June 2025 Paper 2 Q7
7 The point \(A\) lies on the curve with equation
\[y = x^3 + px^2 + qx + 12\](a) Given that \(A\) has coordinates \((-5, 37)\), show that\[5p - q = 30\] [2 marks]
(b) Given that \(A\) is a stationary point, show that\[10p - q = 75\] [3 marks]
(c) Hence find the value of \(p\) and the value of \(q\) [1 mark]
(d) The curve with equation\[y = x^3 + px^2 + qx + 12\]has a second stationary point \(B\)
Find the coordinates of \(B\)
Fully justify your answer.
[3 marks]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(x = -5\) and \(y = 37\) into \(y = x^3 + px^2 + qx + 12\) PI \(37 = -125 + 25p - 5q + 12\) | M1 | 3.1a |
| Completes reasoned argument to show \(5p - q = 30\) There must be at least one intermediate step after \((-5)^3\) and \((-5)^2\) evaluated AG | R1 | 2.1 |
| (2) |
Typical solution
\[37 = (-5)^3 + (-5)^2p + (-5)q + 12\]\[37 = -125 + 25p - 5q + 12\]\[150 = 25p - 5q\]\[5p - q = 30\]| Scheme | Marks | AO |
|---|---|---|
| Differentiates \(x^3\) and 12 correctly | B1 | 1.1b |
| Differentiates \(px^2\) correctly or differentiates \(qx\) correctly and equates their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to zero | M1 | 3.1a |
| Completes reasoned argument by substituting \(x = -5\) into a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain \(10p - q = 75\) AG | R1 | 2.1 |
| (3) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 2px + q\]\[0 = 3 \times (-5)^2 - 10p + q\]\[10p - q = 75\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(p = 9,\ q = 15\) | B1 | 1.1b |
| (1) |
Typical solution
\[5p - q = 30\]\[10p - q = 75\]\[p = 9,\ q = 15\]| Scheme | Marks | AO |
|---|---|---|
| Solves \(3x^2 + 2px + q\ (= 0)\) for their \(p\) and \(q\) Condone \(x^2 + 2px + q\ (= 0)\) | M1 | 3.1a |
| Obtains \(x = -1\) from \(3x^2 + 18x + 15\ (= 0)\) OE | A1 | 1.1a |
| Completes reasoned argument to obtain \((-1, 5)\) Must have obtained \(x = -5\) and \(x = -1\) when solving \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and explained that stationary points occur when \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) OE | R1 | 2.1 |
| (3) | ||
| (9 marks) |
Typical solution
Stationary points occur when \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 18x + 15 = 0\]\[x = -5 \text{ or } -1\]At \(B\), \(x = -1\)
Therefore, stationary point is \((-1, 5)\)