Question 4:
4 | (a) | 00=2⇒k =2
3
ae=k a−k e+k =afor any a
1 2 3
k = 1
1
ea=ke−k a+k =a for any a
1 2 3
k = 1, k = –1, k = 2 so ab=a+b+2AG
1 2 3 | B1
M1
A1
M1
A1
[5] | 3.1a
3.1a
1.1
3.1a
1.1 | Using identity correctly, one
order of a/e, general or specific a.
Comparing coefficients (oe) to
derive the value of k or k .
1 2
Using identity correctly, other
order of a/e, general or specific a.
AG. Solving system of equations
and concluding correctly | eg0e=−k e+k =0 but not e◦e
2 3
(may be seen later)
ege0=ke+k =0 but not e◦e
1 3
Do not award credit for work done
based on assumptions (eg e = –2 or ◦
is associative).
4 | (b) | So the identity element, e, is –2 | B1
[1] | 2.2a
4 | (c) | Because ba=b+a+2=a+b+2=ab, the
operationis commutative over A | B1
[1] | 2.1 | Or stating that a + b + 2 is
symmetrical in a and b. | Simply giving one or more
examples is insufficient for B1.
[6]
4 | (e) | (i) | If ℤ then the arguments in (d) still apply
so there is no change
𝐴𝐴= | B1
[1] | 2.2a | From correct conclusion in 4(d)
only.
4 | (e) | (ii) | If then the arguments in (d)
still apply so there is no change
𝐴𝐴={2𝑚𝑚:𝑚𝑚∈ℤ} | B1
[1] | 2.2a | From correct conclusion in 4(d)
only.
4 | (e) | (iii) | If then the inverse
property is not satisfied (eg 2–1 = –6 ∉A) so
𝐴𝐴={𝑛𝑛:𝑛𝑛 ∈ℤ,𝑛𝑛 ≥−2}
(A,)is not a group | B1
[1] | 2.2a | From correct conclusion in 4(d)
only.
An example does not need to be
given but if given must be
correct. Ignore other irrelevant
comments but if incorrect
statement about any of the other 3
axioms then B0. | Condone slightly incorrect
statements (eg “No element in A has
an inverse”) provided that it is
established that only the inverse
property is not satisfied.