| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve absolute value inequality |
| Difficulty | Challenging +1.2 This is a Further Maths FP1 question requiring case-by-case analysis of an absolute value inequality with a rational function. Students must consider x > 0 and x < 0 separately, handle the discontinuity at |x| = 2, and solve resulting quadratic inequalities. While systematic, it requires careful algebraic manipulation and multiple steps beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02i Represent inequalities: graphically on coordinate plane1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(x<0\): need \(2x-5>\frac{x}{-x-2}\); for \(x\geq 0\): need \(2x-5>\frac{x}{x-2}\); proceeds to find critical values for each | M1 | Must consider both cases for \(x<0\) and \(x\geq 0\); allow "=" for finding CVs |
| For \(x\geq 0\): \(2x-5=\frac{x}{x-2} \Rightarrow 2x^2-10x+10=0 \Rightarrow x=\ldots\) | M1 | Correct method for intersection of line and curve for \(x\) positive |
| \(x=\frac{5\pm\sqrt{5}}{2}\) (oe, awrt 3.62 and awrt 1.38) | A1 | |
| For \(x<0\): \(2x-5=\frac{x}{-x-2} \Rightarrow -2x^2+10=0 \Rightarrow x=\ldots\) | M1 | Correct method for intersection for \(x\) negative |
| \(x=-\sqrt{5}\) only (\(\sqrt{5}\) must be rejected at some stage) | A1 | Must reject positive root |
| Uses graph or other means to identify correct regions; considers asymptotes; identifies at least one correct region e.g. \(-\sqrt{5} | M1 | Asymptotes must be considered; may miss region near \(x=-2\) |
| \(-\sqrt{5} | A1ft | At least one correct interval following through from their solutions |
| \(\left(-\sqrt{5},-2\right)\cup\left(\frac{5-\sqrt{5}}{2},2\right)\cup\left(\frac{5+\sqrt{5}}{2},\infty\right)\) | A1 | Fully correct solution, all three intervals; accept equivalent notation |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $x<0$: need $2x-5>\frac{x}{-x-2}$; for $x\geq 0$: need $2x-5>\frac{x}{x-2}$; proceeds to find critical values for each | M1 | Must consider both cases for $x<0$ and $x\geq 0$; allow "=" for finding CVs |
| For $x\geq 0$: $2x-5=\frac{x}{x-2} \Rightarrow 2x^2-10x+10=0 \Rightarrow x=\ldots$ | M1 | Correct method for intersection of line and curve for $x$ positive |
| $x=\frac{5\pm\sqrt{5}}{2}$ (oe, awrt 3.62 and awrt 1.38) | A1 | |
| For $x<0$: $2x-5=\frac{x}{-x-2} \Rightarrow -2x^2+10=0 \Rightarrow x=\ldots$ | M1 | Correct method for intersection for $x$ negative |
| $x=-\sqrt{5}$ only ($\sqrt{5}$ must be rejected at some stage) | A1 | Must reject positive root |
| Uses graph or other means to identify correct regions; considers asymptotes; identifies at least one correct region e.g. $-\sqrt{5}<x<-2$ or $\frac{5-\sqrt{5}}{2}<x<2$ or $x>\frac{5+\sqrt{5}}{2}$ | M1 | Asymptotes must be considered; may miss region near $x=-2$ |
| $-\sqrt{5}<x<-2$ or $\frac{5-\sqrt{5}}{2}<x<2$ or $x>\frac{5+\sqrt{5}}{2}$ | A1ft | At least one correct interval following through from their solutions |
| $\left(-\sqrt{5},-2\right)\cup\left(\frac{5-\sqrt{5}}{2},2\right)\cup\left(\frac{5+\sqrt{5}}{2},\infty\right)$ | A1 | Fully correct solution, all three intervals; accept equivalent notation |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{55803551-f13d-419f-8b51-31642bd20b6a-08_494_780_258_644}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = \frac { x } { | x | - 2 }$$
Use algebra to determine the values of $x$ for which
$$2 x - 5 > \frac { x } { | x | - 2 }$$
\hfill \mbox{\textit{Edexcel FP1 2021 Q3 [8]}}