Mathematics · May / June 2021 · Paper 41 · Question 5

d

x

d

x

)

−

2

x

(

d

y

d

y

2 e

y

+

3

−

2

x

−

2

x

e

−

2

y

=

2 e

y

+

3

⇒

=

*

d

x

d

x

−

2

x

e

−

2

y

2

Puts

x =

0

into the equation of the curve

⇒

y

=

y

⇒

y

=

1

Attempts tangent at (

0,0 or (

)

)

0,1

y

=

3

x

or

y

= −

5

x

+

1

1 3

y

5

x

1

R

,

Solves

y

=

3

x

with

= −

+ ⇒

=

8 8

(a) Originally scored M1 B1 A1 A1, now B1 M1 A1 A1

d

y

2

B1: Correct differentiation using the chain rule

y

→

2 d

y x

.

d

y

2

−

2

x

−

2

x

M1: Attempts to apply the product rule of differentiation on

−to give

x

e

±

... e

y

y

e

d

x

d

y

d

y

−

2

x

−

2

x

A1: Correct differentiation

2

y

=

e

−

2 e

y

−

3

d

d

x

x

−

2

x

−

2

x

Allow

2 d

y y

e

d

y

2 e

y

d

x

3d

x

=

−

−

A1*: Proceeds to the given answer via an intermediate line equivalent to (

)

d

y

−

2

x

e

−

2

y

=

2 e

y

d

x

correct bracketing.

(b)

B1: Deduces or implies that

x

=

0,

y

= at P . May state or use coordinates of P (0,1)

1

M1: Scored for an attempt to find the equation of the tangent at O or the equation of the tangent at P

−

2

x

d

y

2 e

y

+

3

E.g. Substitutes (

)

0,0 into

=

=

β and states y

=

βx

d

x

−

2

x

e

−

2

y

−

2

x

d

y

2 e

y

+

3

Alternatively substitutes (

)

0,''1'' into

=

→

'' ''

k

and attempts

y

−=

1

'' ''

k x

or equivalent

d

x

−

2

x

e

−

2

y

A1: Achieves a correct equation for either tangent. Look for either

y

=

3

x

or

y

= −

5

x

+ o.e.

1

dM1: Correct attempt at both tangents with an attempt to solve simultaneously.

For the attempt to solve accept

y

=

3

x

,

y

= −

5

x

+ ⇒

1

x

=

...,

y

=

...

1 3

A1: Correct coordinates for

R

=

,

oe

8 8

∫

d d

x

∫