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Date May Example questions Marks available 6 Reference code EXM.1.AHL.TZ0.2
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Express Question number 2 Adapted from N/A

Question

Consider the integral  1 t 1 x + x 2   d x for  t > 1 .

Very briefly, explain why the value of this integral must be negative.

[1]
a.

Express the function  f ( x ) = 1 x + x 2 in partial fractions.

[6]
b.

Use parts (a) and (b) to show that ln ( 1 + t ) ln t < ln 2 .

[4]
c.

Markscheme

The numerator is negative but the denominator is positive. Thus the integrand is negative and so the value of the integral will be negative.     R1AG

[1 mark]

a.

1 x + x 2 = 1 ( 1 + x ) x A 1 + x + B x      M1M1A1

1 A x + B ( 1 + x ) A = 1 , B = 1      M1A1

1 x + x 2 1 1 + x + 1 x      A1

[6 marks]

b.

1 t 1 1 + x + 1 x d x = [ ln ( 1 + x ) ln x ] 1 t = ln ( 1 + t ) ln t ln 2     M1A1A1

Hence  ln ( 1 + t ) ln t ln 2 < 0 ln ( 1 + t ) ln t < ln 2      R1AG

[4 marks]

c.

Examiners report

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

Syllabus sections

Topic 1—Number and algebra » AHL 1.11—Partial fractions
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Topic 1—Number and algebra

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