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.

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

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

$\frac{{1 - 3x}}{{x - 2}} = 2$     (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]