| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Write inequalities from graph |
| Difficulty | Moderate -0.3 Part (a) requires solving 3x - 2√x = 8x - 16 by substitution (u = √x), yielding a quadratic—a standard technique. Part (b) asks students to write inequalities from a shaded diagram, testing understanding of inequality notation and regions. This is straightforward once part (a) is complete, requiring no novel insight, just careful reading of the graph and correct inequality symbols. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.02c Simultaneous equations: two variables by elimination and substitution1.02i Represent inequalities: graphically on coordinate plane |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Sets \(3x - 2\sqrt{x} = 8x - 16\) | B1 | Sets equations equal, achieves correct equation |
| \(2\sqrt{x} = 16 - 5x\) then \(4x = (16-5x)^2 \Rightarrow x = ..\) OR \(5x + 2\sqrt{x} - 16 = 0 \Rightarrow (5\sqrt{x} \pm 8)(\sqrt{x} \pm 2) = 0\) | M1 | Key step: make \(\sqrt{x}\) subject and square, OR recognise quadratic in \(\sqrt{x}\) and factorise |
| \(25x^2 - 164x + 256 = 0\) OR \((5\sqrt{x}-8)(\sqrt{x}+2)=0\) | A1 | Correct intermediate line |
| \((25x-64)(x-4)=0 \Rightarrow x=..\) OR \(\sqrt{x} = \frac{8}{5}, (-2) \Rightarrow x=..\) | M1 | Correct method to find at least one value of \(x\) |
| \(x = \frac{64}{25}\) only | A1 | Rejects \(x=4\); \(x=\frac{64}{25}, 4\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts to solve \(3x - 2\sqrt{x} = 0\) | M1 | e.g. \(3x = 2\sqrt{x} \Rightarrow 9x^2 = 4x\), or \(x^{\frac{1}{2}} = \frac{2}{3}\) |
| \(x = \frac{4}{9}\) | A1 | Correct solution |
| \(y,\ 3x-2\sqrt{x},\ y > 8x-16\ x \ldots \frac{4}{9}\) | B1ft | Consistent region defined using either convention |
## Question 12:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sets $3x - 2\sqrt{x} = 8x - 16$ | B1 | Sets equations equal, achieves correct equation |
| $2\sqrt{x} = 16 - 5x$ then $4x = (16-5x)^2 \Rightarrow x = ..$ **OR** $5x + 2\sqrt{x} - 16 = 0 \Rightarrow (5\sqrt{x} \pm 8)(\sqrt{x} \pm 2) = 0$ | M1 | Key step: make $\sqrt{x}$ subject and square, OR recognise quadratic in $\sqrt{x}$ and factorise |
| $25x^2 - 164x + 256 = 0$ **OR** $(5\sqrt{x}-8)(\sqrt{x}+2)=0$ | A1 | Correct intermediate line |
| $(25x-64)(x-4)=0 \Rightarrow x=..$ **OR** $\sqrt{x} = \frac{8}{5}, (-2) \Rightarrow x=..$ | M1 | Correct method to find at least one value of $x$ |
| $x = \frac{64}{25}$ only | A1 | Rejects $x=4$; $x=\frac{64}{25}, 4$ is A0 |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts to solve $3x - 2\sqrt{x} = 0$ | M1 | e.g. $3x = 2\sqrt{x} \Rightarrow 9x^2 = 4x$, or $x^{\frac{1}{2}} = \frac{2}{3}$ |
| $x = \frac{4}{9}$ | A1 | Correct solution |
| $y,\ 3x-2\sqrt{x},\ y > 8x-16\ x \ldots \frac{4}{9}$ | B1ft | Consistent region defined using either convention |
---12.
\begin{tikzpicture}[>=latex, thick]
% Draw x and y axes
\draw[->] (-1.5, 0) -- (6, 0) node[below] {$x$};
\draw[->] (0, -2.5) -- (0, 4.5) node[left] {$y$};
% Origin label
\node[below left] at (0, 0) {$O$};
% Plot Curve C: y = 0.5 * (x - 2) * sqrt(x)
\draw[domain=0:5.5, smooth, samples=100]
plot (\x, {0.5*(\x-2)*sqrt(\x)})
node[above left] {$C$};
% Plot Line l: y = 1.5x - 4
\draw[domain=0.5:5.5]
plot (\x, {1.5*\x - 4});
% Label for line l
\node[below right] at (1.5, -1.75) {$l$};
% Intersection Point A (calculated at x=4, y=2)
\node[above left=1pt] at (4, 2) {$A$};
\end{tikzpicture}
Figure 3 shows a sketch of the curve $C$ with equation $y = 3 x - 2 \sqrt { x } , x \geqslant 0$ and the line $l$ with equation $y = 8 x - 16$
The line cuts the curve at point $A$ as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Using algebra, find the $x$ coordinate of point $A$.
\item \begin{tikzpicture}[>=latex, thick]
% Define the bounding box dimensions for the gray shading
\def\xmin{-1.5} \def\xmax{6}
\def\ymin{-2.5} \def\ymax{4.5}
% 1. Fill the entire background with gray
\fill[gray!35] (\xmin, \ymin) rectangle (\xmax, \ymax);
% 2. "Cut out" region R by filling it with white
\fill[white] (2, -1) -- (2, 0) --
plot[domain=2:4, smooth, samples=50] (\x, {0.5*(\x-2)*sqrt(\x)}) --
cycle;
% 3. Draw x and y axes ON TOP of the shading
\draw[->] (\xmin, 0) -- (\xmax, 0) node[below] {$x$};
\draw[->] (0, \ymin) -- (0, \ymax) node[left] {$y$};
% Origin label
\node[below left] at (0, 0) {$O$};
% 4. Draw the solid vertical line at the root (x=2)
\draw (2, \ymin) -- (2, \ymax);
% 5. Plot Curve C
\draw[domain=0:5.5, smooth, samples=100]
plot (\x, {0.5*(\x-2)*sqrt(\x)})
node[above left] {$C$};
% 6. Plot Dashed Line l
\draw[dashed, domain=0.8:5.5]
plot (\x, {1.5*\x - 4});
% Label for line l
\node[below right] at (3.2, 0.8) {$l$};
% 7. Labels A and R
\node[above left=1pt] at (4, 2) {$A$};
\node at (2.5, 0.1) {$R$};
\end{tikzpicture}
The region $R$ is shown unshaded in Figure 4. Identify the inequalities that define $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q12 [8]}}