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(x^{2}  10x) [2]
 expand to equal 3x^{3 } 10x^{2}
= 3x^{2}  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 x^{2}  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]
= x^{3  }15x^{2} + 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]
Gradients were 5/2 and 2/5
which are negative reciprocals of each other and therefore the lines must be perpendicular
12. x^{2} + 6x + 2 = 0
a) x^{2} + 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. x^{3}  8x^{2} + 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 (x^{4}  81) [2]
= (x^{2}  9)(x^{2} + 9)
= (x  3)(x + 3)(x^{2} + 9)
17. Working out K[5]
y=1/3x^{3}x^{2}5xk
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
sin^{2}x = 11/36
as 1  sin^{2}x = cos^{2}x, cos^{2}x = 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 a^{2 }+ b^{2} = 18
as it's a square a and b must be equal and therefore half of 18
so a^{2} = 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  a^{2 }= (3n)^{2} + (2n)^{2}  (2 x 3n x 2n x 1/3)
a^{2 }= 13n^{2}  4n^{2}
a^{2 }= 9n^{2}
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_)
x(Original post by Mattyates2000)
x(Original post by elizakbrown)
x(Original post by joshchamp125)
x(Original post by ShonalK)
X(Original post by jjen2603)
X(Original post by jojo41)
X
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AQA Level 2 Further Maths  Unofficial mark scheme
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 1
 20062016 11:46
Last edited by lily628; 24062016 at 00:54.Post rating:5 
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 2
 20062016 11:57
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
 20062016 12:00
(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?
9, wbu? 
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 4
 20062016 12:01

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 5
 20062016 12:03
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 20Post rating:1 
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 6
 20062016 12:03
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 
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 7
 20062016 12:04
(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 
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 8
 20062016 12:07
How many marks do y'all think you got

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 9
 20062016 12:10
(Original post by lily628)
ik that works but isn't 3n^{2 }= 9n^{2} and not just 9n?
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 
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 10
 20062016 12:11
(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 
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 11
 20062016 12:12
(Original post by chelseafc141)
How many marks do y'all think you got 
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 12
 20062016 12:15
(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 (x^{4}  81)
= (x^{2 } 9)(x^{2} + 9)
= (x  3)(x + 3)(x^{2} + 9)
Working out K
y=1/3x^3x^25xk
find dy/dx of the equation (x^{2 } 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
sin^{2}x = 11/36
as 1  sin^{2}x = cos^{2}x, cos^{2}x = 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 a^{2 }+ b^{2} = 18
as it's a square a and b must be equal and therefore half of 18
so a^{2} = 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  a^{2 }= (3n)^{2} + (2n)^{2}  (2 x 3n x 2n x 1/3)
a^{2 }= 13n^{2}  4n^{2}
a^{2 }= 9n^{2}
a = 3n which is the same as one of the other lengths 
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 13
 20062016 12:15
Question 3, did n=12?

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 14
 20062016 12:17
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 
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 15
 20062016 12:19
(Original post by penelopecrux)
I really don't know how u got 80 for the x/y question, i wrote 80 :/
it stated that y was less than zero so you had to use 1/5
therefore it was 16/ 1/5 which is 80 
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 16
 20062016 12:19
(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 
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 17
 20062016 12:20
(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???? 
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 18
 20062016 12:31
For question 2, I got a=3 and b=20

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 19
 20062016 12:34
(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
= 3x^{2}  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 (x^{4}  81)
= (x^{2 } 9)(x^{2} + 9)
= (x  3)(x + 3)(x^{2} + 9)
Perpendicular lines
Gradients were 5/2 and 2/5
which are negative reciprocals of each other and therefore the lines must be perpendicular
Working out K
y=1/3x^3x^25xk
find dy/dx of the equation (x^{2 } 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
sin^{2}x = 11/36
as 1  sin^{2}x = cos^{2}x, cos^{2}x = 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 a^{2 }+ b^{2} = 18
as it's a square a and b must be equal and therefore half of 18
so a^{2} = 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  a^{2 }= (3n)^{2} + (2n)^{2}  (2 x 3n x 2n x 1/3)
a^{2 }= 13n^{2}  4n^{2}
a^{2 }= 9n^{2}
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. 
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 20
 20062016 12:37
(Original post by joshchamp125)
first one was 3xsquared  20x wasn't it? and b was 20 not 20 on the matrices question.
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Updated: June 30, 2016
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