Date | May 2019 | Marks available | 4 | Reference code | 19M.2.SL.TZ1.S_3 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Find | Question number | S_3 | Adapted from | N/A |
Question
Consider the function f(x)=x2e3xf(x)=x2e3x, x∈R.
Find f′(x).
[4]
a.
The graph of f has a horizontal tangent line at x=0 and at x=a. Find a.
[2]
b.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
choosing product rule (M1)
eg uv′+vu′ , (x2)′(e3x)+(e3x)′x2
correct derivatives (must be seen in the rule) A1A1
eg 2x, 3e3x
f′(x)=2xe3x+3x2e3x A1 N4
[4 marks]
a.
valid method (M1)
eg f′(x)=0, ,
a=−0.667(=−23) (accept x=−0.667) A1 N2
[2 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.