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Date May 2018 Marks available 4 Reference code 18M.3.AHL.TZ0.Hca_3
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Show and Show that Question number Hca_3 Adapted from N/A

Question

Find the value of  4 1 x 3 d x .

[3]
a.

Illustrate graphically the inequality  n = 5 1 n 3 < 4 1 x 3 d x < n = 4 1 n 3 .

[4]
b.

Hence write down a lower bound for n = 4 1 n 3 .

[1]
c.

Find an upper bound for n = 4 1 n 3 .

[3]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

4 1 x 3 d x = lim R 4 R 1 x 3 d x       (A1)

Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of  lim x .

Do not award this mark to candidates who use   as the upper limit throughout.

= lim R [ 1 2 x 2 ] 4 R ( = [ 1 2 x 2 ] 4 )      M1

= lim R ( 1 2 ( R 2 4 2 ) )

= 1 32      A1

[3 marks]

a.

      A1A1A1A1

A1 for the curve
A1 for rectangles starting at x = 4
A1 for at least three upper rectangles
A1 for at least three lower rectangles

Note: Award A0A1 for two upper rectangles and two lower rectangles.

sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so

n = 5 1 n 3 < 4 1 x 3 d x < n = 4 1 n 3       AG

[4 marks]

b.

a lower bound is  1 32      A1

Note: Allow FT from part (a).

[1 mark]

c.

METHOD 1

n = 5 1 n 3 < 1 32     (M1)

1 64 + n = 5 1 n 3 = 1 32 + 1 64      (M1)

n = 4 1 n 3 < 3 64 , an upper bound      A1

Note: Allow FT from part (a).

 

METHOD 2

changing the lower limit in the inequality in part (b) gives

n = 4 1 n 3 < 3 1 x 3 d x ( < n = 3 1 n 3 )      (A1)

n = 4 1 n 3 < lim R [ 1 2 x 2 ] 3 R      (M1)

n = 4 1 n 3 < 1 18 , an upper bound     A1

Note: Condone candidates who do not use a limit.

[3 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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Topic 5 —Calculus

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