18M.1.AHL.TZ2.H_2

pestleMathematicsAIHLPaper 118M· sl-2-2-functions-notation-domain-range-and-inverse-as-reflectionsource ↗

Sketch the graphs of $y = \frac{x}{2} + 1$ and $y = \left| {x - 2} \right|$ on the following axes.

[3]

Solve the equation $\frac{x}{2} + 1 = \left| {x - 2} \right|$ .

[4]

Markscheme / solution

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

straight line graph with correct axis intercepts A1

modulus graph: V shape in upper half plane A1

modulus graph having correct vertex and y-intercept A1

[3 marks]

METHOD 1

attempt to solve $\frac{x}{2} + 1 = x - 2$ (M1)

$x = 6$ A1

Note: Accept $x = 6$ using the graph.

attempt to solve (algebraically) $\frac{x}{2} + 1 = 2 - x$ M1

$x = \frac{2}{3}$ A1

[4 marks]

METHOD 2

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

$\frac{{{x^2}}}{4} + x + 1 = {x^2} - 4x + 4$

$0 = \frac{{3{x^2}}}{4} - 5x + 3$

$3{x^2} - 20x + 12 = 0$

attempt to factorise (or equivalent) M1

$\left( {3x - 2} \right)\left( {x - 6} \right) = 0$

$x = \frac{2}{3}$ A1

$x = 6$ A1

[4 marks]

Examiners’ report

[N/A]