Mathematics · May / June 2024 · Paper 31 · Question 9

k

3

d

x

(i)

=

4sin

2

−

1

y =

8sin

cos

o.e.

x

y

y

y

d

±±

x

1

x

+

1

(ii) Attempts either

sin

2

=

or

cos

2

= ±±

1

(both for dM1) M1 dM1

y

y

4

4

d

y

1

1

1

=

=

=

*

d

x

8sin

y

cos

y

+

1

+

1

2

+

1 3

−

x

x

x

x

8

×

×

±±

1

4

4

1

At

, gradient of curve =

2 3 gradient of normal is 2 3

−

x =

2

Point N (base length of triangle) is solution of

(

)

π

π

y

−

= −

2 3

x

−

2

x

=

2

+

(

=

2.3...)

3

6 3

2

1

π

π

π

π

Area =

×

2

+

×

=

+

2

3

3

6 3

36 3

(a)

B1*: Verifies that

k =

2

with no errors seen. (Condone

x =

2

) Do not allow this mark if rounded numbers

π. They may work in degrees and substitute in 60° which is

are seen in their solution for e.g. 3

π

acceptable. Look for as a minimum e.g. (

)

=

4sin

2

−=

1

2

Do not allow

=

4sin

k

k

3

−

x

+

1

If they rearrange the equation to e.g.

sin

1

and substitute in

proceeding to e.g. 3

y

=

x =

2

4

then they must conclude that

k =

2

(or a preamble followed by a minimal conclusion e.g. QED, tick)

(b) (i)

M1:

Differentiates to a form

sin

cos

y or equivalent. Indices do not need to be processed for this mark.

p

y

e.g.

...sin

2 1

−

cos

y

y

d

x

May use the identity

cos2

y

= ±±

1 2sin

2

y

x

=

4sin

2

y

−= −

1 1 2cos2

y

=

q

d

y

Correct lhs and rhs. For d

or 4sin 2y . Note that d

x

y

A1:

y =

8sin

y

cos

y

x =

...

is A0. isw for this mark

d

d

after a correct answer is seen. You may see d

x

y (or allow x′) on an earlier line which is fine for

d

this mark. May be seen at the start of (b)(ii). Indices must be processed for this mark.

Note this is a given answer so full working must be seen.

M1: Attempts to find either sin y (or

sin y ) or cos y (or

2

cos y ) in terms of x (or possibly multiples of

2

±±

1

±±

1

x

x

these) Typically look for

sin

2

y

=

or

sin

y

=

Also allow e.g.

4sin

2

4

4

x

±

May attempt to use

±

sin

2

±

cos

2

= ±

1

cos

2

= ±±

1 sin

2

cos

2

= ±±

1

y

y

y

y

y

4

x

±

1

for this mark. (may be seen within their expression for d

cos

=

±±

1

Ignore

±

...

y

4

dM1: Attempts to find both sin y (or

sin y ) and cos y (or

2

cos y ) or possibly multiples of these in terms of

2

x. (See previous method mark for guidance) It is dependent on the previous method mark.

1

y o.e. in terms of x e.g. d

y

ddM1: Full method to get d

y

x of the form

=

d

sin

cos

d

p

y

x

±±

x

"8"

Condone square root notation which does not fully go over any fraction provided it is implied by further work

which is not the given answer. Condone missing arguments for sin and cos provided the intention is clear or

implied. Condone missing brackets to be recovered before achieving the given answer. If they have

rearranged or simplified incorrectly forcos y before substituting, ddM1 can still be scored.

A1*: Achieves the given answer provided all previous method marks have been scored in (b) and no

errors. Look for

cos y or cos y in x which has not already been simplified to ... 3

2

2

sin y or sin y in x which has not already been simplified to ...

d

y

x (or y′) may appear on an earlier line which is acceptable. Note that if they work backwards by

d

into the given answer there must be a conclusion that d

y

substituting

4sin

2

=

x

=

y

d

x

(c)

M1: Full method for the gradient of the normal using the x or y value. Condone arithmetical slips but the

gradient must be the negative reciprocal of the gradient of the tangent to the curve.

Note that they may use an earlier incorrect line in part (b) for d

x (or even d

x

y

y ).

d

d

Allow use of an earlier line provided it is an attempt at a correct method substituting in x or y and

proceeding from e.g. d

x

=

α

gradient of the normal

= −

α

d

y

May be implied by awrt 3.5

If no method is shown where the gradient of the tangent or normal has

−

come from, then it must be correct.

dM1: Attempts to find the x coordinate of N. It is dependent on the previous method mark. Uses their

π

gradient of the normal with 2, 3

, sets

y =

0

in

−

=

(

−

)

and proceeds to find x. May also

y

y

m x

x

1

1

use y

+ to find the equation for the straight line and then sets

y =

0

and rearranges to find x.

=

mx

c

Do not be concerned by the mechanics of the rearrangement for this mark.

2 3

2

π

π

π

+ π

A1:

+

or exact equivalent such as

which is in the form

π

+

π

2

a

b

3

36 3

3

108

Pearson Education Limited. Registered company number 872828

with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom