Question 4:
4 | (a) | (i) | 1071 ÷ 67 = 1566 so q = 15 and r = 66
67 | B1
[1] | 1.1 | BC
(ii) | 1071 ≡ 66 (mod 67) | B1
[1] | 1.1 | FT (a) (i)’s answer
(b) | Suppose hcf(715, 1071) = h, so h | 715 and h | 1071
⇒ h | 715a + 1071b
In this case, a = 358, b = –239 give h | 1
⇒ h = 1 and 715, 1071 are co-prime | M1
M1
A1
B1
[4] | 3.1a
2.1
1.1
2.2a | (for all integers a, b)
Must show understanding of conclusion relative to question’s
demand. (Condone correct conclusion of primality from h = ±1.)
(a) | ∂z 0.096y ∂z 0.096
= 0.9 – − 2xy2 = – 2x2y
∂x x2 ∂y x
∂z ∂z
Setting both = 0 and = 0 and expressing
∂x ∂y
(say) y in terms of x from the second of these
2
∂z 0.0960.048 0.048
In = 0, 0.9−  −2x  =0
∂x x2  x3   x3 
⇒ x5 = 0.01024 ⇒ x = 0.4
⇒ y = 0.75
and Cost = £4500 | M1 A1
M1
M1
A1
A1
A1
[7] | 1.1 1.1
1.1
3.1a
1.1
1.1
3.2a | Attempt at both first p.d.s; at least one correct
Must include an attempt to do something with at least one of the
∂z 0.048
two equations: = 0 ⇒ y = *
∂y x3
Substituting into the other equation and solving attempt
FT from 1st variable found in clearly s.o.i. relationship (e.g.*)
CAO From correct z = 0.45
∂2z 0.192y ∂2z
= − 2y2 and =− 2x2
∂x2 x3 ∂y2
= 1.125 = −0.32
Change in y will decrease z since 2nd-p.d. < 0 | M1
A1 A1
B1 | 3.3
3.4 1.1
3.5a | Attempt at the two second partial differentials
Correct value of z & value or sign statement for a suitable z
xx yy
Correct answer for correct reason
(FT + and − if values are not offered)
ve ve
NB z > 0 ⇒ the SV is a min. in the x-direction, while
xx
z < 0 ⇒ the SV is a max. in the y-direction
yy
Alternative method | M1
A1 | Adjusting values of x with y fixed; then y with x fixed
Correct 1st set of results
When with y = 0.75, x = 0.39, 0.4, 0.41
z = 0.450 059, 0.45, 0.450 054
and when x = 0.4, y = 0.74, 0.75, 0.76 | A1 | Correct 2nd set of results
z = 0.449 984, 0.45, 0.449 984
Change in y will decrease z | B1 | Correct answer for correct reason
When with y = 0.75, x = 0.35, 0.4, 0.45
z = 0.451 808, 0.45, 0.451 094
and when x = 0.4, y = 0.7, 0.75, 0.8
z = 0.449 6, 0.45, 0.449 6
M1
A1
Adjusting values of x with y fixed; then y with x fixed
Correct 1st set of results
(a) | (i) | B1
B1
B1
[3] | 1.1
1.1
1.1 | R and C correct
3 3
Any other row or column correct
All correct
(ii) | Closed, since no new elements in the table
Since ◯ involves only complex multiplication, which
is known to be Associative, this axiom holds
1−i 3 is the identity (see the Cayley table)
−1+i 3 is self-inverse
( )
and ± 3 +i form an inverse-pair | B1
B1
B1
B1
[4] | 2.4
2.1
1.1
2.5 | Don’t accept unqualified statements like “Table shows …”
Or showing, either separately or together, that
( )
2
(z1 ◯ z2) ◯ z3 = 1 1+i 3 z z z = z1 ◯ (z2 ◯ z3)
16 1 2 3
All (non-identity) elements shown to have an inverse (and not
just stated that this is so)
(iii) | 1−i 3 has order 1
−1+i 3 has order 2
( )
± 3 +i have order 4 | B1
[1] | 1.1
(b) | {1−i 3, −1+i 3} | B1
[1] | 2.2a
(i) | It has an (i.e. at least one) element of order 4 | B1
[1] | 2..2a | Or any equivalent, longer reason
[1]