Date | May 2009 | Marks available | 6 | Reference code | 09M.1.sl.TZ2.7 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Solve | Question number | 7 | Adapted from | N/A |
Question
Let f(x)=√3e2xsinx+e2xcosx , for 0≤x≤π . Solve the equation f(x)=0 .
Markscheme
e2x(√3sinx+cosx)=0 (A1)
e2x=0 not possible (seen anywhere) (A1)
simplifying
e.g. √3sinx+cosx=0 , √3sinx=−cosx , sinx−cosx=1√3 A1
EITHER
tanx=−1√3 A1
x=5π6 A2 N4
OR
sketch of 30∘ , 60∘ , 90∘ triangle with sides 1, 2, √3 A1
work leading to x=5π6 A1
verifying 5π6 satisfies equation A1 N4
[6 marks]
Examiners report
Those who realized e2x was a common factor usually earned the first four marks. Few could reason with the given information to solve the equation from there. There were many candidates who attempted some fruitless algebra that did not include factorization.