Mathematics · Oct / Nov 2020 · Paper 31 · Question 6
YMA01
As printed in the paper
The question as text
6.. y P C1 β O α x C2 Figure 3 Figure 3 shows a sketch of curve C1 with equation y = 5e x−1 + 3 and curve C2 with equation y = 10 − x2 The point P lies on C1 and has y coordinate 18
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
5e
−
+
3
=
18
⇒
e
−
=
3
x =
⇒
x
=
ln3 1
+ or e
3e
⇒
x
=
ln3e
x
−+
1
2
Sets
5e
3
=
10
−
x
and proceeds to find and use a suitable function. Eg
6(b)[3]
(
)
2
x
−
1
f ( )
x
=
7
−
x
−
5e
Attempts f(1.1335)
0.001
and f(1.1345)
0.007
=
= −
Correct values with reason(change of sign and continuous) and conclusion, hence
α is 1.134to 3dp
6(c)[3]
x
= −
7
−
5e
−−
3 1
= −
2.628388
2
β
= −
2.620330
(a)
x =
x−+
1
x−=
1
M1
For attempting to proceed from
5e
3
=
18
to
e
...
or e
...
x =
A1
x =
ln3 1
+ or for e
3e
1
x =
ln3e
x =
ln3e
A1
Accept
(b)
B1
Sets the equations equal to each other and finds a suitable function which is then used.
(
)
2
x
−
1
Suitable functions are
f ( )
x
= ±
7
−
x
−
5e
x
−
1
,
g( )
x
= ±
x
−
7
−
5e
or
h( )
x
= ±
x
−−
x =
1.1335
x =
1.1345
M1
Substitutes
and
, or suitable values for a tighter interval (either side of 1.133634..), into a
suitable function and obtains one correct value to 1sf rounded or truncated.
FYI for the + options above
f(1.1335)
=
0.001
and f(1.1345)
= −
0.007
g(1.1335)
= −
0.0005
and g(1.1345)
=
0.003
h(1.1335)
0.0002
and g(1.1345)
0.001
= −
=
A1
Requires both values to be correct (1sf rounded or truncated), a reason (change of sign and continuous)
and a minimal conclusion
(c)
For an attempt at substituting 3
− into the iterative equation.
M1
This is implied by the sight of the expression or awrt 2.63
−
A1
awrt 2.628388
−
β =
awrt
−
2.620330
A1
condone −2.62033
.............................................................................................................................................................................
It is possible in part (b) to score B0 M1 A1 by comparing y values for
y
=
5e
x
−
1
+
3
and
y
=
10
−
x
2
at
x =
1.1345
For the A1 apply the similar criteria as for the main scheme with values to 3d.p..
x =
1.1335
y
=
5e
x
−
1
+
3
=
8.714
y
==
10
−
x
2
=
8.715
At
,
and
1.1335
1.1335
x
−
1
At
x =
1.1345
,
y
=
5e
+
3
=
8.720
and
y
=
10
−
x
2
=
8.713
1.1345
1.1345
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