Mathematics · May / June 2021 · Paper 21 · Question 4
YMA01
As printed in the paper
Show the answer and mark scheme
4(a)[4]
x
2
+
=
8192
+
...
8
13
1
13
2
13
3
x
x
x
+
2
12
+
2
11
+
2
10
+
...
1
8
2
8
3
8
13
x
(
)
2
+
=
8192
+
6656
x
+
2496
x
2
+
572
x
3
+
...
8
x
4(b)[3]
=
0.012
5
⇒
x
=
0
.
1
8
13
2.0125
≈
8192
+
665.6
+
24.96
+
0.572
=
8883.132
cao
As all the terms in the expansion are positive, the truncated series will give
4(c)[1]
an underestimate of the actual value.
Notes:
(a)
13
B1: For the correct constant term of 8192 in their expression. Do not accept
2
M1: Correct binomial coefficients linked with correct powers of x at least twice in the x, x2 and x 3
terms in their expansion. Accept alternative notation, such as 13
C or just 13 etc, for the coefficients.
1
(The power 1 need not be seen.) The powers of 2 and the 8’s may be missing or incorrect.
A1: Any two of the final three terms correct. (See also note below.)
A1: All three of the final terms correct.
Note: Students may attempt to take out a common factor of
2 first. In such cases the B mark is for
13
correct constant term in their final expansion, M follows as above, correct coefficients linked to
13
39
correct powers of x. The first A can be awarded as above or SC
8192 1
+
x
+
x
2
+
16
128
2048
(oe), final A as scheme
(b)
B1: Identifies x = 0.1 needs to be substituted into the expression.
M1: Substitutes their x = 0.1 into their answer to (a)
A1: Correct answer only, must be to 3 dp.
(c)
B1: States or implies underestimate with correct explanation of why the answer found is an
underestimate. Must refer to terms in the expansion being positive or equivalent wording. Do not
accept just “as adding more terms” without reference to the terms being nonnegative.
If all that is done is a comparison with the true value, then B0.
1
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