| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |quadratic| compared to linear: algebraic inequality |
| Difficulty | Standard +0.8 This requires systematic case analysis of the modulus inequality, solving two quadratic inequalities (one for each case), factorizing quadratics, and carefully combining solution sets. While the individual algebraic steps are A-level standard, the multi-case structure and need to track where the expression inside the modulus changes sign makes this moderately challenging, above average difficulty. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2x^2 + x - 3 \geq 0\); \(2x^2 + x - 3 = 3(1-x) \Rightarrow 2x^2 + 4x - 6 = 0\) | M1 | Assume \(2x^2 + x - 3 \geq 0\) and obtain a 3TQ |
| \(2x^2 + 4x - 6 \Rightarrow x^2 + 2x - 3 = (x+3)(x-1) = 0\); \(x = -3, 1\) | A1 | Correct CVs from correct equation |
| \(2x^2 + x - 3 \leq 0\); \(-2x^2 - x + 3 = 3(1-x) \Rightarrow 2x^2 - 2x = 0\) | M1 | Assume \(2x^2 + x - 3 \leq 0\) and obtain a 2 or 3TQ |
| \(2x(x-1) = 0\); \(x = 0, 1\) | A1 | Correct CVs from correct equation |
| \(x < -3 \quad 0 < x < 1 \quad x > 1\) | dM1A1A1 (7) | dM1: form 3 distinct inequalities with 3 CVs (must have scored both M marks); A1 all 3 correct CVs used correctly; A1 inequalities fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((2x^2 + x - 3)^2 > 9(1-x)^2\) | ||
| \(4x^4 + 4x^3 - 20x^2 + 12x > 0\) | M1A1 | Square both sides and collect terms to obtain quartic with 4 or 5 terms; A1 correct quartic |
| \(x(x+3)(x-1)(x-1) > 0\) | M1 | Factorise quartic |
| CVs: \(x = 0, -3, 1\) | A1 | 3 correct CVs |
| Then as main scheme |
## Question 5 (Inequality):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2x^2 + x - 3 \geq 0$; $2x^2 + x - 3 = 3(1-x) \Rightarrow 2x^2 + 4x - 6 = 0$ | M1 | Assume $2x^2 + x - 3 \geq 0$ and obtain a 3TQ |
| $2x^2 + 4x - 6 \Rightarrow x^2 + 2x - 3 = (x+3)(x-1) = 0$; $x = -3, 1$ | A1 | Correct CVs from correct equation |
| $2x^2 + x - 3 \leq 0$; $-2x^2 - x + 3 = 3(1-x) \Rightarrow 2x^2 - 2x = 0$ | M1 | Assume $2x^2 + x - 3 \leq 0$ and obtain a 2 or 3TQ |
| $2x(x-1) = 0$; $x = 0, 1$ | A1 | Correct CVs from correct equation |
| $x < -3 \quad 0 < x < 1 \quad x > 1$ | dM1A1A1 (7) | dM1: form 3 distinct inequalities with 3 CVs (must have scored both M marks); A1 all 3 correct CVs used correctly; A1 inequalities fully correct |
**ALT (Squaring both sides):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(2x^2 + x - 3)^2 > 9(1-x)^2$ | | |
| $4x^4 + 4x^3 - 20x^2 + 12x > 0$ | M1A1 | Square both sides and collect terms to obtain quartic with 4 or 5 terms; A1 correct quartic |
| $x(x+3)(x-1)(x-1) > 0$ | M1 | Factorise quartic |
| CVs: $x = 0, -3, 1$ | A1 | 3 correct CVs |
| Then as main scheme | | |\begin{enumerate}
\item Use algebra to find the set of values of $x$ for which
\end{enumerate}
$$\left| 2 x ^ { 2 } + x - 3 \right| > 3 ( 1 - x )$$
[Solutions based entirely on graphical or numerical methods are not acceptable.]
\includegraphics[max width=\textwidth, alt={}, center]{0d44aec7-a6e8-47fc-a215-7c8c4790e93f-21_2647_1840_118_111}\\
\hfill \mbox{\textit{Edexcel F2 2021 Q5 [7]}}