Mathematics · May / June 2025 · Paper 41 · Question 2

→

2

→

×

B1:

2

y

4 d

y x

o.e. For example, accept

2

y

2

2 d

y x

d

y

M1:

Attempts the product rule on 6xy

−

. Look for

±

α

x

±

β

y

d

x

Don’t be concerned by the signs here unless they state and use an incorrect product rule

d

y

d

y

A1:

Fully correct differentiation

4

y

−

6

x

−

6

y

=

14e

2

x

−

1

o.e. Allow unsimplified versions

d

x

d

x

4 d

y y

−

6 d

x y

−

6

y x

d

=

14e

2

x

−

1

d

x

4

yy

' 6

−

xy

' 6

−

y

=

14e x

2

−

1

You may see versions like

or

d

y

d

y

d

y

this will be A0 unless the d

=

−

−

−

2

x

−

1

=

acceptable. If they write

4

y

6

x

6

y

14e

0

d

x

d

x

d

x

subsequently disappears and is ignored.

1

M1:

Attempts to substitute in

x =

with their positive y value to find a value for the gradient of the tangent. It

2

is dependent upon having exactly two terms in d

y

x one each from the differentiation of

2y and 6xy . This

d

can be implied by a correct value for their d

y

x . Allow slips if they attempt to make d

y

x the subject before

d

d

1

d

y

6

y

+

14e

2

x

−

1

d

y

24

+

14

x =

=

→

=

=

substituting in

with their positive y value. FYI

2

d

x

4

y

−

6

x

d

x

16

−

3

Uses the value for their d

y

x with the point 1 ,

M1:

where a is their positive y coordinate to find the

2 a

d

equation of the tangent. If they have two positive y values, use of either one would be OK for this mark. It

must follow some attempt at differentiating the equation for C. If the form y

=

mx

+

c

is used the method

must proceed as far as c = …

A1:

38

x

−

13

y

+

33

=

0

(or any integer multiple of this). The order of the terms is not important

2

d

V

t = ±

k

d

4

d

V

=

π

→

=

π

V

r

3

4

r

2

3

d

r

d

r

±

k

=

λ

r

2

×

d

t

d

r

±

k

d

r

1

=

∝

✓ *

π

d

t

4

r

2

d

t

r

2

States or uses d

V

or d

V

t =

t = − (We don’t know if k is positive or negative).

B1:

k

k

d

d

Ignore any reference to units even if they are incorrect. E.g. d

V

t =

k

cm / sec

3

d

d

V

=

π

2

B1:

States or uses

4

r

o.e.

d

r

Uses a correct chain rule e.g. d

V

d

V

d

r

or equivalent with d

V

d

V

M1:

=

×

t = ±and

k

r = λ

r

2

d

t

d

r

d

t

d

d

A1*: Completes the proof by looking for the following three elements

d

r

±

k

= π

d

t

4

r

2

d

r

1

∝

d

t

r

2

All previous marks must have been awarded

Examples of acceptable reasons and acceptable conclusions are;

d

r

k

c

E.g. I

= −

= −

, hence proven (Note that c cannot be k)

d

t

4

π

r

2

r

2

d

r

k

k

d

r

1

=

,as

∝

E.g. II

is a constant

π

2

π

2

d

t

4

r

4

d

t

r

d

r

k

= −π

Eg. III

as k and π are constants, it is true

d

t

4

r

2

d

r

k

k

1

k

1

E.g. IV

= −

= −

×

or

−

×

✓

d

t

4

π

r

2

4

π

r

2

π

2

4

r

Reason

E.g.V.

Conclusion

R