A-LevelMathematicsDifferentiation9709/22 · Feb/Mar 2024 · Q69 Marks
The diagram shows the curve with parametric equations x = 1 + √t, y = (lnt+2)(lnt-3), for 0 < t < 25. The curve crosses the x-axis at the points A and B and has a minimum point M. [Diagram]
- (a)Show that dy/dx = (4lnt-2)/√t[4]
- (b)Find the exact gradient of the curve at B.[2]
- (c)Find the exact coordinates of M.[3]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2024). The mark or mark type is shown in brackets.
- [M1] (a) Use product rule to find dy/dt [Or chain rule if expression expanded before differentiation.]
- [A1] (a) Obtain dy/dt = 2lnt/t - 1/t
- [M1] (a) Divide attempt at dy/dt by attempt at dx/dt
- [A1] (a) Confirm dy/dx = (4lnt-2)/√t [Answer given – necessary detail needed.]
- [B1] (b) State or imply t = e³ at point B
- [B1] (b) Obtain gradient 10e⁻¹·⁵ or exact equivalent
- [M1] (c) Equate first derivative to zero and obtain value for lnt or t
- [A1] (c) Obtain x-coordinate 1+e¹ [Or exact equivalent.]
- [A1] (c) Obtain y-coordinate -25/4 [Or equivalent.]
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