June 2022 Paper 2 Q16
16 The equation of a curve is
\(y = 6x^4 + 8x^3 - 21x^2 + 12x - 6\).
Determine
- The coordinates of the stationary points on the curve.
- The nature of the stationary points on the curve.
- The \(x\)-coordinate of the non-stationary point of inflection on the curve.
\(y = 6x^4 + 8x^3 - 21x^2 + 12x - 6\). [3]
| Scheme | Marks | AO | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 24x^3 + 24x^2 - 42x + 12\) | M1 | 3.1a | ||||||||
| their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 | 1.1 | ||||||||
| \(\mathrm{f}(k)\) evaluated, where \(k\) is a factor of \(\pm 12\) or \(\pm\frac{a}{12}\), where \(a = 1, 2, 3, 4\) or 6 | M1 | 2.1 | ||||||||
| \((x + 2)(4x^2 - 4x + 1)\) or \((2x - 1)(2x^2 + 3x - 2)\) | M1 | 3.1a | ||||||||
| \(x = -2\) and \(x = \frac{1}{2}\) and no others | A1 | 1.1 | ||||||||
| \(\left(\frac{1}{2}, -\frac{31}{8}\right)\) and \((-2, -82)\) and no others | A1 | 1.1 | ||||||||
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 72x^2 + 48x - 42\) | M1* | 1.1 | ||||||||
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 150\) when \(x = -2\) so minimum value or eg
| A1 | 1.1 | ||||||||
eg
| M1 | 3.1a | ||||||||
| inflection at \(\left(\frac{1}{2}, -\frac{31}{8}\right)\) | A1 | 3.2a | ||||||||
| their \(72x^2 + 48x - 42 = 0\) | M1dep* | 1.1 | ||||||||
| \(x = -\frac{7}{6}\) isw | A1 | 1.1 | ||||||||
| [12] |
Notes
M1: allow one sign or coefficient error; must be four terms
M1: at least two terms correct
M1: may be implied by \(x = -2\) seen unsupported or \((x + 2)\) identified as factor
M1: by inspection or long division; allow one sign error or one coefficient error in trinomial
may be implied by \(x = \frac{1}{2}\) seen unsupported or \((2x - 1)\) oe identified as factor
A1: may see \(x = \frac{1}{2}\) (repeated)
A0 for \(x = -2\) (repeated)
M1*: allow one sign or one coefficient error, FT their \(\frac{\mathrm{d}y}{\mathrm{d}x}\);
allow M1 for \(12x^2 + 8x - 7\)
A1: NB test indecisive at \(x = \frac{1}{2}\)
A0 for just eg \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\) so minimum
award M1A1 for consideration of gradient either side of −2, values must be correct to at least 2sf for A1
M1: or eg
| \(x\) | 0 | (½) | 1 |
| \(y\) | −6 | (−3.875) | −1 |
| \(x\) | 0 | (½) | 1 |
| \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) | −42 | (0) | 78 |
A1: values in table must be correct
A1: ignore calculation of associated \(y\)-value
allow any correct decimals to 3 sf or more
| Scheme | Marks | AO |
|---|---|---|
 | M1 B1 A1 | 1.1 1.1 1.1 |
| [3] |
Notes
M1: curve with a minimum in 3rd quadrant and stationary point of inflection in 4th quadrant and no other stationary points
B1: \((0, -6)\) identified as \(y\)-intercept (intercept must be below the \(x\)-axis and above \(-20\))
A1: correct curve with intercepts at \((-a, 0)\) and \((b, 0)\), where \(-3 \lt a \lt -2.6\) and \(0.8 \lt b \lt 1.2\); minimum at \((-2, y)\) where \(-90 \lt y \lt -80\) and inflection for \(0 \lt x \lt 1\) and \(y\) is between the \(x\)-axis and the \(y\)-intercept