# AQA Level 2 Further Maths - Unofficial mark scheme Watch

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**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}-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

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**

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**joshchamp125**)x

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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?

what did you get for the value of K, the minimum point?

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(Original post by

I didn't even do that obtuse angle question

what did you get for the value of K, the minimum point?

**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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#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

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

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#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

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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(Original post by

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

**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

^{2 }= 9n

^{2}and not just 9n?

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#9

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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(Original post by

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

**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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(Original post by

How many marks do y'all think you got

**chelseafc141**)How many marks do y'all think you got

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#12

(Original post by

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

= -80

= 39

= (x

= (x - 3)(x + 3)(x

y=1/3x^3-x^2-5x-k

find dy/dx of the equation (x

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

(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

(x - 3)(x - 7)(x + 2)

Sinx = root 11/ 6

sin

as 1 - sin

therefore cosx = +/- 5/6

as the angle was obtuse, it's -5/6

the hypotenuse of the two right angled triangle was 3root2 which is equal to root18

therefore a

as it's a square a and b must be equal and therefore half of 18

so a

and a = 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 2root3

this added together was 9 + 3root3

which can be factorised to 3(3 + root3)

use the cosine rule - a

a

a

a = 3n which is the same as one of the other lengths

**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^3-x^2-5x-k

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 = 11/36as 1 - sin

^{2}x = cos^{2}x, cos^{2}x = 25/36therefore 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}= 18as it's a square a and b must be equal and therefore half of 18

so a

^{2}= 9and 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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#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

Posted from TSR Mobile

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(Original post by

I really don't know how u got -80 for the x/y question, i wrote 80 :/

**penelopecrux**)I really don't know how u got -80 for the x/y question, i wrote 80 :/

^{-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

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(Original post by

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

**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

(Original post by

for y

it stated that y was less than zero so you had to use -1/5

therefore it was 16/ -1/5 which is -80

**lily628**)for y

^{-2}, it has two possible solutions: 1/5 and -1/5it 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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#19

(Original post by

thought i might as well get one going...

if anyone can remember more questions / answers / how marks a question was then please say

= 3x

a = 3

b = 20

= 12th term

= 3/5

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

= 39

= (x

= (x - 3)(x + 3)(x

Gradients were -5/2 and 2/5

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

y=1/3x^3-x^2-5x-k

find dy/dx of the equation (x

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

(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

(x - 3)(x - 7)(x + 2)

Sinx = root 11/ 6

sin

as 1 - sin

therefore cosx = +/- 5/6

as the angle was obtuse, it's -5/6

the hypotenuse of the two right angled triangle was 3root2 which is equal to root18

therefore a

as it's a square a and b must be equal and therefore half of 18

so a

and a = 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 2root3

this added together was 9 + 3root3

which can be factorised to 3(3 + root3)

use the cosine rule - a

a

a

a = 3n which is the same as one of the other lengths

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

**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^3-x^2-5x-k

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 = 11/36as 1 - sin

^{2}x = cos^{2}x, cos^{2}x = 25/36therefore 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}= 18as it's a square a and b must be equal and therefore half of 18

so a

^{2}= 9and 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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(Original post by

first one was 3xsquared - 20x wasn't it? and b was -20 not 20 on the matrices question.

**joshchamp125**)first one was 3xsquared - 20x wasn't it? and b was -20 not 20 on the matrices question.

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