17N.1.AHL.TZ0.H_6

pestleMathematicsAIHLPaper 117N· sl-2-2-functions-notation-domain-range-and-inverse-as-reflectionsource ↗

Sketch the graph of $y = \frac{{1 - 3x}}{{x - 2}}$ , showing clearly any asymptotes and stating the coordinates of any points of intersection with the axes.

N17/5/MATHL/HP1/ENG/TZ0/06.a

[4]

Hence or otherwise, solve the inequality $\left| {\frac{{1 - 3x}}{{x - 2}}} \right| < 2$ .

[5]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

N17/5/MATHL/HP1/ENG/TZ0/06.a/M

correct vertical asymptote A1

shape including correct horizontal asymptote A1

$\left( {\frac{1}{3},{\text{ }}0} \right)$ A1

$\left( {0,{\text{ }} - \frac{1}{2}} \right)$ A1

Note: Accept $x = \frac{1}{3}$ and $y = - \frac{1}{2}$ marked on the axes.

[4 marks]

METHOD 1

N17/5/MATHL/HP1/ENG/TZ0/06.b/M

$\frac{{1 - 3x}}{{x - 2}} = 2$ (M1)

$\Rightarrow x = 1$ A1

$- \left( {\frac{{1 - 3x}}{{x - 2}}} \right) = 2$ (M1)

Note: Award this M1 for the line above or a correct sketch identifying a second critical value.

$\Rightarrow x = - 3$ A1

solution is $- 3 < x < 1$ A1

METHOD 2

$\left| {1 - 3x} \right| < 2\left| {x - 2} \right|,{\text{ }}x \ne 2$

$1 - 6x + 9{x^2} < 4({x^2} - 4x + 4)$ (M1)A1

$1 - 6x + 9{x^2} < 4{x^2} - 16x + 16$

$5{x^2} + 10x - 15 < 0$

${x^2} + 2x - 3 < 0$ A1

$(x + 3)(x - 1) < 0$ (M1)

solution is $- 3 < x < 1$ A1

METHOD 3

$- 2 < \frac{{1 - 3x}}{{x - 2}} < 2$

consider $\frac{{1 - 3x}}{{x - 2}} < 2$ (M1)

Note: Also allow consideration of “>” or “=” for the awarding of the M mark.

recognition of critical value at $x = 1$ A1

consider $- 2 < \frac{{1 - 3x}}{{x - 2}}$ (M1)

Note: Also allow consideration of “>” or “=” for the awarding of the M mark.

recognition of critical value at $x = - 3$ A1

solution is $- 3 < x < 1$ A1

[5 marks]

Examiners’ report

[N/A]