# AQA Level 2 Further Maths - Unofficial mark schemeWatch

Announcements
#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) 
- expand to equal 3x3 - 10x2
= 3x2 - 20x

2. Find a and b in the 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

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

5. Circle Theorem Q
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
m = 8
p = -1

7. State the range of integers for x2 - 20x + 96 < 0
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 
√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
= x3 - 15x2 + 75x - 125

10. What is x/y?
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
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 
Complete the square = (x + 3)2 -9 + 2
h = 3 , k = -7
b) Find the minimum point of the curve 
= (-3, -7)
c) Find x 
(x + 3)2 = 7
x + 3 = +/-√7
x = +/-√7 - 3

13. 
x = 121

14. x3 - 8x2 + x + 42
a) (x - 3) is a factor, 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 a = -1

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

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

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

17. Working out K
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
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 
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) 
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 
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:
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(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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5
3 years ago
#2
I didn't even do that obtuse angle question what did you get for the value of K, the minimum point?
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#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?
0
3 years ago
#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
0
3 years ago
#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
1
3 years ago
#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
0
#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?
0
3 years ago
#8
How many marks do y'all think you got
0
3 years ago
#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
0
#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
0
#11
(Original post by chelseafc141)
How many marks do y'all think you got
no clue tbh, hoping for 40ish?
0
3 years ago
#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 :/
0
3 years ago
#13
Question 3, did n=12?
0
3 years ago
#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
0
#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 0
#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?
0
3 years ago
#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????
0
3 years ago
#18
For question 2, I got a=3 and b=-20
0
3 years ago
#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.
0
#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
0
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