Date | November 2016 | Marks available | 2 | Reference code | 16N.1.SL.TZ0.T_14 |
Level | Standard Level | Paper | Paper 1 (with calculator from previous syllabus) | Time zone | Time zone 0 |
Command term | Find | Question number | T_14 | Adapted from | N/A |
Question
The equation of a curve is y=12x4−32x2+7.
The gradient of the tangent to the curve at a point P is −10.
Find dydx.
Find the coordinates of P.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
2x3−3x (A1)(A1) (C2)
Note: Award (A1) for 2x3, award (A1) for −3x.
Award at most (A1)(A0) if there are any extra terms.
[2 marks]
2x3−3x=−10 (M1)
Note: Award (M1) for equating their answer to part (a) to −10.
x=−2 (A1)(ft)
Note: Follow through from part (a). Award (M0)(A0) for −2 seen without working.
y=12(−2)4−32(−2)2+7 (M1)
Note: Award (M1) substituting their −2 into the original function.
y=9 (A1)(ft) (C4)
Note: Accept (−2, 9).
[4 marks]