Question 1:
1 | (a) | Even (or odd)-numbered places sum to (a + b)  no. of
blocks
Odd (or even)-numbered places sum to (a + b)  no. of
blocks
Difference is 0, and 11 | 0  11 | N | M1
A1
A1 | 2.1
1.1
2.2a | Or considering a single block:
𝑎−𝑏+𝑏−𝑎 = 0 (terms must be in this order, may
be embedded but must be identifiable, eg ‘Consider
abba…’)
Considering all blocks
𝑎−𝑏+𝑏−𝑎 +𝑎−𝑏+𝑏−𝑎+⋯ = 0 (at least
two blocks and …) or a clear explanation in words
M1A0A1 is possible
SCB1 Each block “abba” is of the form | For fully correct working using method not asked for
1001a + 110b = 11(91a + 10b) is divisible by 11  N
is
[3]
(b
) | Each block “cddc” = 𝑐𝑛3+ⅆ𝑛2 +ⅆ𝑛+𝑐(𝑛0) or
𝑐𝑛𝑘+ⅆ𝑛𝑘−1 +ⅆ𝑛𝑘−2 +𝑐𝑛𝑘−3
= (𝑛+1)(𝑛2 +𝛼𝑛+1)𝑐 +𝑛(𝑛+1)ⅆ or
= 𝑛𝑘−3((𝑛+1)(𝑛2 +𝛼𝑛+1)𝑐 +𝑛(𝑛+1)ⅆ)
(𝑟)(𝑛+1)((𝑛2 −𝑛+1)𝑐 +𝑛 ⅆ) (r an integer) 
M always a multiple of (n + 1) | M1
M1
A1
[3] | 2.1
3.1a
2.4 | Expressing each block in base n (could be seen as
part of the expansion of M)
Attempt to factor (n + 1) out of their polynomial in n;
if using the Factor theorem, c−ⅆ +ⅆ −𝑐 must be
seen
Correct answer from fully correct working.
If using the Factor theorem, c−ⅆ +ⅆ −𝑐 = 0 must
be seen
Questio
n | Answer | Marks | AO | Guidance
 z 4 x
= − x + 2 y
 y 2 y
 z  z
At P, = 6 and = 8
 x  y
Tangent plane eqn. is z−20=6(x−1)+8(y−4)
6x + 8y – z = 18 | A1
A1
M1dep*
A1 | 1.1
1.1
1.1
1.1 | Both correct
Substituting coordinates correctly in both partial
derivatives (can be implied by a correct tangent plane
equation)
Attempt at plane equation using this form and their
numerical gradients
Equation in the form 𝑎𝑥 +𝑏𝑦+𝑐𝑧 = ⅆ
(or ⅆ = 𝑎𝑥 +𝑏𝑦+𝑐𝑧)
ALT for the last two marks
1 0 −6 1
𝒏 = (0)×(1) = (−8) and 𝒏⋅( 4 ) = −18
6 8 1 20 | M1dep* | Finding the normal vector to the plane and k in
𝒏⋅𝒓 = 𝑘
6x + 8y – z = 18 | A1 | Equation in the form 𝑎𝑥 +𝑏𝑦+𝑐𝑧 = ⅆ
(or ⅆ = 𝑎𝑥 +𝑏𝑦+𝑐𝑧)
[5]
ALT for the last two marks
1 0 −6 1
𝒏 = (0)×(1) = (−8) and 𝒏⋅( 4 ) = −18
6 8 1 20
Finding the normal vector to the plane and k in
𝒏⋅𝒓 = 𝑘
Questio
n | Answer | Marks | AO | Guidance
 their 259a + 22  7 (mod 67)
their a  24 (mod 67) or a = 67n + 24
 x = 259(67n + 24) + 22
x = 17353n + 6238
hcf(7, 37) = 1 or hcf(3, 37) = 1 or … | M1
M1
A1
B1 | 3.1a
1.1
1.1
2.4
ALT I - 1st 2-marks (first relationship sorted)
x  1 (mod 7)  x  22 (mod 7) | M1 | Setting up a relationship between two “bases”
x  22 (mod 7  37 = 259) | M1 | Attempt to relate result to the third “base”
ALT II - Using the Chinese Remainder Theorem
x = 2479 m + 259 l + 469 k ((mod 7×67×37)) | M1
2479 m  1 (mod 7)  m  1 (mod 7) | M1 | Calculating m in explicit form
259 l  7 (mod 67)  l  29 (mod 67) | M1 | Calculating l in explicit form
469 k  22 (mod 37)  k  29 (mod 37) | M1 | Calculating k in explicit form
x = 2479 × 1 + 259 × 29 + 469 × 29 = 23591
 6238 (mod 17353) | A1 | Substituting m, l, and k into x and stating equivalence
(mod 17353)
 6238 (mod 17353)
A solution exists because 7, 67, 37 are co-prime or | B1 | One justification explicitly stated
hcf(7, 67) = 1 or hcf(7, 37) = 1 or …
[6]
(a) | Considering a = 0 and/or b = 0
(p, q, r) = (1, −2, 32 ) when a = 0
(p, q, r) = (−2, 52 , −3) when b = 0 | M1
A1
A1
[3] | 1.1
1.1
1.1 | Soi by subsequent working. Allow |a|=0 and |b|=0.
Not 𝒂×𝒃 = 𝟎 or 𝒂⋅𝒃 = 0
Or when |a|=0
Or when |b|=0
And no others.
If A0A0, SCB1 if values for (p, q, r) are ‘ungrouped’
but the two sets (1, −2, 32 ) and (−2, 52 , −3) can be
distinguished www
9
E. g. = − 2 − q and q an integer  ( − 2) | 9
− 2 
 ( − 2 = 3, 1, −1, −3  ) one of  = 5, 3, 1, −1
 = 5 gives (p, q, r) = (3, −5, 2)
Other cases visibly rejected: since || > 1 and
 = 3 gives r = 1 25 (and p = 5, q = −11) | M1
A1
A1
A1
[7] | 1.1
2.2a
1.1
3.2a | Method for determining values of  in a second case
At least one of the possible values of ;
candidates may note that  must be odd when
considering q  ℤ
www
Selecting  = 5
Must be correct cases rejected for correct reasons
If (M1A1)M0M0, SCB1  = 5 www
SCB1 (p, q, r) = (3, −5, 2) www
(a) | Solving 10000 = 20000(1 – 10000k)
 k = 10 or 0.00005
2 0 0 0 | M1
A1
[2] | 1.1
1.1 | AG
If M0, SCB1 for verifying that k = 0.00005
with 10000 substituted
(b) | P converges to 10049 (as there are no part animals)
n | B1
[1] | 2.2a | Condone 10050, if 10049.75… or 10049.8 is seen,
either as the value of any P , k >1, or obtained by
k
equating P and P
n n+1
i | 𝑃 = 2𝑃 (1−0.00005𝑃 )−2400 (for n  0)
𝑛 + 1 𝑛 𝑛 | B1
[1] | 3.3
ii | P decreases (monotonically) (each year)
n | B1
[1] | 2.5 | Accept ‘Population will decrease’.
Not ‘Population will decline’.
Ignore any statement on long-term behaviour, such as
‘P approaches a limiting value’ or ‘P → 6000’
n n
iii | x=2x(1−0.00005x)−2400
 0.0001𝑥2 −𝑥 +2400 = 0  x = 6000, 4000
Reject 4000, so 6000
since when first equilibrium point is reached, P does not
n
continue to decrease | M1
A1
A1
[3] | 3.4
1.1
2.3 | Setting up “fixed-term” equation and attempt to solve
Correct answer chosen and other rejected (the word
‘decrease’ may be seen in part (c) (ii)).
Condone selecting 6000 with justification and not
explicitly rejecting 4000.
‘P = 10000 and P decreases, so in the long term
0 n
P → 6000’
n
Accept ‘4000 < 6000’ as justification only if P is
n
stated as decreasing in part (c) (ii).
NB M0 for non-algebraic method
iv | Solving 10000 = 20000(1 – 10000k) – 2400
 k = 0.000038 oe | M1
A1
[2] | 3.5c
1.1 | NB M0A0 for answer with no working
Questio
n | Answer | Marks | AO | Guidance
(a) |  z
= 2 x + a y …
 x
 z
… and = 3 y 2 + a x
 y
 z  z 2  2  2
= = 0  y = − x  0 = 3 − x + a x
 x  y a a
a 3
 x = 0 or x = −
1 2
 (x, y, z) = (0, 0, 0) or
 a 3 a 2 a 6 
− , − ,
1 2 6 4 3 2 | M1*
A1
M1dep*
M1
A1
A1
[6] | 1.1
1.1
3.1a
1.1
1.1
1.1 | Both first partial derivatives attempted
𝑘 𝑥 +𝑎𝑦 and 𝑘 𝑦+𝑎𝑥, 𝑘 ,𝑘 ≠ 0
1 2 1 2
Both correct
Both set to zero and substituting for one variable
Solving for the second variable.
Condone sign errors.
AG for SP at O shown. Allow verification that
(0,0,0) is a SP.
Condone (0,0) or ‘the origin’
Second SP correct
Questio
n | Answer | Marks | AO | Guidance
2 a
| H | = =
a 6 y
2 a 2 a
= = −a2 at O OR =a2 at other SP
a 0 a a2
At O, | H | < 0  a saddle point
At other SP, | H | > 0 and f >0  a (local) Minimum
xx | M1FT
M1FT
A1
A1
[5] | 2.1
1.1
2.2a
2.2a | Determinant of Hessian matrix attempted (with non-
𝜕2𝑧 𝜕2𝑧
zero 𝑎𝑛ⅆ )
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
2 𝑎
Can be implied by sight of | | below
𝑎 𝑎2
𝑎3 𝑎2
FT only ( ,− ,…) in part (a)
12 6
Attempt to evaluate at least one case (with non-zero
𝜕2𝑧 𝜕2𝑧
𝑎𝑛ⅆ )
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
𝑎3 𝑎2
FT only ( ,− ,…) in part (a)
12 6
Correct conclusion
Correct conclusion
Condone missing coordinates of SP if y-coordinate is
given
𝑎3 𝑎2
FT only ( ,− ,…) in part (a)
12 6
| H | < 0  a saddle point
Questio
n | Answer | Marks | AO | Guidance
| H | = 0 so the nature of the SP cannot be determined (by
this method) | B1
[2] | 2.4 | Or origin is a saddle point with justification, eg
‘When x=0, 𝑧 = 𝑦3 has a point of inflection at the
origin
When y=0, 𝑧 = 𝑥2 has a minimum point at the origin
So the origin is a saddle point’
Or diagrams of 𝑧 = 𝑦3 and 𝑧 = 𝑥2 seen
(a) | d ( ) 12 ( ) − 12
x 3 + 1 = 12 x 3 + 1 .3 x 2
d x
2 2
So (𝐼 =)[ √𝑥3 +1] = 4
2 3 0 3 | M1
A1
[2] | 1.1
1.1 | Attempt at differentiation by the chain rule
1
𝑘 𝑥2(𝑥3 +1)− 2 soi
Needs to be via 2 ∫𝑓′(𝑥)ⅆ𝑥 (not via substitution)
3
cao (Intermediate step must be seen)
(b) | x 2
I =  x n − 2 . d x = [𝑎 𝑥𝑛 − 2.√𝑥3 +1]−𝑏∫(𝑛−
n
x 3 + 1
2).𝑥𝑛 − 3√𝑥3 +1 d𝑥
2
𝑎 = 𝑏 =
3
𝑥3+1
√𝑥3 +1 =
√𝑥3+1
3𝐼 = 3×2𝑛 − 1 −2(𝑛−2)(𝐼 +𝐼 )
𝑛 𝑛 𝑛 − 3
( 2 n − 1 ) I = 3  2 n − 1 − 2 ( n − 2 ) I
n n − 3 | M1
A1
M1*
M1dep*
A1
[5] | 3.1a
1.2
1.1
1.1
1.1 | Use of integration by parts with appropriate splitting
so that result of part (a) can be used
First-stage of integration by parts correct
Valid preparation for second integral in I forms
k
( ) x 3 + 1
23 2 n − 2  3 − 0 − 23 ( n − 2 )  x n − 3 . d x
x 3 + 1
Writing in terms of I and I
n n-3
AG fully shown (showing correct substitution of
limits at some point) www
(c) | 𝑥3+1
(𝐼 = ∫𝑥5. d𝑥 =) 𝐼 +𝐼
√𝑥3+1 8 5
n = 5  9𝐼 = 𝑎−6𝐼 oe
5 2
n = 8  15𝐼 = 𝑏 −12𝐼 oe
8 5
9 I = 3  1 6 − 6 I  I = 4 09
5 2 5
1 5 I = 3  1 2 8 − 1 2 I  I = 992 I = 1 14 95 2 o r 2 6 24 25
8 5 8 45 | B1
M1
A1
[3] | 3.1a
1.1
1.1 | Must use given reduction formula for n = 5 and n =
4
8 (𝑎,𝑏 ≠ 0). I can only be or their numerical
2
3
value from part (a). I can be numerical in the
5
expression for I .
8
cao from full working
(a) | i | (By Lagrange’s theorem) o(H) | o(G)
so (o(H) =) 2, 3, 4 or 6 | B1
B1
[2] | 2.4
2.2a | Allow ‘By Lagrange’s Theorem’ only if their orders
below are correct
Ignore inclusion of 1 and/or 12
ii | (g is the generator, so) g12 = e
(o(H) = 2) {e, g6}
(o(H) = 3) {e, g4, g8}
(o(H) = 4) {e, g3, g6, g9}
(o(H) = 6) {e, g2, g4, g6, g8, g10 } | B1
B1
B1
B1
B1
[5] | 3.1a
1.1
1.1
1.1
1.1 | Generator element and identity element must be
(implicitly) defined
Allow G ={e, g, g2, g3, …, g11}
Ignore {e}, H
SCB1 For an example of a cyclic group of order 12,
completely defined and subgroups all correct (eg
(ℤ ,+))
12
iii | All powers of g commute | B1
[1] | 2.4 | Commutative because 𝑔𝑘𝑔𝑙 = 𝑔𝑙𝑔𝑘 ( 1 ≤ 𝑘,𝑙 ≤ 11,
𝑘 ≠ 𝑙) (an example is sufficient)
“Cyclic  abelian” is insufficient
i | (a, b) where a  {0, 1, 2} and b  {0, 1, 2, 3} | B1
[1] | 3.1a | All 12 elements may be listed (and no extras)
ii | (1, 1) generates J so mapping this to g creates the required
isomorphism | B1
[1] | 2.4 | J is a cyclic group generated by (1,1). J and G are
both cyclic groups of order 12, so isomorphic.
Or an explicit bijective mapping of each element of
G onto an element of J. Condone one error but not in
mapping one generator onto the other generator.
(c) | i | m = 2 and n = 6 | B1
[1] | 3.1a | And no extras.
Accept 2, 6 (but not 6, 2)
ii | (a, b) where a  ℤ and b  ℤ
2 6 | B1
[1] | 1.1 | All 12 elements may be listed (and no extras)
iii | (𝑥 ,𝑦 )⊕(𝑥 ,𝑦 ) = (𝑥 + 𝑥 ,𝑦 + 𝑦 )
1 1 2 2 1 2 2 1 6 2
= (𝑥 + 𝑥 ,𝑦 + 𝑦 ) = (𝑥 ,𝑦 )⊕(𝑥 ,𝑦 )
2 2 1 2 6 1 2 2 1 1
𝑥 ,𝑥 ∈ ℤ and 𝑦 ,𝑦 ∈ ℤ
1 2 2 1 2 6 | B1
[1] | 2.5 | Or Both ‘additions’ (in ℤ and ℤ ) are commutative
2 6
and the two components do not interfere with each
other (ie components are independent, separate,…).
Condone ‘individual’
G is cyclic while K is not cyclic eg (1,2), (1,4) cannot be
generated by (1,1) or any other element/max order possible
for an element of K is (LCM(2,6)=)6
G has one self-inverse element, g6, while K has at least
two/three ie (1,0), (0,3) and (1, 3) (at least two given)
G has a generator (g) while K does not have a generator
i.e. (1,2), (1,4) cannot be generated by (1,1) or any other
element/ max order possible for an element of K is
(LCM(2,6)=)6 | A1
[2] | 2.4 | Convincingly concluded.
Not just ‘G and K have different structure’
PMT
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