Question 5:
5 | a | O A  O B = 0 requires OA // OB …
… and (3 −8 …) can never be a multiple of (1 2 …) | M1
A1
[2] | 2.1
2.4 | Clear use of this result (parallellism)
Properly explained
b | i | (3i – 8j + tk)  (i + 2j – 2k) = (16 – 2t)i + (6 + t)j + 14k
Area OAB = 12 t h e ir V e c t o r P r o d u c t
= 12 5 t 2 − 5 2 t + 4 8 8 | M1
A1
M1
A1
[4] | 1.1
1.1
1.1
1.1 | Either of the i- or j- components correct
All correct
Attempted (must have t’s involved)
12 ( 1 6 − 2 t ) 2 + ( 6 + t ) 2 + 1 4 2 is fine here
ii | Differentiating their answer, or by completing the square
t = 5.2 | M1
A1
[2] | 3.1a
1.1 | Leading to a linear equation of the form 𝑎𝑡 +𝑏 = 0
with non-zero a and b
a | A = 0.96A + 40 and A = 1000
n + 1 n 0 | B1
[1] | 3.3 | or A = 0.96A + 40 and A = 1000
n n-1 0
b | i | 80, i.e. 8% of adults lost linked to the 0.92
(noting that the 1000 and 40 remain the same as before)
A = 0.92A + 40 and A = 1000
n + 1 n 0 | B1
[1] | 2.1 | AG
ii | PS from A = A = a is A = 500
n + 1 n n
CS is A = C  0.92n
n
Soln. from GS: A = C  0.92n + 500 with A = 1000
n 0
A = 500  0.92n + 500
n | B1
B1
M1
A1
[4] | 1.1
1.1
1.1
1.1
iii | A decreases, stabilising at a constant 500
n
i.e. A → 500
n | B1
[1] | 3.4 | Not required to say “monotonic decreasing”, since
only long-term behaviour asked-for
c | i | PS is A = 250
n
General solution is of the form A = D   n + E   n +
n
“250”
where  and  are positive and less than 1
Hence both  n and  n → 0
A → ‘250’
n | B1
M1
A1
A1
[4] | 1.1
3.1a
2.4
1.1 | May be implied by following answer
Accept statement that  and  are both positive and
less than 1
If calculated,  and  must be correct
Note that = 12 ( 0 . 9 + 0 . 4 1 )  = 0.77 and
= 1(0.9− 0.41)= 0.13
2
665 665
(𝐷 =375+ ≈894.27776 and 𝐸 =375− ≈
2√0.41 2√0.41
−144.27776)
FT from PS
SC1 X = 0.9X – 0.1X + 50, so X = 250
B1M1A0A1 is possible
ii | A = INT(0.9A – 0.1A + 50)
n + 1 n n – 1 | B1
[1] | 3.5c | (Note that this gives A → 249)
n