Mathematics · May / June 2024 · Paper 31 · Question 5

(

)

(

)

6

2

6 2

+

1

−

2 6

−

2

x

−

x

x

g( )

x

=

g ( )

′

x

=

2

1

x

+

(

)

2

2

x

+

1

10

States (

)

g ( )

′

x

=

>

0

hence increasing *

(

)

2

2

x

+

1

6

−

2

x

y

=

2

xy

+

y

=

6

x

−

2

2

xy

−

6

x

= −−

y

2

2

1

x

+

−−

2

−−

x

2

y

−

1

g

( )

x

=

So

x

=

o.e.

2

−

6

2

−

6

y

x

Domain 0

<

x

<

3

Range fg is fg

<

2

+

5ln3

(a)

M1:

Sets 2

and rearranges to ln

(may be implied by further work)

+

5ln

x

=

22

x =

...

Alternatively, may attempt to find

f

−

1

( )x

and substitute 22 into the expression. i.e.

Maybe implied by awrt 55.

Condone sign slips only in their rearrangement.

4e (must be simplified and exact) isw once a correct answer is seen.

A1:

(b)

M1:

Either

(

)

(

)

A

2

x

+

1

−

B

6

x

−

2

o.e. where A and

(

2

+

1

)

2

x

B are both positive. Do not award this mark if the quotient rule is applied the wrong way round.

1

2

1

(6

2)(2

1)

−

(6

2)(2

1)

−

(2

1)

−

x

−

x

+

±

C

x

−

x

+

±

D

x

+

5

3

−

=

3

−

5(2

+

1)

−

1

and attempts to use the chain rule proceeding to the

x

2

+

1

x

form

(2

+

1)

−

2

E

x

In all methods invisible brackets can be recovered or implied by further work.

Correct unsimplified e.g. (

)

(

)

6 2

+

1

−

2 6

−

2

x

x

−

2

−

1

A1:

or

−

2(6

x

−

2)(2

x

+

1)

+

6(2

x

+

1)

(

)

2

2

x

+

1

g ( )

′

=

x

10

′

e.g. writes g ( )x

as

and deduces > 0 with minimal conclusion e.g. “it is an increasing

(

2

1

)

2

x +

function” or “hence increasing” Condone use of

They may consider just the denominator separately, but they would still need to simplify the

numerator to 10 and state e.g. (

)

2

2

x +

1

>

0

o.e. with a minimal conclusion

Do not accept the conclusion g ( )x

′

is an increasing function if stated instead of g( )x .

−

2

−

1

Note that they cannot just state e.g. “

−

2(6

−

2)(2

+

1)

+

6(2

+

1)

>

0

so increasing” as it is not

x

x

x

clear from the expression that this will be the case.

Do not withhold this mark if invisible brackets are recovered but A0* if there are errors in the

method.

(c)

M1:

Attempts to change the subject. Look for cross multiplication with an attempt to collect terms

achieving the form 2

±

±

6

= ±±

2

or 2

±

±

6

= ±±

2

. May be implied by further work.

xy

x

y

xy

y

x

β

β

If they start from their

=

α

±

+ look for an alternative rearrangement to

α=

y

2

1

±

x

y

β

α=

2

y

+

1

)

x

±

−−

2

+

2

x

x

−

1

( )

−

1

( )

−

1

( )

A1:

g

x

=

or equivalent such as

g

x

=

or may be seen as

g

x

=

2

x

−

6

6

−

2

x

5

−

1

( )

g

x

=

−

0.5

6

−

2

x

2

x

+

g

1

...

g

1

...

=

Accept

g

−

1

:

x

→

Condone

−=

or

−=

y

=

but do not allow just

y

6

−

2

x

B1:

Correct domain 0

o.e. e.g. 0

or ]0, 3[ but not 0

< or

<

x

<

3

<

x

∩

x

<

3

<

y

3

0

<

t

+

2

Note: It is also acceptable to define

g− in any variable e.g. as

1

g ( )

−

1

=

, 0

<<

t

t

6

−

2

t

the variable is used consistently to score M1A1B1. If another variable is used other than x it must

be fully defined e.g.

g ( )

−

1

=

...

not just

g

−=

1

...

t

(d)

M1:

2

+

5ln3

found. Allow awrt 7.5 instead of 2

+

5ln3

for this mark only. Do not be concerned by the

direction of the inequality if seen.

Fully correct. (

) fg

A1:

−∞<

<

2

+

5ln3

o.e. 5ln3 must be exact. Do not be too concerned by

alternative notation to fg as long as it is not x. There are many alternatives including

or (

)

−∞

, 2

+

5ln3

but not

−∞

, 2

+

5ln3

as requires a strict inequality at the upper bound. Do not

allow f

2

5ln3

or g

2

5ln3

<

+

<

+

d

1

y

1

(

)

(

)

−

=

4

−

7

=

2 4

−

7

(see notes)