Higher June 2025 Paper 2 Q17
17 \(y = 4x^3 + 5x^2 + 2x\)
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
(b) Find the coordinates of the turning points on the graph with equation \(y = 4x^3 + 5x^2 + 2x\)
Show clear algebraic working. (4)
Show clear algebraic working. (4)
| Scheme | Marks |
|---|---|
| Two of \(12x^2 + 10x + 2\) | M1 |
| \(12x^2 + 10x + 2\) | A1 |
| (2) |
Notes
M1: for differentiating 2 or 3 terms correctly
A1: for all 3 terms correct
| Scheme | Marks |
|---|---|
\((3x + 1)(4x + 2)\;(= 0)\) or \((6x + 2)(2x + 1)(= 0)\) or \(2(3x + 1)(2x + 1)\;(= 0)\) or \((3x + 1)(2x + 1)(= 0)\) \(\dfrac{-10 \pm \sqrt{10^2 - 4 \times 12 \times 2}}{2 \times 12}\) or \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 6 \times 1}}{2 \times 6}\) or \(12\left[\left(x + \dfrac{10}{24}\right)^2 - \left(\dfrac{10}{24}\right)^2\right] + 2(= 0)\) oe or \(6\left[\left(x + \dfrac{5}{12}\right)^2 - \left(\dfrac{5}{12}\right)^2\right] + 1(= 0)\) oe | M1 |
| \(-\dfrac{1}{2}, -\dfrac{1}{3}\) | A1 |
\((y =)4 \times \left(-\dfrac{1}{2}\right)^3 + 5\left(-\dfrac{1}{2}\right)^2 + 2\left(-\dfrac{1}{2}\right)\left(= -\dfrac{1}{4}\right)\) or \((y =)4 \times \left(-\dfrac{1}{3}\right)^3 + 5\left(-\dfrac{1}{3}\right)^2 + 2\left(-\dfrac{1}{3}\right)\left(= -\dfrac{7}{27}\right)\) | M1 |
Working required Answer: \(\left(-\dfrac{1}{2}, -\dfrac{1}{4}\right)\) \(\left(-\dfrac{1}{3}, -\dfrac{7}{27}\right)\) | A1 |
| (4) | |
| (6 marks) |
Notes
M1: ft dep on M1 for a correct method to solve their 3 term quadratic equation (with at least 2 correct coefficients) 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 \(\dfrac{-10 \pm \sqrt{100 - 96}}{24}\)) Derivative must be a 3 term quadratic for this M mark
NB Can be implied by answers of \((x =)-\dfrac{1}{2}\) and \((x =)-\dfrac{1}{3}\)
A1: oe dep on previous M1
Allow –0.33(333) or \(-0.\dot{3}\) for correct \(x\) values
M1: ft dep on previous M1 for substituting at least one \(x\) value into \(y\)
NB Can be implied by one correct value of \(y\)
A1: oe dep on M1 for correct coordinates \((-0.5, -0.25)\), \((-0.33, -0.25(9\ldots))\)