AS June 2019 Q2

EdexcelAS paperCurrent spec6 marksInequalities

2. A student was set the following problem.

Use algebra to find the set of values of \(x\) for which
\(\dfrac{x}{x - 24} \gt \dfrac{1}{x + 11}\)

The student’s attempt at a solution is written below.

\(x(x - 24)(x + 11)^2 \gt (x + 11)(x - 24)^2\)
\(x(x - 24)(x + 11)^2 - (x + 11)(x - 24)^2 \gt 0\)
\((x - 24)(x + 11)\left[x(x + 11) - x - 24\right] \gt 0\)	Line 3
\((x - 24)(x + 11)\left[x^2 + 10x - 24\right] \gt 0\)
\((x - 24)(x + 11)(x + 12)(x - 2) \gt 0\)
\(x = 24,\ x = -11,\ x = -12,\ x = 2\)
\(\left\{x \in \mathbb{R} : -12 \lt x \lt -11\right\} \cup \left\{x \in \mathbb{R} : 2 \lt x \lt 24\right\}\)	Line 7

There are errors in the student’s solution.

(a) Identify the error made

(i) in line 3

(ii) in line 7 (2)

(b) Find a correct solution to this problem. (4)

Mark scheme (a)

Scheme	Marks	AO
(a)(i) Line 3: Allow any of either bracketing error \(-24\) should be \(24\) in the square brackets \(x(x + 11) - x - 24\) should be \(x(x + 11) - (x - 24)\) \(x(x + 11) - x - 24\) should be \(x(x + 11) - x + 24\)	B1	2.3
(a)(ii) Line 7: Allow any of either should be \(\{x \in \mathbb{R} : x \lt -12 \text{ or } -11 \lt x \lt 2 \text{ or } x \gt 24\}\) they have found the regions where the inequality is \(\lt 0\) they have reversed the inequality	B1	2.3
	(2)

Notes

(a)(i)

B1: See scheme

Note: Give B0 for contradictory reasons

(a)(ii) Way 1

B1: See scheme

Note: Give B0 for contradictory reasons

Note: Allow “Should be \(x \lt -12,\ -11 \lt x \lt 2,\ x \gt 24\)”

Note: Do not allow

“Should be \(x \lt -12 \cap -11 \lt x \lt 2 \cap x \gt 24\)”
They have found where \(x \lt 0\) and not where \(x \gt 0\)
“There should be 3 inequalities and not 2 inequalities”
“The sign is the wrong way around”

Mark scheme (b)

Scheme	Marks	AO
Way 1 \((x - 24)(x + 11)[x(x + 11) - (x - 24)] \gt 0\) \((x - 24)(x + 11)[x^2 + 10x + 24] \gt 0\)	M1	1.1b
\((x - 24)(x + 11)(x + 6)(x + 4) \gt 0\) Critical values \(x = -11, -6, -4, 24\)	A1	1.1b
\(\left\{x \in \mathbb{R} : x \lt -11\right\} \cup \left\{x \in \mathbb{R} : -6 \lt x \lt -4\right\} \cup \left\{x \in \mathbb{R} : x \gt 24\right\}\)	M1 A1	2.2a 2.5
	(4)
	(6 marks)

Notes

Way 1

M1: Uses brackets {to correct the error made on line 3}, forms a 3TQ and uses a correct method of solving a 3TQ to give \(x = \ldots\)

A1: All four correct critical values for \(x\)

M1: Deduces that the 2 “outsides” and the “middle interval” are required

A1: Exactly 3 correct intervals. Their answer must be given in set notation. Accept equivalent set notation. E.g. Allow

\(\{x \in \mathbb{R} : x \lt -11 \text{ or } -6 \lt x \lt -4 \text{ or } x \gt 24\}\)
\(\{x \lt -11 \text{ or } -6 \lt x \lt -4 \text{ or } x \gt 24\}\)
\(\{x \lt -11 \cup -6 \lt x \lt -4 \cup x \gt 24\}\)
\(\mathbb{R} - \left([-11, -6] \cup [-4, 24]\right)\)

Note: Give final A0 for \(\{x \in \mathbb{R} : x \lt -11\} \cap \{x \in \mathbb{R} : -6 \lt x \lt -4\} \cap \{x \in \mathbb{R} : x \gt 24\}\)

Note: Allow A1 for \(\{x \in \mathbb{R} : x \lt -11,\ -6 \lt x \lt -4,\ x \gt 24\}\)

(b) Way 2

Scheme	Marks	AO
\(\dfrac{x}{x - 24} \gt \dfrac{1}{x + 11} \Rightarrow \dfrac{x}{x - 24} - \dfrac{1}{x + 11} \gt 0 \Rightarrow \dfrac{x(x + 11) - (x - 24)}{(x - 24)(x + 11)} \gt 0\)	M1	1.1b
\(\Rightarrow \dfrac{x^2 + 10x + 24}{(x - 24)(x + 11)} \gt 0 \Rightarrow \dfrac{(x + 6)(x + 4)}{(x - 24)(x + 11)} \gt 0\) Critical values \(x = -11, -6, -4, 24\)	A1	1.1b
\(\left\{x \in \mathbb{R} : x \lt -11\right\} \cup \left\{x \in \mathbb{R} : -6 \lt x \lt -4\right\} \cup \left\{x \in \mathbb{R} : x \gt 24\right\}\)	M1 A1	2.2a 2.5
	(4)

M1: Gathers terms on one side and puts over a common denominator. Simplifies the numerator to \(x(x + 11) - (x - 24)\) {and thereby corrects the error made in line 3}, forms a 3TQ and uses a correct method of solving a 3TQ to give \(x = \ldots\)

A1: See Way 1

M1: See Way 1

A1: See Way 1

(b) Way 3

Scheme	Marks	AO
Considering \(x \lt -11\) \(\dfrac{x}{x - 24} \gt \dfrac{1}{x + 11} \Rightarrow x^2 + 11x \gt x - 24 \Rightarrow x^2 + 10x + 24 \gt 0\) gives \(x \lt -6\) or \(x \gt -4\). Hence \(x \lt -11\)	M1	1.1b
Considering \(-11 \lt x \lt 24\) \(\dfrac{x}{x - 24} \gt \dfrac{1}{x + 11} \Rightarrow x^2 + 11x \lt x - 24 \Rightarrow x^2 + 10x + 24 \lt 0\) gives \(-6 \lt x \lt -4\). Hence \(-6 \lt x \lt -4\) Considering \(x \gt 24\) \(\dfrac{x}{x - 24} \gt \dfrac{1}{x + 11} \Rightarrow x^2 + 11x \gt x - 24 \Rightarrow x^2 + 10x + 24 \gt 0\) gives \(x \lt -6\) or \(x \gt -4\). Hence \(x \gt 24\)	A1	1.1b
Overall, \(\left\{x \in \mathbb{R} : x \lt -11\right\} \cup \left\{x \in \mathbb{R} : -6 \lt x \lt -4\right\} \cup \left\{x \in \mathbb{R} : x \gt 24\right\}\)	M1 A1	2.2a 2.5
	(4)

M1: Considers each of the intervals \(x \lt -11,\ -11 \lt x \lt 24,\ x \gt 24\) separately and evaluates which parts (if any) of these regions satisfy the original inequality

A1: Obtains a correct inequality statement for each of the intervals \(x \lt -11,\ -11 \lt x \lt 24,\ x \gt 24\)

M1: See Way 1

A1: See Way 1