Question 16:
--- 16(a) ---
16(a) | Chooses an appropriate
technique to differentiate accept
any evidence of product rule or
quotient rule | 3.1a | M1 | dy
=−e−x(sinx+cosx)+e−x(cosx−sinx)
dx
=−2e−xsinx
Differentiates fully correctly | 1.1b | A1
Obtains fully correct simplified
answer. | 1.1b | A1
Uses their result from (a) in the
Be−xsinx
form of
showing an understanding of the
fundamental theorem of calculus
Condone missing constant. | 3.1a | M1 | ∫( −2e−xsinx ) dx=e−x(sinx+cosx)+k
∴∫( e−xsinx ) dx=− 1 e−x(sinx+cosx)+c
2
1
Obtains e−x( sinx+cosx )
B | 2.1 | R1F
∫ π ( e−xsinx ) dx=− 1 e−x(sinx+cosx) π
 
2 0
0
1
=− [e−π(sinπ+cosπ)
2
−e0(sin0+cos0)]
e−π+1
=
2
Deduces correct limits and
substitutes correctly | 2.2a | A1
Obtains correct exact value from
correct answer in part(b)
CSO | 1.1b | A1
--- 16
(c)(ii) ---
16
(c)(ii) | Substitutes correct limits for A
2
Into their 1   e−x( sinx+cosx )  π
B 0
Or
Writes A =±∫ 2π( e−xsinx ) dx
2
π
and uses the substitution
u = x−π | 1.1a | M1 | ∫ 2π ( e−xsinx ) dx=− 1 e−x(sinx+cosx)2π
 
2 π
π
e−π+1
=− e−π
2
e−π +1
e−π
Area =
2
e−π +1
e−π
A
2 = 2 =e−π
A e−π +1
1
2
Obtains correct exact area
for ±A
2
CSO
Or
Makes complete substitution
A =±∫ π( e −(u+π) sin ( u+π ) ) du
2
0 | 1.1b | A1
Forms required ratio using their
exact A1 and A2, may be
unsimplified
Or
e−π
Extracts factor of and uses
sin ( u+π )=−sinu
π
To obtain A =−e−π∫ e−usinudu
2
0 | 1.1a | M1
Completes rigorous argument,
with correct limits and negatives
handled correctly
CSO | 2.1 | R1
--- 16
(c)(iii) ---
16
(c)(iii) | Deduces that the areas form a
geometric series
Accept any indication of this
being geometric series | 2.2a | B1 | a e−π +1 1
= ×
1−r 2 1−e−π
1+eπ
=
( )
2 eπ −1
A
Uses 1
1−e−π | 3.1a | M1
Obtains a value for the geometric
series first term using their (c)(i) | 1.1a | B1F
Completes rigorous argument to
achieve required result in correct
form.
CSO
AG | 2.1 | R1
Total | 16
PPMMTT
MARK SCHEME – A-LEVEL MATHEMATICS – 7357/1 – JUNE 2019
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