Date | May 2010 | Marks available | 14 | Reference code | 10M.2.hl.TZ2.11 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find, Hence, Sketch, and Solve | Question number | 11 | Adapted from | N/A |
Question
The function f is defined by
f(x)=(x3+6x2+3x−10)12, for x∈D,
where D⊆R is the greatest possible domain of f.
(a) Find the roots of f(x)=0.
(b) Hence specify the set D.
(c) Find the coordinates of the local maximum on the graph y=f(x).
(d) Solve the equation f(x)=3.
(e) Sketch the graph of |y|=f(x), for x∈D.
(f) Find the area of the region completely enclosed by the graph of |y|=f(x)
Markscheme
(a) solving to obtain one root: 1, – 2 or – 5 A1
obtain other roots A1
[2 marks]
(b) D=x∈[−5, −2]∪[1, ∞) (or equivalent) M1A1
Note: M1 is for 1 finite and 1 infinite interval.
[2 marks]
(c) coordinates of local maximum −3.73−2−√3, 3.22√6√3 A1A1
[2 marks]
(d) use GDC to obtain one root: 1.41, – 3.18 or – 4.23 A1
obtain other roots A1
[2 marks]
(e)
A1A1A1
Note: Award A1 for shape, A1 for max and for min clearly in correct places, A1 for all intercepts.
Award A1A0A0 if only the complete top half is shown.
[3 marks]
(f) required area is twice that of y=f(x) between – 5 and – 2 M1A1
answer 14.9 A1 N3
Note: Award M1A0A0 for ∫−2−5f(x)dx=7.47… or N1 for 7.47.
[3 marks]
Total [14 marks]
Examiners report
This was a multi-part question that was well answered by many candidates. The main difficulty was sketching the graph and this meant that the last part was not well answered.