A-LevelMathematicsDifferentiation9709/23 · May/June 2021 · Q68 Marks
The diagram shows the curve with equation y = (lnx)² – 2 lnx. The curve crosses the x-axis at the points A and B, and has a minimum point M. [Figure X.X]
- (a)Find the exact value of the gradient of the curve at each of the points A and B.[6]
- (b)Find the exact x-coordinate of M.[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [B1] Obtain x=1
- [B1] Obtain x=e²
- [*M1] Differentiate to obtain at least one correct term
- [A1] Obtain correct first derivative (2lnx)/x - 2/x
- [DM1] Substitute at least one of their x-values corresponding to y = 0 to find gradient
- [A1] Obtain gradient -2 [at A] and gradient 2e⁻² at [B]
- [M1] Equate first derivative to zero
- [A1] Obtain x=e
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