Question 1:
1 | a | z
z = 1z−1.−6x = 1 z−1.−4y
x 2 y 2
 z 2 y
z 3x = − = 1 63
=− =−11 and
x z  y z | M1
A1
M1
A1 | 1.1
1.1
1.1
1.1 | Attempt at partial differentiation (at least one case)
𝑎𝑥 ×𝑧−1 or 𝑏𝑦×𝑧−1
At least one correct (any form)
Substituting values into their partial derivatives to get
numerical “gradients” (can be implied by a correct
answer)
Both correct
Alternative method
 z
2 z  z = − 6 x 2 z = − 4 y
 x  y | M1
A1 | Squaring and use of implicit differentiation (at least
one case)
At least one correct (any form)
 z 2 y
z 3x = − = 1 63
=− =−11 and
x z  y z | M1
A1 | Substituting values to get numerical “gradients”
Both correct
[4]
b | z – 3 = − 1 1 (x – 11) + 1 63 (y + 8)
 33x – 16y + 3z = 500 | M1
A1
[2] | 1.1
1.1 | Eqn. for tangent plane used
Must have numerical “gradients” involved
CAO (any non-zero integer multiple)
a | If h | 7n + 4 and h | 8n + 5 then h | 7(8n + 5) – 8(7n + 4)
i.e. h | 3
 h = 1 or 3 | M1
A1
B1 | 2.1
1.1
2.2a | A linear combination of the two numbers attempted
Correct from choice of 7, −8 (or −7, 8)
inclusion of either/both of h = −1, −3 → B0
Alternative method
hcf(7n+4, 8n+5)=hcf(7n+4, n+1)=hcf(7(n+1)-(7n+4), n+1) | M1
hcf(3, n+1) | A1
h = 1 or 3
B1
[3]
M1
A1
M1
A1
Substituting values to get numerical “gradients”
Both correct
b | 7n + 4  0 (mod 3)
 n  2 (mod 3) | M1
A1
[2] | 3.1a
1.1 | Solving attempt at this, or 8n + 5  0 (mod 3), or both
(if a list is given, at least n = 2, 5, 8 seen and no errors)
Condone n  −1 (mod 3) instead