Mathematics · Oct / Nov 2022 · Paper 31 · Question 4

10

10

y

1

−

x =

2

d

x

110 ln10

=

y

d

y

2

d

y

d

x

1

x =

1

=

d

d

y

1

10 ln10

y

2

d

y

2

=

(

)

d

x

2

x

+

1 ln10

Notes:

(a)

M1: Correctly undoes the logarithm and attempts to rearrange to make x the subject.

A1: Correct expression for x. Apply isw after a correct answer.

(b)

M1: Attempts to differentiate 10y . Accept

α

10

y

→

β

10

y

ln1

0

(ignore any extra terms). Do not be concerned

about the lhs for this mark.

M1: Applies reciprocal d

y

d

x

x =

1 d

. Variables must be consistent.

d

y

A1: Correct answer. Accept equivalents in terms of x, e.g. with

10

log(2

x+ isw after a correct answer.

1)

Alt for (b): Allow use of change of base formula.

(

)

ln 2

x

+

1

M1:

y

=

(effectively a B mark via this method)

ln10

d

y

1

1

M1:

=

×

×

.

.

.

Attempts to differentiate using the chain rule.

d

x

ln10

2

x

+

1

A1: Correct answer.

d

y

1

d

y

1

Note: Candidates who write

..

.

or

.

..

with no ln 10 term can score M0M1A0 under

=

×

=

×

d

x

2

x

1

d

x

10

y

+

the Alt.

Alt 2: Implicit Differentiation

d

y

M1: As main scheme, attempts to differentiate as part of their work, so

α

10

y

→

β

10

y

ln10 d

x

M1: Makes d

y

x the subject. A1: Correct answer.

d

4 1

−

3

2