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Date May 2009 Marks available 7 Reference code 09M.1.hl.TZ1.9
Level HL only Paper 1 Time zone TZ1
Command term Draw and Find Question number 9 Adapted from N/A

Question

(a)     Let a>0a>0 . Draw the graph of y=|xa2|y=xa2 for axa on the grid below.

 

 

(b)     Find k such that 0a|xa2|dx=ka0|xa2|dx .

Markscheme

(a)

    A1A1

Note: Award A1 for the correct x-intercept,

A1 for completely correct graph.

 

(b)     METHOD 1

the area under the graph of y=|xa2| for axa , can be divided into ten congruent triangles;     M1A1

the area of eight of these triangles is given by 0a|xa2|dx and the areas of the other two by a0|xa2|dx     M1A1

so, 0a|xa2|dx=4a0|xa2|dxk=4     A1     N0

METHOD 2

use area of trapezium to calculate     M1

0a|xa2|dx=a×12(3a2+a2)=a2     A1

and area of two triangles to obtain     M1

a0|xa2|dx=2×12(a2)2=a24     A1

so, k = 4     A1     N0

METHOD 3

use integration to find the area under the curve

0a|xa2|dx=0ax+a2dx     M1

=[x22+a2x]0a=a22+a22=a2     A1

and

a0|xa2|dx=a20x+a2dx+aa2xa2dx     M1

=[x22+a2x]a20+[x22a2x]aa2=a28+a24+a22a22a28+a24=a24     A1

so, k = 4     A1     N0

[7 marks]

 

Examiners report

Most candidates attempted this question but very often produced sketches lacking labels on axes and intercepts or ignored the domain of the function. For part (b) many candidates attempted to use integration to find the areas but seldom considered the absolute value. A small number of candidates used geometrical methods to determine the areas, showing good understanding of the problem.

Syllabus sections

Topic 6 - Core: Calculus » 6.5 » Definite integrals.

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