FP2 June 2015 Q1

EdexcelOld spec7 marksInequalities

(a) Use algebra to find the set of values of \(x\) for which \[x + 2 \gt \frac{12}{x + 3}\] (6)

(b) Hence, or otherwise, find the set of values of \(x\) for which \[x + 2 \gt \frac{12}{|x + 3|}\] (1)

Mark scheme (a)

Scheme	Marks
\((x + 2)(x + 3)^2 - 12(x + 3) = 0\) OR \(\dfrac{(x + 3)(x + 2) - 12}{(x + 3)} \gt 0\)	M1
\((x + 3)\left(x^2 + 5x - 6\right) = 0 \qquad (x + 3)(x + 6)(x - 1) = 0\)
CVs: \(-3,\ -6,\ 1\)	B1,A1,A1
\(-6 \lt x \lt -3,\quad x \gt 1\) OR: \(x \in (-6, -3) \cup (1, \infty)\)	dM1A1
	(6)

Notes

M1: Mult through by \((x + 3)^2\) and collect on one side or use any other valid method (NOT calculator)
Eg work from \(\dfrac{(x + 3)(x + 2) - 12}{(x + 3)} \gt 0\)
NB: Multiplying by \((x + 3)\) is not a valid method unless the two cases \(x \gt -3\) and \(x \lt -3\) are considered separately or \(-3\) stated to be a cv

B1: for -3 seen anywhere

A1A1: other cvs (A1A0 if only one correct)

dM1: obtaining inequalities using their critical values and no other numbers. Award if one correct inequality seen or any valid method eg sketch graph or number line seen

A1: correct inequalities and no extras. Use of \(\leqslant\) or \(\geqslant\) scores A0. May be written in set notation.

No marks for candidates who draw a sketch graph and follow with the cvs without any algebra shown. Those who use some algebra after their graph may gain marks as earned (possibly all)

(Corrected from the printed mark scheme: the two cases are printed as \(x \gt 3\) and \(x \lt 3\); they are \(x \gt -3\) and \(x \lt -3\).)