O-LevelAdditional MathematicsTrigonometry4037/11 · May/June 2011 · Q55 Marks
On the diagram above, sketch the curve y = 1 + 3sin2x for 0° ≤ x ≤ 180°. [Figure 5.1]
- (i)On the diagram above, sketch the curve y = 1 + 3sin2x for 0° ≤ x ≤ 180°. [Figure 5.1][3]
- (ii)On the diagram above, sketch the curve y = |1 + 3sin2x| for 0° ≤ x ≤ 180°. [Figure 5.2][1]
- (iii)Write down the number of solutions of the equation |1 + 3sin2x| = 1 for 0° ≤ x ≤ 180°.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- [1] B1 for shape of sine curve, period 90°, amplitude 3, shifted up by 1.
- [1] B1 for curve with maximum 4, minimum -2, completing one cycle (peak to peak or trough to trough) within 0° to 180° range.
- [1] B1 for all correct: passing through (0,1), (45,4), (90,1), (135,-2), (180,1).
- [1] √B1 for reflecting the negative part of their curve from (i) about the x-axis.
- [1] √B1 for 5 solutions, following through from their graph.
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