Mathematics · Oct / Nov 2020 · Paper 31 · Question 8

g ( )

′

x

=

3e

sec2

x

+

2e

sec2 tan 2

x

x

′

3e

x

(

)

g ( )

x

=

0

⇒

sec2

x

3

+

2tan 2

x

=

0

tan 2

x

1.5

x

0.491

= −

⇒

= −

d

x

1

x

=

ln(sin )

y

⇒

=

×

cos

y

d

y

sin

y

or d

y

sin

y

2

2

x

Attempts

cos

y

=

1 sin

−

y

=

1 e

−

=

d

x

cos

y

x

d

y

sin

y

e

Hence

=

=

d

x

cos

y

2

x

1 e

−

(i)(a)

3

x

3

x

M1

Correct attempt at the product rule

g ( )

′

x

=

A

e

sec2

x

+

B

e

sec2 tan 2

x

x

A

cos2

x

e x

3

−

B

e

3

x

sin 2

x

For use of the quotient rule look for

cos

2

2

x

3

x

3

x

A1

g ( )

′

x

=

3e

sec2

x

+

2e

sec2 tan 2

x

x

Allow in any form and then isw

3cos2

x

e

3

x

+2

e

3

x

sin 2

x

For use of the quotient rule look for

cos 2

2

x

(i)(b)

Sets g ( )

′

x

=

0

⇒ and takes out / factorises out

3e

x

sec2

x to identify a linear factor in tan2x

M1

e

3

x

For the quotient rule they should be factorising out

to leave a linear factor in cos2x and sin 2x

cos 2

2

x

dM1

Correct order of operations to x = ...

sin 2

x

tan 2

x

=

For quotient rule approach they must use

(oe correct work)

cos2

x

You may need to check their answer if no method is shown. Accept awrt 2.s.f. for their tan 2x = …. in either

radians or degrees.

A1

x

=

awrt

−

0.491

only in the range. If extra solutions arise from trying to solve sec2x = 0 then A0.

(ii)

d

x

1

d

y

x

B1

=

×

cos

y

OR

e

=

cos

y x

via e

x

=

sin

y

d

y

sin

y

d

M1

For one of the two operations needed to complete the proof

x look for an attempt using

Either an attempt to get cos y in terms of e

sin

2

y

+

cos

2

y

= with sin y being

1

replaced by ex . Alternatively allow use of arcsin

Or taking the reciprocal and making d

y

x the subject (variables must be consistent)

d

Applies both operations to obtain d

y

x

dM1

x in terms of just e

d

x

d

y

e

e

x

1

A1

x =

or

Allow

or states

f ( )

x =

1 e x

−

2

following a correct

(

)

d

cos a c

r

sin

e

x

2

x

(

)

2

1 e

−

1

−

ex

d

y

e

x

expression for

=

or similar.

d

x

cos

y

Alt:

(

)

B1

x

=

ln(si

n )

y

⇒

y =

arcsin

e

x

d

y

1

d

y

1

M1

⇒

=

×

.

..

or

⇒

=

×

e

x

d

x

d

x

(

)

2

1 (. .

−

.

)

2

1

−

ex

dM1

both of these

e

x

A1

(

)

2

1

−

e

x