Solutions June 2015 paper 2 (AS1051)
Section A
1. (a) (seen in sample class test)
i. y is an upper bound for A precisely when 8a 2 A; a ≤ y. 2
ii. u is the least upper bound of A precisely when u is an upper
bound o
...
Solutions June 2015 paper 2 (AS1051)
Section A
1. (a) (seen in sample class test)
i. y is an upper bound for A precisely when 8a 2 A; a ≤ y. 2
ii. u is the least upper bound of A precisely when u is an upper
bound of A, and for any y another upper bound of A, we have
u ≤ y. 2
(b) (unseen) In the interval X we have cos(x) > 1=p2 precisely when
−π=4 < x < π=4. The least upper bound of A is therefore π=4. 4
2. (unseen) Using the auxiliary equation
λ2 − 5λ + 4 = (λ − 4)(λ − 1)
we determine that the solution for the homogeneous equation is
A
n = 4nA + B:
Now we can try An = a7n, substituting we get 3
(a7n+2) − 5(a7n+1) + 4(a7n) = 7n
a(72 − 5 × 7
Section A
1. (a) (seen in sample class test)
i. y is an upper bound for A precisely when 8a 2 A; a ≤ y. 2
ii. u is the least upper bound of A precisely when u is an upper
bound of A, and for any y another upper bound of A, we have
u ≤ y. 2
(b) (unseen) In the interval X we have cos(x) > 1=p2 precisely when
−π=4 < x < π=4. The least upper bound of A is therefore π=4. 4
2. (unseen) Using the auxiliary equation
λ2 − 5λ + 4 = (λ − 4)(λ − 1)
we determine that the solution for the homogeneous equation is
A
n = 4nA + B:
Now we can try An = a7n, substituting we get 3
(a7n+2) − 5(a7n+1) + 4(a7n) = 7n
a(72 − 5 × 7
[Show More]
.png)
.png)
 V1-V2.png)
