June 2022 Paper 3 Q8
8 The curves \(y = \mathrm{h}(x)\) and \(y = \mathrm{h}^{-1}(x)\), where \(\mathrm{h}(x) = x^3 - 8\), are shown below.
The curve \(y = \mathrm{h}(x)\) crosses the \(x\)-axis at B and the \(y\)-axis at A.
The curve \(y = \mathrm{h}^{-1}(x)\) crosses the \(x\)-axis at D and the \(y\)-axis at C.
Determine the equation of the circle. Give your answer in the form \((x - a)^2 + (y - b)^2 = r^2\), where \(a\), \(b\) and \(r^2\) are constants to be determined. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(y = x^3 - 8\) \(x^3 = y + 8\) | M1 | 1.1a |
| \(\sqrt[3]{(x + 8)}\) oe isw | A1 | 2.2a |
| [2] |
Notes
M1: Attempt to re-arrange
A1: Ignore labelling of this expression eg \(\mathrm{fh}(x) = \sqrt[3]{(x+8)}\) scores 2
| Scheme | Marks | AO |
|---|---|---|
| A \((0, -8)\) | B1 | 1.1 |
| B \((2, 0)\) | B2 | 2.2a |
| C \((0, 2)\) | B1 | 1.1 |
| D \((-8, 0)\) | B1 | 1.1 |
| [5] |
Notes
B1: Condone lack of brackets if meaning is clear
B1: C: FT their B
B1: D: FT their A
| Scheme | Marks | AO |
|---|---|---|
| Midpoint is \((1, -4)\) and Gradient of AB is 4 | B1 | 1.1 3.1a |
| Gradient of perpendicular bisector is \(-\dfrac{1}{4}\) | B1 | 2.2a |
| Equation \(y + 4 = -\dfrac{1}{4}(x - 1)\) | M1 | |
| \(y = -\dfrac{1}{4}x - 3\dfrac{3}{4}\) | A1 | 1.1 |
| [4] |
Notes
B1: FT their A and B. May be implied by later work
B1: FT -ve reciprocal of their 4
M1: Or using \(y = mx + c\) and attempting to evaluate \(c\)
Must be using their midpoint and their \(-\dfrac{1}{4}\)
A1: Final answer
| Scheme | Marks | AO |
|---|---|---|
| Either \((0 - a)^2 + (2 - b)^2 = r^2\) using C \((0 - a)^2 + (-8 - b)^2 = r^2\) using A \((2 - a)^2 + (0 - b)^2 = r^2\) using B \((-8 - a)^2 + (0 - b)^2 = r^2\) using D Setting up any 2 of the above equations | M1 | 3.1a |
| Attempting to solve to find \(a\) or \(b\) | M1 | 3.1a |
| Centre \((-3, -3)\) oe | A1 | 1.1 |
| Using their centre and another point (A, B, C or D) to find the radius | M1 | 2.1 |
| \((x + 3)^2 + (y + 3)^2 = 34\) cao | A1 | 2.2a |
| [5] |
Notes
M1: May solve all 4 which also earns 3rd method mark
Or
| Scheme | Marks | AO |
|---|---|---|
| Intersection of any 2 of \(y = x\), \(x = -3\), \(y = -3\), \(y = -\dfrac{1}{4}x - 3\dfrac{3}{4}\), \(y = -4x - 15\) | M1 | 3.1a |
| Attempting to solve to find intersection point | M1 | 3.1a |
| \((-3, -3)\) | A1 | 1.1 |
| Using their centre and another point (A, B, C or D) to find the radius | M1 | 2.1 |
| \((x + 3)^2 + (y + 3)^2 = 34\) | A1 | 2.2a |
M1: Identifying 2 perp bisectors of 2 chords eg BD and AC, or AB and CD