A-LevelMathematicsDifferentiation9709/21 · Oct/Nov 2021 · Q58 Marks
The diagram shows the curve with parametric equations x = ln(2t + 3), y = (2t-3)/(2t+3). [Figure 5.1] The curve crosses the y-axis at the point A and the x-axis at the point B.
- (a)Show that dy/dx = 6/(2t+3).[4]
- (b)Find the gradient of the curve at A.[2]
- (c)Find the gradient of the curve at B.[2]
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 (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [B1] Obtain dx/dt = 2/(2t+3)
- [M1] Use quotient rule, or equivalent, to find dy/dt
- [A1] Obtain dy/dt = (2(2t+3) - 2(2t - 3))/((2t+3)²) OE
- [A1 AG] Divide to confirm dy/dx = 6/(2t+3)
- [M1] Attempt to find value of t corresponding to x=0
- [A1] Obtain t=-1 and hence gradient is 6
- [M1] Attempt to find value of t corresponding to y=0
- [A1] Obtain t = 3/2 and hence gradient is 1
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