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# AQA Level 2 Further Maths - Unofficial mark scheme

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1. 23/6/16 - Hey guys, I'll definitely be making a markscheme for paper 2 because it will be my last one (although i may be a bit tired after watching the ref results lol) so stay around for a link

thought i might as well get one going...
if anyone can remember more questions / answers / how marks a question was then please say

1. Differentiate x(x2 - 10x) [2]
- expand to equal 3x3 - 10x2
= 3x2 - 20x

2. Find a and b in the matrices[2]
a = 3
b = - 20

3. Nth term
a) Which term equates to 1/2? [2]
= 12th term
b) What is the limiting factor of n as n -> infinity [1]
= 3/5

4. Equation of a circle
a) Which is the centre of the circle? [1]
= (-5,8)
b) What was the radius of the circle [1]
= √10

5. Circle Theorem Q[3]
Angle at the circumference was 3x so angle at the centre must've been 6x
The other angle in the centre was 2x + 48
6x + 2x + 48 = 360
so 8x = 312
= 39

6. Find m and p[4]
m = 8
p = -1

7. State the range of integers for x2 - 20x + 96 < 0[3]
factorise to get (x - 12)(x -8) < 0
test out values < 8
test out values > 12
test out values 8 < x < 12
= 8 < x < 12

8. Find x when √125 + √20 = √80 - √x [3]
√125 = 5√5 and √20 = 2√5, so the total is 7√5
√80 = 4√5
so √x = 3√5
= √45

9. Expand (x - 5)3[3]
= x3 - 15x2 + 75x - 125

10. What is x/y?[3]
x was 16y was +/- 1/5
but as it stated it was less than 0
it has to be -1/5
16 / -1/5
= -80

11. Perpendicular lines[3]
which are negative reciprocals of each other and therefore the lines must be perpendicular

12. x2 + 6x + 2 = 0
a) x2 + 6x + 2 = (x + h)2 + k, find h and k [2]
Complete the square = (x + 3)2 -9 + 2
h = 3 , k = -7
b) Find the minimum point of the curve [1]
= (-3, -7)
c) Find x [1]
(x + 3)2 = 7
x + 3 = +/-√7
x = +/-√7 - 3

13. [3]
x = 121

14. x3 - 8x2 + x + 42
a) (x - 3) is a factor, prove a = -1 [2]
(x - 3) is a factor so you could substitute 3 in for x
then you had to equate the equation to zero
then you would rearrange to get a = -1

b) factorise the cubic [3]
(x - 3)(x - 7)(x + 2)

15. Rationalise the Denominator of 6/√7 + 2[3]
= 6√7 + 12 / 3
= 2√7 - 4

16. Factorise (x4 - 81) [2]
= (x2 - 9)(x2 + 9)
= (x - 3)(x + 3)(x2 + 9)

17. Working out K[5]
y=1/3x3-x2-5x-k
find dy/dx of the equation (x2 - 2x - 3)
then factorise (x - 3)(x + 1)
so stationary points are when x = 3 and -1
substitute 3 into the original equation and rearrange to get k
k = 9

18. Sin / Cos question[4]
Sinx = root 11/ 6
sin2x = 11/36
as 1 - sin2x = cos2x, cos2x = 25/36
therefore cosx = +/- 5/6
as the angle was obtuse, it's -5/6

19. Trapezium Question
a. Prove the sides = 3 [2]
the hypotenuse of the two right angled triangle was 3root2 which is equal to root18
therefore a2 + b2 = 18
as it's a square a and b must be equal and therefore half of 18
so a2 = 9
and a = 3

b. Prove the perimeter of the trapezium is 3(3 + √3) [4]
we know that the sides of the square are 3, so 3+3+3 = 9
you had to use sin and tan to get the other two sides which were root3 and 2√3
this added together was 9 + 3√3
which can be factorised to 3(3 + √3)

20. Prove the triangle is isosceles [4]
Use the cosine rule - a2 = (3n)2 + (2n)2 - (2 x 3n x 2n x 1/3)
a2 = 13n2 - 4n2
a2 = 9n2
a = 3n which is the same as one of the other lengths

Total marks for paper: 70

CREDIT:
Spoiler:
Show
(Original post by jazz_xox_)
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(Original post by Mattyates2000)
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(Original post by elizakbrown)
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(Original post by joshchamp125)
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(Original post by ShonalK)
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(Original post by jjen2603)
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(Original post by jojo41)
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2. I didn't even do that obtuse angle question

what did you get for the value of K, the minimum point?
3. (Original post by jazz_xox_)
I didn't even do that obtuse angle question

what did you get for the value of K, the minimum point?
dw it wasn't too many marks !!
9, wbu?
4. (Original post by lily628)
dw it wasn't too many marks !!
9, wbu?
I didn't even get that far - in the end I put -3 but I knew it was wrong
5. for the x/y question I got -80

for the circle theorems I got 39 degrees

for the last question I got w=3n (by the cosine rule) so it is isosceles

for the 2nd question on matrices, i got 3 and -20
6. For the last question:

This is what I did...

w^2 = 3n^2 + 2n^2 - 2n*3n*1/3
w^2 = 9n + 4n - 6n^2*1/3
w^2 = 9n +4n - 2n^2
w = 3n + 2n - 2n
w = 3n --which is the same as the other line meaning it is an isoscelese triangle
7. (Original post by Mattyates2000)
For the last question:

This is what I did...

w^2 = 3n^2 + 2n^2 - 2n*3n*1/3
w^2 = 9n + 4n - 6n^2*1/3
w^2 = 9n +4n - 2n^2
w = 3n + 2n - 2n
w = 3n --which is the same as the other line meaning it is an isoscelese triangle
ik that works but isn't 3n2 = 9n2 and not just 9n?
8. How many marks do y'all think you got
9. (Original post by lily628)
ik that works but isn't 3n2 = 9n2 and not just 9n?
hmmm... that is a good point

I guess you can keep the squared and then square root them:

w^2 = 3n^2 + 2n^2 - 2n*3n*1/3
w^2 = 3n^2 +2n^2 - 2n^2
w = 3n + 2n - 2n
w = 3n --which is the same as the other line meaning it is an isoscelese triangle
10. (Original post by Mattyates2000)
hmmm... that is a good point

I guess you can keep the squared and then square root them:

w^2 = 3n^2 + 2n^2 - 2n*3n*1/3
w^2 = 3n^2 +2n^2 - 2n^2
w = 3n + 2n - 2n
w = 3n --which is the same as the other line meaning it is an isoscelese triangle
ah true cheers
11. (Original post by chelseafc141)
How many marks do y'all think you got
no clue tbh, hoping for 40ish?
12. (Original post by lily628)
thought i might as well get one going...

1. differentiation q
2. matrices.
3. n = 1/2 question
3b. Limiting factor of n to infinity was 5/3

Random Qs

x/y
= -80

Circle Theorem Q
= 39

Factorise (x4 - 81)
= (x2 - 9)(x2 + 9)
= (x - 3)(x + 3)(x2 + 9)

Working out K
y=1/3x^3-x^2-5x-k
find dy/dx of the equation (x2 - 2x - 3)
then factorise (x - 3)(x + 1)
so stationary points are when x = 3 and -1
substitute 3 into the original equation and rearrange to get k
k = 9

Factorising the cubic
a) prove a = -1
(x - 3) is a factor so you could substitute 3 in for x
then you had to equate the equation to zero
then you would rearrange to get 1 = -1

b) factorise the cubic
(x - 3)(x - 7)(x + 2)

Sin / Cos question
Sinx = root 11/ 6
sin2x = 11/36
as 1 - sin2x = cos2x, cos2x = 25/36
therefore cosx = +/- 5/6
as the angle was obtuse, it's -5/6

Trapezium Question
a. Prove the sides = 3
the hypotenuse of the two right angled triangle was 3root2 which is equal to root18
therefore a2 + b2 = 18
as it's a square a and b must be equal and therefore half of 18
so a2 = 9
and a = 3

b. Prove the perimeter of the trapezium is 3(3 + root3)
we know that the sides of the square are 3, so 3+3+3 = 9
you had to use sin and tan to get the other two sides which were root3 and 2root3
this added together was 9 + 3root3
which can be factorised to 3(3 + root3)

Last question (triangle)
use the cosine rule - a2 = (3n)2 + (2n)2 - (2 x 3n x 2n x 1/3)
a2 = 13n2 - 4n2
a2 = 9n2
a = 3n which is the same as one of the other lengths
I really don't know how u got -80 for the x/y question, i wrote 80 :/
13. Question 3, did n=12?
14. For the trapezium question, where you had to prove the perimeter I messed up my trigonometry because I drew the special triangle wrong I ended up close to the answer 3(5 + root 3) will I get any marks?

Posted from TSR Mobile
15. (Original post by penelopecrux)
I really don't know how u got -80 for the x/y question, i wrote 80 :/
for y-2, it has two possible solutions: 1/5 and -1/5
it stated that y was less than zero so you had to use -1/5
therefore it was 16/ -1/5 which is -80
16. (Original post by BobbiBlue)
For the trapezium question, where you had to prove the perimeter I messed up my trigonometry because I drew the special triangle wrong I ended up close to the answer 3(5 + root 3) will I get any marks?

Posted from TSR Mobile
i can't remember how many marks it was, but if you got some of the trigonometry right and stated the sides of the square then maybe?
17. (Original post by lily628)
for y-2, it has two possible solutions: 1/5 and -1/5
it stated that y was less than zero so you had to use -1/5
therefore it was 16/ -1/5 which is -80

oh noooo, so i would lose 1 mark right????
18. For question 2, I got a=3 and b=-20
19. (Original post by lily628)
thought i might as well get one going...
if anyone can remember more questions / answers / how marks a question was then please say

1. Differentiation Q
= 3x2 - 20

2. Matrices
a = 3
b = 20

3. Nth term
a) Which term equates to 1/2?
= 12th term
b) What is the limiting factor of n as n -> infinity
= 3/5

6. What is x/y?
x was 16
y was +/- 1/5 but as it stated it was less than 0 it has to be -1/5
16 / -1/5
= -80

5. Circle Theorem Q
= 39

Factorise (x4 - 81)
= (x2 - 9)(x2 + 9)
= (x - 3)(x + 3)(x2 + 9)

Perpendicular lines
which are negative reciprocals of each other and therefore the lines must be perpendicular

Working out K
y=1/3x^3-x^2-5x-k
find dy/dx of the equation (x2 - 2x - 3)
then factorise (x - 3)(x + 1)
so stationary points are when x = 3 and -1
substitute 3 into the original equation and rearrange to get k
k = 9

Factorising the cubic
a) prove a = -1
(x - 3) is a factor so you could substitute 3 in for x
then you had to equate the equation to zero
then you would rearrange to get 1 = -1

b) factorise the cubic
(x - 3)(x - 7)(x + 2)

Sin / Cos question
Sinx = root 11/ 6
sin2x = 11/36
as 1 - sin2x = cos2x, cos2x = 25/36
therefore cosx = +/- 5/6
as the angle was obtuse, it's -5/6

Trapezium Question
a. Prove the sides = 3
the hypotenuse of the two right angled triangle was 3root2 which is equal to root18
therefore a2 + b2 = 18
as it's a square a and b must be equal and therefore half of 18
so a2 = 9
and a = 3

b. Prove the perimeter of the trapezium is 3(3 + root3)
we know that the sides of the square are 3, so 3+3+3 = 9
you had to use sin and tan to get the other two sides which were root3 and 2root3
this added together was 9 + 3root3
which can be factorised to 3(3 + root3)

Last question (triangle)
use the cosine rule - a2 = (3n)2 + (2n)2 - (2 x 3n x 2n x 1/3)
a2 = 13n2 - 4n2
a2 = 9n2
a = 3n which is the same as one of the other lengths

OTHERS
There was a question about stating the range? 8 < x < 12
There was a completing the square question
a question asked you to rationalise the denominator

Total marks for paper: 70

first one was 3xsquared - 20x wasn't it? and b was -20 not 20 on the matrices question.
20. (Original post by joshchamp125)
first one was 3xsquared - 20x wasn't it? and b was -20 not 20 on the matrices question.
ah can't really remember clearly ty

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