| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Write inequalities from graph |
| Difficulty | Moderate -0.8 This question requires students to write inequalities from a graph showing a quadratic curve and a line. While it involves multiple inequalities and understanding of regions, it's a relatively straightforward application of reading graphs and translating visual information into algebraic form. The quadratic and line equations can be determined from given points using standard methods, making this easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02i Represent inequalities: graphically on coordinate plane |
| Answer | Marks |
|---|---|
| M1 | Attempts to find the gradient of equation of line l with points \((2,4)\) and \((6,0)\) and substitutes either \((2,4)\) or \((6,0)\) into \(y - y_1 = m(x - x_1)\) to obtain an equation of line l |
| A1 | \(y = -x + 6\) |
| M1 | A complete method to find the constant \(a\) in \(y = ax(6 - x)\) or the constants \(a, b\) in \(y = ax^2 + bx\), \(a = -\frac{1}{2}\), \(b = 3\) |
| A1 | Equation of the curve \(C\) is \(y = \frac{1}{2}x(6 - x)\) or \(y = -\frac{1}{2}x^2 + 3x\) |
| B1 | Fully defines the region R: \(x \ge 0\), \(y \le -x + 6\), \(y \ge \frac{1}{2}x(6 - x)\) |
**Working out equation of line l:**
Gradient is given by: $\frac{4}{-4} = -1$
$y - 4 = -1(x - 2) \Rightarrow y = -x + 6$
Given that $C$ is a quadratic function it has equation $y = ax^2 + bx$ (no $c$ as the y intercept is 0)
Substituting in $(2,4)$ and $(6,0)$ gives: $4 = 4a + 2b, 0 = 36a + 6b$
Solving simultaneously, this gives: $a = -\frac{1}{2}, b = 3$
Therefore $C: y = -\frac{1}{2}x^2 + 3x$
The region $R$ is defined by $x \ge 0$, $y \le -x + 6$, $y \ge \frac{1}{2}x(6 - x)$
| M1 | Attempts to find the gradient of equation of line l with points $(2,4)$ and $(6,0)$ and substitutes either $(2,4)$ or $(6,0)$ into $y - y_1 = m(x - x_1)$ to obtain an equation of line l |
| --- | --- | --- |
| A1 | $y = -x + 6$ |
| M1 | A complete method to find the constant $a$ in $y = ax(6 - x)$ or the constants $a, b$ in $y = ax^2 + bx$, $a = -\frac{1}{2}$, $b = 3$ |
| A1 | Equation of the curve $C$ is $y = \frac{1}{2}x(6 - x)$ or $y = -\frac{1}{2}x^2 + 3x$ |
| B1 | Fully defines the region R: $x \ge 0$, $y \le -x + 6$, $y \ge \frac{1}{2}x(6 - x)$ |
**(5) marks total for Question 7**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{802e56f7-5cff-491a-b90b-0759a9b35778-09_928_1093_258_497}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a curve $C$ with equation $y = \mathrm { f } ( x )$ and a straight line $l$. The curve $C$ meets $l$ at the points $( 2,4 )$ and $( 6,0 )$ as shown.
The shaded region $R$, shown shaded in Figure 1, is bounded by $\mathrm { C } , l$ and the $y$-axis. Given that $\mathrm { f } ( x )$ is a quadratic function in $x$, use inequalities to define region $R$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q7 [5]}}