| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Moderate -0.3 This is a straightforward rational inequality requiring standard algebraic manipulation: bringing terms to one side, combining fractions over a common denominator, factoring the numerator, and analyzing sign changes using a sign chart. While it requires careful attention to the domain (x ≠ 0, -3) and testing intervals, it follows a well-established procedure taught in Further Pure modules with no novel insight required. Slightly easier than average due to being a routine textbook-style question. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\frac{x-4}{(x+3)} - \frac{5}{x(x+3)} \leq 0\) | M1 | Collects expressions to one side |
| \(\frac{x^2 - 4x - 5}{x(x+3)} \leq 0\) | M1A1 | M1: Attempt common denominator; A1: Correct single fraction |
| \(x = 0, -3\) | B1 | Correct critical values |
| \(x^2 - 4x - 5 \Rightarrow (x-5)(x+1) = 0 \Rightarrow x = \ldots\) | M1 | Attempt to solve quadratic to obtain the other 2 critical values |
| \(x = -1, 5\) | A1 | Correct critical values |
| \(-3 < x \leq -1,\ 0 < x \leq 5\) or \((-3,-1] \cup (0,5]\) | dM1A1A1 | M1: Attempts two inequalities using 4 cv's in ascending order; A1: All 4 cv's in inequalities correct; A1: Both intervals fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(x^2(x+3)(x-4) \leq 5x(x+3)\) | M1 | Multiplies both sides by \(x^2(x+3)^2\), must be positive multiplier |
| \(x^3(x+3) - 4x^2(x+3) - 5x(x+3) \leq 0\) | M1A1 | M1: Collects to one side; A1: Correct inequality |
| \(x = 0, -3\) | B1 | Correct critical values |
| \(x(x+3)(x-5)(x+1) = 0 \Rightarrow x = \ldots\) | M1 | Attempt quartic; allow \(x\) and \(x+3\) cancelled to get other cv's |
| \(x = -1, 5\) | A1 | Correct critical values |
| \(-3 < x \leq -1,\ 0 < x \leq 5\) or \((-3,-1] \cup (0,5]\) | dM1A1A1 | As Way 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(x = 0, -3\) | B1 | Correct critical values (vertical asymptotes) |
| \(\frac{x-4}{x+3} = \frac{5}{x(x+3)}\) | M1 | Eliminate \(y\) |
| \(x(x-4) = 5\) | M1A1 | M1: Obtains quadratic equation; A1: Correct quadratic |
| \(x^2 - 4x - 5 = 0 \Rightarrow x = -1, 5\) | M1A1 | M1: Solves quadratic; A1: Correct critical values |
| \(-3 < x \leq -1,\ 0 < x \leq 5\) or \((-3,-1] \cup (0,5]\) | M1A1A1 | As above |
# Question 2:
## Way 1:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\frac{x-4}{(x+3)} - \frac{5}{x(x+3)} \leq 0$ | M1 | Collects expressions to one side |
| $\frac{x^2 - 4x - 5}{x(x+3)} \leq 0$ | M1A1 | M1: Attempt common denominator; A1: Correct single fraction |
| $x = 0, -3$ | B1 | Correct critical values |
| $x^2 - 4x - 5 \Rightarrow (x-5)(x+1) = 0 \Rightarrow x = \ldots$ | M1 | Attempt to solve quadratic to obtain the **other** 2 critical values |
| $x = -1, 5$ | A1 | Correct critical values |
| $-3 < x \leq -1,\ 0 < x \leq 5$ or $(-3,-1] \cup (0,5]$ | dM1A1A1 | M1: Attempts two inequalities using 4 cv's in ascending order; A1: All 4 cv's in inequalities correct; A1: Both intervals fully correct |
## Way 2:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $x^2(x+3)(x-4) \leq 5x(x+3)$ | M1 | Multiplies both sides by $x^2(x+3)^2$, must be positive multiplier |
| $x^3(x+3) - 4x^2(x+3) - 5x(x+3) \leq 0$ | M1A1 | M1: Collects to one side; A1: Correct inequality |
| $x = 0, -3$ | B1 | Correct critical values |
| $x(x+3)(x-5)(x+1) = 0 \Rightarrow x = \ldots$ | M1 | Attempt quartic; allow $x$ and $x+3$ cancelled to get other cv's |
| $x = -1, 5$ | A1 | Correct critical values |
| $-3 < x \leq -1,\ 0 < x \leq 5$ or $(-3,-1] \cup (0,5]$ | dM1A1A1 | As Way 1 |
## Way 3:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $x = 0, -3$ | B1 | Correct critical values (vertical asymptotes) |
| $\frac{x-4}{x+3} = \frac{5}{x(x+3)}$ | M1 | Eliminate $y$ |
| $x(x-4) = 5$ | M1A1 | M1: Obtains quadratic equation; A1: Correct quadratic |
| $x^2 - 4x - 5 = 0 \Rightarrow x = -1, 5$ | M1A1 | M1: Solves quadratic; A1: Correct critical values |
| $-3 < x \leq -1,\ 0 < x \leq 5$ or $(-3,-1] \cup (0,5]$ | M1A1A1 | As above |
**[Total: 9]**
---\begin{enumerate}
\item Use algebra to find the set of values of $x$ for which
\end{enumerate}
$$\frac { x - 4 } { ( x + 3 ) } \leqslant \frac { 5 } { x ( x + 3 ) }$$
\hfill \mbox{\textit{Edexcel F2 2017 Q2 [9]}}