Mathematics · Oct / Nov 2025 · Paper 21 · Question 3
YMA01
As printed in the paper
The question as text
3. y x O Figure 1 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 1 shows a sketch of the curve with equation y = 3 × 2–x The point P(k, 300 000) lies on the curve.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
300000
= ×
3 2
2
=
100000
−=
k
log2
( )
k = −
16.61
3(b)(i)[3]
Strip width = 1.5
1.5 4.243
+
+
+
+
+
= 6.63
0.023
2(1.5
0.530
0.188
0.066)
2
7
−
0.5
3(b)(ii)[2]
= 1
1
−
x
+
x
2
d
x
2 d
x
×
''6.63''
+
×
''6.63''
=
4.42
3
3
−
0.5
7
(a)
M1: Correct use of logs leading to an expression or value for x or x
−.
Alternatively applies the correct use of logs leading to an expression for k or k
There are lots of variations, examples of which include
log100000
1
ln100000
(
)
, (
)
and (
)
±
k
=
±
x
=
log 100000
±
x
=
2
ln 2
log2
−
k
300000
=
3 2
×
log300000
=
log3
−
k
log2
k
=
...
Condone a slip on the number of zeros (4 or 6 zeros can still score this method
mark)
It cannot be scored if they incorrectly combine the 3 and the 2 and work from
6
−=
k
300 000
A1: AWRT −16.61 following M1. ISW after a correct answer
Note that the answer only is M0, A0. The demand is that the candidates use
logs
B1: Correct strip width. E.g. States or uses h = 1.5. Condone it being unprocessed.
E.g. 7
−−
( 0.5)
o.e.
5
Can be implied by a correct h value within an attempted trapezium rule which
does not need to be perfect. A correct application would be
1.5 4.243
+
+
+
+
+
or
0.023
2(1.5
0.530
0.188
0.066)
2
+
+
+
+
+
0.75 4.243
0.023
2(1.5
0.530
0.188
0.066)
M1:
Correct application of the trapezium rule
'' '' 4.243
h
+
+
+
+
+
0.023
2(1.5
0.530
0.188
0.066)
2
Condone slips on values but there must be the correct number of them and
they must be in the correct place within the formula.
Condone a missing trailing bracket, but hard to decipher bracketing can be
awarded this mark only if followed by awrt 6.6. Answers only score M0
A1:
awrt 6.63 (Answer in full is 6.6255). ISW after a correct answer
Note that the calculator answer for this integral is 6.087
(b)(ii) Attempts to (b)(ii) must use their answer to (b)(i) but could use an
−
0.5
x
additional attempt at the trapezium rule on
with h = 1.5 or smaller
2 d
x
7
Attempts to find 2
or alternatively 1
1
M1:
3 ×
×
+
×
following
''6.63''
''6.63''
''6.63''
3
3
through on their 6.63
Note that 1
3 ×
+
=4.24 is M0 as 2.03 is the calculator answer for
''6.63'' 2.03
0.5
−
x
2 d
x
7
A1:
4.42. Allow 4.417 or awrt 4.42 if the full answer is used.
Note that the calculator answer for the sum of the two integrals is 4.058...
=
3
+
2
−
+
f( )
x
4
x
13
x
10
x
8
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