Moderate -0.8 This is a straightforward application of differentiation to find where a function is decreasing. Students differentiate the cubic, set dy/dx < 0, solve a quadratic inequality using the formula or factoring, and state the interval. It's more routine than average since it follows a standard algorithm with no conceptual challenges or multi-step reasoning required.
Attempt to solve \(12x^2 + 14x - 6 = 0\) or any multiple e.g. \(6x^2 + 7x - 3 = 0\) (accept with \(< 0\) or \(> 0\) etc shown)
M1
May see QF or \((2x+3)(3x-1)\) or completing the square. \((2x \pm 3)(3x \pm 1)\) M1; \(k(2x \pm 3)(3x \pm 1)\) M1. If using QF a correctly quoted formula followed by a slip in substitution scores M1, but if formula isn't quoted and there are errors in substitution then M0.
\(x = -\frac{3}{2}\) and \(x = \frac{1}{3}\) only identified
A1
If these values are stated incorrectly with no method shown for solving the quadratic, then the last three marks are lost.
\(-\frac{3}{2} < x < \frac{1}{3}\)
A1
Accept strict or non-strict inequalities. Accept \(x > -\frac{3}{2}\) and \(x < \frac{1}{3}\) or \(x \in \left(-\frac{3}{2}, \frac{1}{3}\right)\) with strict or non-strict inequalities but NOT \(x > -\frac{3}{2}\) or \(x < \frac{1}{3}\)
Total: [5]
## Question 13:
$\frac{dy}{dx} = 12x^2 + 14x - 6$ | **M1** | Expression of the form $\alpha x^2 + \beta x + \gamma$ where $\alpha, \beta, \gamma \in \mathbb{R}$ with at least two terms of the correct form
$\frac{dy}{dx} = 12x^2 + 14x - 6$ (fully correct) | **A1** | Fully correct derivative
Attempt to solve $12x^2 + 14x - 6 = 0$ or any multiple e.g. $6x^2 + 7x - 3 = 0$ (accept with $< 0$ or $> 0$ etc shown) | **M1** | May see QF or $(2x+3)(3x-1)$ or completing the square. $(2x \pm 3)(3x \pm 1)$ M1; $k(2x \pm 3)(3x \pm 1)$ M1. If using QF a correctly quoted formula followed by a slip in substitution scores M1, but if formula isn't quoted and there are errors in substitution then M0.
$x = -\frac{3}{2}$ **and** $x = \frac{1}{3}$ only identified | **A1** | If these values are stated incorrectly with no method shown for solving the quadratic, then the last three marks are lost.
$-\frac{3}{2} < x < \frac{1}{3}$ | **A1** | Accept strict or non-strict inequalities. Accept $x > -\frac{3}{2}$ **and** $x < \frac{1}{3}$ or $x \in \left(-\frac{3}{2}, \frac{1}{3}\right)$ with strict or non-strict inequalities but NOT $x > -\frac{3}{2}$ or $x < \frac{1}{3}$
**Total: [5]**
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13 Determine the range of values of $x$ for which $y = 4 x ^ { 3 } + 7 x ^ { 2 } - 6 x + 8$ is a decreasing function.
\hfill \mbox{\textit{OCR MEI AS Paper 2 2024 Q13 [5]}}