| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |quadratic| compared to linear: algebraic inequality |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to work with modulus functions, rational expressions, and solve a compound inequality. Part (a) requires setting equations equal, clearing denominators, and showing a repeated root (discriminant = 0). Part (b) requires careful case analysis for |x| and interpreting the tangency result to determine solution regions. The modulus complicates the algebra significantly, and the 'hence' connection requires insight about what the tangency means for the inequality across different domains. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 - 8x = \frac{4x}{4-x} \Rightarrow (x^2-8x)(4-x) = 4x \Rightarrow x(4x-32-x^2+8x-4)=0\) | M1 | Attempts intersection by setting equal, cross-multiplies and factorises \(x\) out or cancels |
| (so \(x=0\) or) \(x^2 - 12x + 36 = 0\) | A1 | Correct quadratic; may be implied by solutions of \(0,6\) seen from cubic (by calculator) |
| \(\Rightarrow x(x-6)^2 = 0 \Rightarrow x = \ldots\) | M1 | Solves the quadratic to find roots |
| Meet at \((6, -12)\) | A1 | Obtains correct point where curves meet |
| Touch at \((6,-12)\) as repeated root | B1 | Correct reason given; accept "\((x-6)^2=0\) therefore touches", or discriminant \(=0\) shown with conclusion, or gradients equal at both points shown with conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{dx}(x^2-8x) = 2x-8\) and \(\frac{d}{dx}\left(\frac{4x}{4-x}\right) = \frac{4(4-x)-4x(-1)}{(4-x)^2} = \frac{16}{(4-x)^2}\) | M1, A1 | Attempts derivatives of both curves; both derivatives correct |
| \(2x-8 = \frac{16}{(4-x)^2} \Rightarrow (x-4)^3 = 8 \Rightarrow x = \ldots\) | M1 | Sets derivatives equal and solves to find \(x\) where gradients agree |
| Meet at \((6,-12)\) | A1 | Obtains correct point |
| e.g. \(6^2-6\times9=-12\) and \(\frac{4\times6}{4-6}=-12\), so curves meet at tangent at \((6,-12)\) | B1 | Correct value checked in both curves with conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2-8x = \frac{4x}{4+x} \Rightarrow x(x-8)(4+x)-4x=0 \Rightarrow x(x^2-4x-36)=0 \Rightarrow x=\ldots\) | M1 | Attempts intersection of other branch \(\frac{4x}{4- |
| \(x = (0),\ 2\pm2\sqrt{10}\); critical value is \((0\) and\()\ 2-2\sqrt{10}\) | A1 | Correct value of \(2-2\sqrt{10}\) identified (no need to see second root rejected) |
| Other C.V.s are \(0,\ \pm4\) | B1 | Both \(0\) and \(\pm4\) identified as critical values |
| Extremes are \(x < 2-2\sqrt{10}\) and \(x > 6\) or any two suitable ranges | M1 | Forms at least two suitable ranges from critical values (allow \(\leqslant\) instead of \(<\)) |
| Solution is \(x < 2-2\sqrt{10},\ -4 < x < 0,\ 4 < x < 6,\ x > 6\) | A1, A1 | A1: at least two correct ranges; A1: fully correct answer |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - 8x = \frac{4x}{4-x} \Rightarrow (x^2-8x)(4-x) = 4x \Rightarrow x(4x-32-x^2+8x-4)=0$ | M1 | Attempts intersection by setting equal, cross-multiplies and factorises $x$ out or cancels |
| (so $x=0$ or) $x^2 - 12x + 36 = 0$ | A1 | Correct quadratic; may be implied by solutions of $0,6$ seen from cubic (by calculator) |
| $\Rightarrow x(x-6)^2 = 0 \Rightarrow x = \ldots$ | M1 | Solves the quadratic to find roots |
| Meet at $(6, -12)$ | A1 | Obtains correct point where curves meet |
| Touch at $(6,-12)$ as repeated root | B1 | Correct reason given; accept "$(x-6)^2=0$ therefore touches", or discriminant $=0$ shown with conclusion, or gradients equal at both points shown with conclusion |
**Alt method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(x^2-8x) = 2x-8$ and $\frac{d}{dx}\left(\frac{4x}{4-x}\right) = \frac{4(4-x)-4x(-1)}{(4-x)^2} = \frac{16}{(4-x)^2}$ | M1, A1 | Attempts derivatives of both curves; both derivatives correct |
| $2x-8 = \frac{16}{(4-x)^2} \Rightarrow (x-4)^3 = 8 \Rightarrow x = \ldots$ | M1 | Sets derivatives equal and solves to find $x$ where gradients agree |
| Meet at $(6,-12)$ | A1 | Obtains correct point |
| e.g. $6^2-6\times9=-12$ and $\frac{4\times6}{4-6}=-12$, so curves meet at tangent at $(6,-12)$ | B1 | Correct value checked in both curves with conclusion |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2-8x = \frac{4x}{4+x} \Rightarrow x(x-8)(4+x)-4x=0 \Rightarrow x(x^2-4x-36)=0 \Rightarrow x=\ldots$ | M1 | Attempts intersection of other branch $\frac{4x}{4-|x|}$ with $x^2-8x$; allow any attempt at solving $\frac{4x}{4+x}=x^2-8x$ that reaches a value for $x$ |
| $x = (0),\ 2\pm2\sqrt{10}$; critical value is $(0$ and$)\ 2-2\sqrt{10}$ | A1 | Correct value of $2-2\sqrt{10}$ identified (no need to see second root rejected) |
| Other C.V.s are $0,\ \pm4$ | B1 | Both $0$ and $\pm4$ identified as critical values |
| Extremes are $x < 2-2\sqrt{10}$ and $x > 6$ or any two suitable ranges | M1 | Forms at least two suitable ranges from critical values (allow $\leqslant$ instead of $<$) |
| Solution is $x < 2-2\sqrt{10},\ -4 < x < 0,\ 4 < x < 6,\ x > 6$ | A1, A1 | A1: at least two correct ranges; A1: fully correct answer |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0d458344-42cb-48d1-90b3-e071df8ea7bb-08_693_987_116_482}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C _ { 1 }$ with equation
$$y = \frac { 4 x } { 4 - | x | }$$
and the curve $C _ { 2 }$ with equation
$$y = x ^ { 2 } - 8 x$$
For $x > 0 , C _ { 1 }$ has equation $y = \frac { 4 x } { 4 - x }$
\begin{enumerate}[label=(\alph*)]
\item Use algebra to show that $C _ { 1 }$ touches $C _ { 2 }$ at a point $P$, stating the coordinates of $P$
\item Hence or otherwise, using algebra, solve the inequality
$$x ^ { 2 } - 8 x > \frac { 4 x } { 4 - | x | }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2022 Q3 [11]}}