2. Use algebra to find the set of values of \(x\) for which
$$\frac { x - 2 } { 2 ( x + 2 ) } \leqslant \frac { 12 } { x ( x + 2 ) }$$
"
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Question 2:
Answer Marks
Guidance
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 Marks
Guidance
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 Marks
Guidance
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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# 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]}}