IB Mathematics SL 2.6 Modelling skills AI SL Paper 1- Exam Style Questions- New Syllabus
Question
The cross-section of an arched entrance into the ballroom of a hotel is in the shape of a parabola. This cross-section can be represented by part of the graph of \( h(x) = -1.6x^2 + 4.48x \), where \( h(x) \) is the height of the archway, in metres, at a horizontal distance, \( x \) metres, from the point O, in the bottom corner of the archway.
(a) Calculate an equation for the axis of symmetry of the parabola that represents the archway. [2]
(b) To prepare for an event, a square-based crate that is 1.6 m wide and 2.0 m high is to be moved through the archway into the ballroom. The crate must remain upright while it is being moved. Calculate whether the crate will fit through the archway. Justify your answer. [3]
▶️ Answer/Explanation
Markscheme
(a) Given: \( h(x) = -1.6x^2 + 4.48x \). \[ \begin{aligned} \text{Axis of symmetry: } x &= -\frac{b}{2a} \\ a = -1.6, \, b = 4.48: \quad x &= -\frac{4.48}{2 \times (-1.6)} = \frac{4.48}{3.2} = 1.4 \end{aligned} \] Answer: \( x = 1.4 \) M1 A1 [2 marks]
(b) METHOD 1
Crate: 1.6 m wide, 2.0 m high. Center at \( x = 1.4 \). Edges: \( x = 1.4 – 0.8 = 0.6 \), \( x = 1.4 + 0.8 = 2.2 \). \[ h(x) = -1.6x^2 + 4.48x \] At \( x = 0.6 \): \[ h(0.6) = -1.6 \times 0.36 + 4.48 \times 0.6 = -0.576 + 2.688 = 2.112 \, \text{m} \] At \( x = 2.2 \): \[ h(2.2) = -1.6 \times 4.84 + 4.48 \times 2.2 = -7.744 + 9.856 = 2.112 \, \text{m} \] Height 2.112 m > 2.0 m. A1 The archway is tall enough. R1 A1
METHOD 2
Solve: \( -1.6x^2 + 4.48x \geq 2.0 \). \[ -1.6x^2 + 4.48x – 2.0 = 0 \] \[ \begin{aligned} \Delta &= 4.48^2 – 4 \times (-1.6) \times (-2.0) = 20.0704 – 12.8 = 7.2704 \\ x &= \frac{-4.48 \pm \sqrt{7.2704}}{2 \times (-1.6)} \approx 0.5574, 2.2426 \end{aligned} \] Width: \( 2.2426 – 0.5574 \approx 1.6852 \, \text{m} > 1.6 \, \text{m}\). A1 The archway is wide enough. R1 A1
Answer: The crate will fit through the archway. [3 marks]
Total Marks: 5