A-LevelMathematicsDifferentiation9709/22 · Feb/Mar 2018 · Q711 Marks

The diagram shows part of the curve defined by the parametric equations x = t² + 4t, y = t³ – 3t². The curve has a minimum point at M and crosses the x-axis at the point P. [Figure 7.1]

(i)Find the gradient of the curve at P.[4]
(ii)Find the coordinates of the point M.[3]
(iii)The value of the gradient of the curve at the point with parameter t is denoted by m. Show that 3t² – (2m + 6)t – 4m = 0 and hence find the set of possible values of m for points on the curve.[4]

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 (Feb/Mar 2018). The mark or mark type is shown in brackets.

[M1] Obtain expression for dy/dx with numerator quadratic, denominator linear Or equivalent where separate derivatives evaluated first when t=3
[A1] Obtain (3t² - 6t) / (2t + 4)
[B1] Identify t = 3 at P
[A1] Obtain 9/10 or equivalent
[M1] Equate first derivative to zero and obtain non-zero value of t
[A1] Obtain t = 2
[A1] Substitute to obtain (12, -4)
[B1] Equate expression for gradient to m and rearrange to confirm 3t² - (2m + 6)t – 4m = 0 AG; necessary detail needed
[M1] Attempt solution of quadratic inequality or equation resulting from discriminant
[A1] Obtain critical values −√72–9 and √72-9
[A1] Conclude m ≤ −√72–9, m ≥ √72-9 Or exact equivalents

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.