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Date	May Specimen paper	Marks available	5	Reference code	SPM.2.AHL.TZ0.9
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	9	Adapted from	N/A

Question

Consider the graphs of $y = \frac{{{x^2}}}{{x - 3}}$ and $y = m\left( {x + 3} \right)$ , $m \in \mathbb{R}$ .

Find the set of values for $m$ such that the two graphs have no intersection points.

Markscheme

METHOD 1

sketching the graph of $y = \frac{{{x^2}}}{{x - 3}}$ ( $y = x + 3 + \frac{9}{{x - 3}}$ ) M1

the (oblique) asymptote has a gradient equal to 1

and so the maximum value of $m$ is 1 R1

consideration of a straight line steeper than the horizontal line joining (−3, 0) and (0, 0) M1

so $m$ > 0 R1

hence 0 < $m$ ≤ 1 A1

METHOD 2

attempting to eliminate $y$ to form a quadratic equation in $x$ M1

${x^2} = m\left( {{x^2} - 9} \right)$

$\Rightarrow \left( {m - 1} \right){x^2} - 9m = 0$ A1

EITHER

attempting to solve $- 4\left( {m - 1} \right)\left( { - 9m} \right) < 0$ for $m$ M1

attempting to solve ${x^2}$ < 0 ie $\frac{{9m}}{{m - 1}} < 0\,\left( {m \ne 1} \right)$ for $m$ M1

THEN

$\Rightarrow 0 < m < 1$ A1

a valid reason to explain why $m = 1$ gives no solutions eg if $m = 1$ ,

$\left( {m - 1} \right){x^2} - 9m = 0 \Rightarrow - 9 = 0$ and so 0 < $m$ ≤ 1 R1

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » AHL 2.13—Rational functions

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