Mathematics · May / June 2023 · Paper 31 · Question 10

3

3

d

y

−

y

(

)

2

3

y

−

3

d

x

6

y

2

−

12

y

−

18

2

y

2

−

4

y

−

6

=

=

o.e

d

y

(

)

(

)

2

2

9

y

−

1

3

y

−

1

P and Q are where d

x

2

y =

0

or where

2

y

−

4

y

−

6

=

0

d

(

)(

)

Solves

2

y

2

−

4

y

−

6

=

0

2

y

−

3

y

+

1

=

0

y

=

3, 1

−

2

y

2

+

6

1 and 3

..

Subs

y = −

in

x

=

x

=

3

y

−

3

4

4

Achieves

x = −

and

x =

3

Notes

(a)

Ay

(3

y

−

3

)

−

B

M1

Attempts the quotient rule. Condone slips on the coefficients look for

(3

3)

y

−

A,B > 0. Allow a product rule attempt:

(

)(

d

x

(

)

)

(

)

(

)

2

−

1

−

1

2

−

2

x

=

2

y

+

6

3

y

−

3

=

Ay

3

y

−

3

+

2

y

+

6

×−

B

3

y

−

3

d

y

Correct differentiation which may be unsimplified. Allow if the d

y is missing or called d

x

A1

d

(

)

(

)

−

1

(

)

−

2

this mark. By product rule

4

y

3

y

−

3

+

2

y

2

+

6

×−

3 3

y

−

3

Condone missing brackets if

recovered.

dM1

Requires an attempt to get a single fraction with some attempt to simplify.

For the quotient rule look for a simplification of the numerator with like terms collected giving a

3TQ.

Attempts via the product rule will require a correct method to put as a single fraction.

or exact simplified equivalent such as (

)(

)

d

2

2

4

6

2

y

−

3

y

+

1

x

y

−

y

−

A1

=

isw after a correct

d

y

3

y

2

−

6

y

+

3

(

)

2

3

y

−

1

simplified answer. Common factor 3 must have been cancelled. Must be seen in part (a). A0 if

called d

y

x but allow A1 if LHS is not stated.

d

Attempts at d

y

x can score the first 3 marks if correct. Allow use of x in place of y for the Ms.

d

(b)

Indicates P and Q are where d

x

2

2

y

4

y

6

0

B1

y =

0

or where their

−

−

=

(which may be the

d

denominator of d

y

x if they found this instead).

d

Solves their 3TQ from an attempt at d

x

(or denominator of their d

y

M1

y =

0

x =0), usual rules.

d

d

2

y

+

6

2

dM1

Substitutes both their solutions to

2

y

−

4

y

−

6

=

0

into

x

=

. Condone slips if the attempt

3

y

−

3

is clear. At least one should be correct if no method is shown.

4

A1cso Achieves

x = −

and

x =

4

only. Must be equations not just values but isw after correct equations

3

seen as long as no contrary work is shown (such as giving horizontal lines). Accept equivalents. Must

have come from a correct derivative though allow from an isw form if a numerical factor was lost in the

numerator. Must be exact.

Answers from no working score 0/4 as the question instructs use of part (a), so must see the attempt at

setting d

x

y =

0

d

2

2

y

+

6

d

x

d

x

2

Alt (a)

x

=

3

xy

−

3

x

=

2

y

+

6

3

x

+

3

y

−

3

=

4

y

3

y

−

3

d

y

d

y

d

x

4

y

−

3

x

=

(

)

d

y

3

y

1

−

States that P and Q are where d

x

(b) First

0

or where 4

y

−

3

x

=

0

y =

d

2 marks.

4

2

y

2

+

6

y

=

4

y

2

−

4

y

=

2

y

2

+

6

as main scheme

3

3

y

−

3

2

y

2

+

6

2

y

2

8

d

x

2

8

x

=

=

+

+

=

−

(

)

Alt II (a)

3

y

−

3

3

3

d

y

3

(

)

3

y

−

1

2

3

y

−

1

(

)

2

d

x

2

y

−

1

−

8

2

y

2

4

y

6

−

−

=

=

oe

d

y

(

)

2

(

)

2

3

y

−

1

3

y

−

1

Notes

(a)

C

M1

Attempts long division or other method to achieve

Ay

+

B

+

oe and differentiates.

3

y

−

3

A1

Correct differentiation.

dM1

Attempts to get a single fraction and simplifies numerator to 3TQ or uses difference of squares to

factorise.

A1

Correct answer.

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