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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+3x10)12, for xD,

where DR 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 xD.

(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 3.7323, 3.2263     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 25f(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.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Local maximum and minimum values.
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