Mathematics · Oct / Nov 2022 · Paper 31 · Question 6
YMA01
As printed in the paper
The question as text
6. y = 2 + 3 sin x cos x + sin x Show that d y x = a x + b x + c tan sec d sec x + 2 sin x where a, b and c are integers to be found. (6)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
(
)
x
+
si
n
x
=
−
s n
i
x
+
cos
x
d
x
(
)(
) (
)(
)
d
y
cos
x
+
sin
x
3cos
x
−−
sin
x
+
cos
x
2
+
3sin
x
=
(
)
d
x
cos
x
+
sin
x
2
2
2
3cos
x
+
3sin cos
x
x
+
2sin
x
+
3sin
x
−
2cos
x
−
3sin cos
x
x
=
cos
2
x
+
2sin cos
x
x
+
sin
2
x
3
+
2sin
x
−
2cos
x
=
1 2sin cos
+
x
x
3
+
2sin
x
−
2cos
x
sec
x
3sec
x
+
2sec sin
x
x
−
2
=
×
=
1 2s
+
i
n co
x
s
x
se
c
x
s c
e
x
+
2s n
i
x
2tan
3sec
2
x
+
x
−
=
sec
x
+
2sin
x
Notes:
B1: Correct differentiation of
c
o
s
x
+
si
n
x →
−
si
n
x
+
c
o
s
x
seen somewhere in the proof. This can be scored
if seen in workings for quotient rule, or even in the denominator of an incorrect attempt at u’/v’.
M1: Differentiates using the quotient rule or product rule. For quotient rule look for
(
)(
) (
)(
)
cos
x
+
sin
x
±
a
cos
x
−
±
sin
x
±
cos
x
2
+
3sin
x
(
)
cos
x
+
sin
x
2
For product rule look for (
) (
1
) (
) (
2
)(
)
cos
x
+
sin
x
−
±
a
cos
x
±
cos
x
+
s
in
x
−
±
c
os
x
±
s n
i
x
2
+
3sin
x
A1: Fully correct derivative.
M1: Expands numerator or denominator and applies
= or other appropriate correct
sin
2
x
+
cos
2
x
1
Pythagorean identity at least once in the proof.
M1: Attempts to multiply through by sec x in numerator and denominator (to achieve the 2sin x term). Not
dependent and may be scored before the previous M.
A1: Correct answer. Terms may be in different order. Allow minor slips in notation (e.g. a single missing x)
and recovery from missing brackets if the intent is clear, but A0 for persistent incorrect notation throughout
(e.g no x’s in the trig terms).
(i) 22
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