Edexcel FP2 2017 June — Question 2 9 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2017
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeRational inequality algebraically
DifficultyStandard +0.3 This is a standard rational inequality requiring algebraic manipulation: finding common denominator, combining fractions, factoring the numerator, and analyzing sign changes. While it's Further Maths (FP2), the technique is routine and methodical with no novel insight required, making it slightly easier than average.
Spec1.02g Inequalities: linear and quadratic in single variable

2. Use algebra to find the set of values of \(x\) for which $$\frac { x - 2 } { 2 ( x + 2 ) } \leqslant \frac { 12 } { x ( x + 2 ) }$$ "

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{x-2}{2(x+2)} - \frac{12}{x(x+2)} \leq 0\)M1 Collects expressions to one side
\(\frac{x^2-2x-24}{2x(x+2)} \leq 0\)M1A1 M1: Attempt common denominator. A1: Correct single fraction
\(x = 0, -2\)B1 Correct critical values
\(x^2-2x-24 \Rightarrow (x+4)(x-6)(=0) \Rightarrow x = \ldots\)M1 Attempt to solve quadratic as far as \(x = \ldots\)
\(x = -4, 6\)A1 Correct critical values. May be seen on a sketch
\(-4 \leq x < -2,\ 0 < x \leq 6\) with \(\leq\) or \(<\) throughoutdM1A1 M1: Attempt two inequalities using 4 CVs in ascending order (dependent on at least one previous M). A1: All 4 CVs in inequalities correct
\(-4 \leq x < -2,\ 0 < x \leq 6\) or \([-4,-2) \cup (0,6]\)A1cao (9) Inequality signs correct. Set notation may use \(\cup\) or "or" but not "and"
Alternative 1: Multiply both sides by \(x^2(x+2)^2\)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^2(x-2)(x+2) \leq 24x(x+2)\)M1 Both sides \(\times x^2(x+2)^2\), must be a positive multiplier containing \(x^2(x+2)^2\)
\(x^3(x+2)-2x^2(x+2)-24x(x+2) \leq 0\)M1A1 M1: Collects to one side. A1: Correct inequality
\(x = 0, -2\)B1 Correct critical values
\(x^4-28x^2-48x = 0 \Rightarrow x(x+2)(x-6)(x+4)=0 \Rightarrow x = \ldots\)M1 Attempt to solve quartic; can cancel \(x\) and \((x+2)\) and solve a quadratic
\(x = -4, 6\)A1 Correct critical values
\(-4 \leq x < -2,\ 0 < x \leq 6\)dM1A1 As above
\([-4,-2) \cup (0,6]\)A1cao (9) As above
Alternative 2: Using a sketch graph
AnswerMarks Guidance
Answer/WorkingMark Guidance
Draw graphs of \(y = \frac{x-2}{2(x+2)}\) and \(y = \frac{12}{x(x+2)}\)
CVs \(x = 0, -2\)B1 Vertical asymptotes of graphs
\(\frac{x-2}{2(x+2)} = \frac{12}{x(x+2)}\)M1 Eliminate \(y\)
\(x(x-2) = 24\)M1A1 M1: Obtains quadratic. A1: Correct equation
\(x^2-2x-24 \Rightarrow (x+4)(x-6)=0 \Rightarrow x = \ldots\)M1 Attempt to solve quadratic
CVs \(x = -4, 6\)A1 Correct critical values
\(-4 \leq x < -2,\ 0 < x \leq 6\)dM1, A1, A1cao (9) As main scheme
# Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x-2}{2(x+2)} - \frac{12}{x(x+2)} \leq 0$ | M1 | Collects expressions to one side |
| $\frac{x^2-2x-24}{2x(x+2)} \leq 0$ | M1A1 | M1: Attempt common denominator. A1: Correct single fraction |
| $x = 0, -2$ | B1 | Correct critical values |
| $x^2-2x-24 \Rightarrow (x+4)(x-6)(=0) \Rightarrow x = \ldots$ | M1 | Attempt to solve quadratic as far as $x = \ldots$ |
| $x = -4, 6$ | A1 | Correct critical values. May be seen on a sketch |
| $-4 \leq x < -2,\ 0 < x \leq 6$ with $\leq$ or $<$ throughout | dM1A1 | M1: Attempt two inequalities using 4 CVs in ascending order (dependent on at least one previous M). A1: All 4 CVs in inequalities correct |
| $-4 \leq x < -2,\ 0 < x \leq 6$ or $[-4,-2) \cup (0,6]$ | A1cao **(9)** | Inequality signs correct. Set notation may use $\cup$ or "or" but not "and" |

**Alternative 1: Multiply both sides by $x^2(x+2)^2$**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2(x-2)(x+2) \leq 24x(x+2)$ | M1 | Both sides $\times x^2(x+2)^2$, must be a positive multiplier containing $x^2(x+2)^2$ |
| $x^3(x+2)-2x^2(x+2)-24x(x+2) \leq 0$ | M1A1 | M1: Collects to one side. A1: Correct inequality |
| $x = 0, -2$ | B1 | Correct critical values |
| $x^4-28x^2-48x = 0 \Rightarrow x(x+2)(x-6)(x+4)=0 \Rightarrow x = \ldots$ | M1 | Attempt to solve quartic; can cancel $x$ and $(x+2)$ and solve a quadratic |
| $x = -4, 6$ | A1 | Correct critical values |
| $-4 \leq x < -2,\ 0 < x \leq 6$ | dM1A1 | As above |
| $[-4,-2) \cup (0,6]$ | A1cao **(9)** | As above |

**Alternative 2: Using a sketch graph**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Draw graphs of $y = \frac{x-2}{2(x+2)}$ and $y = \frac{12}{x(x+2)}$ | | |
| CVs $x = 0, -2$ | B1 | Vertical asymptotes of graphs |
| $\frac{x-2}{2(x+2)} = \frac{12}{x(x+2)}$ | M1 | Eliminate $y$ |
| $x(x-2) = 24$ | M1A1 | M1: Obtains quadratic. A1: Correct equation |
| $x^2-2x-24 \Rightarrow (x+4)(x-6)=0 \Rightarrow x = \ldots$ | M1 | Attempt to solve quadratic |
| CVs $x = -4, 6$ | A1 | Correct critical values |
| $-4 \leq x < -2,\ 0 < x \leq 6$ | dM1, A1, A1cao **(9)** | As main scheme |

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2. Use algebra to find the set of values of $x$ for which

$$\frac { x - 2 } { 2 ( x + 2 ) } \leqslant \frac { 12 } { x ( x + 2 ) }$$

"\\

\hfill \mbox{\textit{Edexcel FP2 2017 Q2 [9]}}