Date | May 2016 | Marks available | 3 | Reference code | 16M.1.sl.TZ2.15 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 15 | Adapted from | N/A |
Question
Consider the function f(x)=x3−3x2+2x+2 . Part of the graph of f is shown below.
Find f′(x) .
There are two points at which the gradient of the graph of f is 11. Find the x-coordinates of these points.
Markscheme
(f′(x)=)3x2−6x+2 (A1)(A1)(A1) (C3)
Note: Award (A1) for 3x2, (A1) for −6x and (A1) for +2.
Award at most (A1)(A1)(A0) if there are extra terms present.
11=3x2−6x+2 (M1)
Note: Award (M1) for equating their answer from part (a) to 11, this may be implied from 0=3x2−6x−9 .
(x=)−1,(x=)3 (A1)(ft)(A1)(ft) (C3)
Note: Follow through from part (a).
If final answer is given as coordinates, award at most (M1)(A0)(A1)(ft) for (−1,−4) and (3,8) .
Examiners report
Question 15: Differential calculus.
Many candidates correctly differentiated the cubic equation. Most candidates were unable to use differential calculus to find the point where a cubic function had a specified gradient.
Question 15: Differential calculus.
Many candidates correctly differentiated the cubic equation. Most candidates were unable to use differential calculus to find the point where a cubic function had a specified gradient.