Higher June 2024 Paper 1R Q19
19 A curve \(C\) has equation \(\;y = x^3 - 8x^2 - 12x + 5\)
Curve \(C\) has exactly two stationary points, one at point \(A\) and one at point \(B\) such that
\(x\) coordinate of point \(A \gt x\) coordinate of point \(B\)
Find the coordinates of point \(A\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 3x^2 - 16x - 12\) | M1 |
| \(\text{``}{3x^2 - 16x - 12}\text{''} = 0\) | M1ft |
eg \((3x + 2)(x - 6)\; (= 0)\) or \((x =)\, \dfrac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 3 \times (-12)}}{2 \times 3}\) or \(3\left[\left(x - \dfrac{8}{3}\right)^2 - \dfrac{64}{9}\right] - 12\; (= 0)\) | M1ft |
| eg \(6^3 - 8 \times 6^2 - 12 \times 6 + 5\; (= -139)\) | M1ft |
| Working required Answer: (6, –139) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for differentiation with at least 2 terms correct
M1ft: (dep on previous M1) for their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)
M1ft: (dep on 1st M1)
for the correct \(x\) value (of 6) – ignore other \(x\) value
OR
for solving their 3 term quadratic equation using any correct method
(if factorising, allow brackets which expanded give 2 out of 3 terms correct)
(if using formula allow one sign error and some simplification – allow as far as \(\dfrac{16 \pm \sqrt{256 + 144}}{6}\))
(if completing the square then as far as shown on LHS)
The award of this mark implies the previous M mark
M1ft: (dep on 1st M1) for \(x = 6\) substituted into correct equation for curve \(C\)
OR
(dep on 1st M1 and 2 values for \(x\)) for their greatest \(x\) value substituted into correct equation for curve \(C\)
(ignore any attempt to substitute their least \(x\) value)
A1: (dep on M2) cao