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Mathematics · May / June 2025 · Paper 11 · Question 3

YMA01

As printed in the paper

The question as text

3. Find 2 x 3 d x 3 x writing your answer in simplest form. (5)

Read automatically from the paper; the printed page above is exact.

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3

x

3

x

3

5

3

1

(

3

)

2

x

+

1

2

2

2

d

x

=

×

x

2

+ ×

2

x

2

+ ×

3

x

2

+

c

3

5

3

1

3

x

5

3

1

2

4

2

2

2

=

x

+

x

+

6

x

+

c

15

3

M1: For an attempt to multiply out the numerator and divide at least two terms by the denominator with

at least one power correct. Allow if there are only two terms from squaring.

dM1: For correctly achieving two correct powers following integration (need not be simplified).

A1: Previous dM must have been gained (so two correct powers after integrating), at least one correct

1

5

1 ...

algebraic term, allow unsimplified. E,g, allow

x Allow ( )

with one correct term inside the

3

2

5

3

2

bracket.

A1: Two correct simplified algebraic terms. May have the 1

3 factored out.

A1: Fully correct and simplified including the + c, accepting equivalent simplified forms. Ignore

spurious extra integral signs or dx‘s if they have the correct expression. May have the 1

ISW if they multiply through their answer by a constant after the integration.

Note: If you see attempts at parts or substitution that look worthy of merit then send to review.

(

)

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