² − 7x − 3 = 0. Answer: x² − 7x − 3 = 0. The distractors: x² + 7x − 3 = 0 moves the 7x across the equals sign without changing its sign; x² − 7x + 3 = 0 makes that same slip on the constant instead, leaving the 3 positive as it crosses; x² − 7x − 9 = 0 squares the expression term by term, squaring the 3 to give 9 as though squaring √(7x + 3) gave 7x + 9, which is the (a + b)² = a² + b² mistake dressed as a square root.