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AQA level 2 Further maths GCSE

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I-got-9
2. QUESTION=X^4+81
Simplify=(x^2+9)(x^2-9)
Simplify-further-blah
Thanks
4. (Original post by vishantu_success)
QUESTION=X^4+81
Simplify=(x^2+9)(x^2-9)
Simplify-further-blah
oh i put ((x-3)^2(x+3)^2)
5. (Original post by vishantu_success)
QUESTION=X^4+81
Simplify=(x^2+9)(x^2-9)
Simplify-further-blah
That's one mark, second mark for (x + 3)(x - 3)(x^2 + 9)

(x^2 + 9) can't be simplified further.
6. For the last question did you have to use the cosine rule....

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. hey guys, making an unofficial mark scheme lol
any help with remembering the questions would be appreciated

http://www.thestudentroom.co.uk/show....php?t=4176536
8. The hardest one was working out k. Does anyone have the workings to see how many marks I would get? The original equation was y=1/3x^3-x^2-5x-k
9. (Original post by Mattyates2000)
For the last question did you have to use the cosine rule....

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
Yeah, exactly that. Good question, with nice numbers.
10. (Original post by Ishan_2000)
That's one mark, second mark for (x + 3)(x - 3)(x^2 + 9)

(x^2 + 9) can't be simplified further.
Thats what i got!!!
11. (Original post by daisie)
The hardest one was working out k. Does anyone have the workings to see how many marks I would get? The original equation was y=1/3x^3-x^2-5x-k
Differentiate first to get: x^2 - 2x - 3 (it was 3x not 5x, I think)

Then, make this equal to 0, as it was a minimum point, so has a gradient of 0, so:
x^2 - 2x - 3 = 0
Factorise: (x - 3)(x + 1) = 0
Solve: x = 3, -1.

Sub in x = 3 into the equation, as that was where y was, and you should end up with k - 9 = 0 (y = 0, when x = 3, as the x-axis was a tangent to the minimum point.)

k = 9 (4 marks)
12. (Original post by daisie)
The hardest one was working out k. Does anyone have the workings to see how many marks I would get? The original equation was y=1/3x^3-x^2-5x-k
I think the answer was 9... you had to do dy/dx = 0 to find x and then you knew that why was 0 so you sub y = 0 and x = (can't remember) into
y=1/3x^3-x^2-5x-k and then you get +9 I think

13. These are some random answers for random questions that me and a few others got, they're not all right but if you got anything similar or recall writing something similar then woo
14. (Original post by ihatehannah)
-80
Fs i got -80/sqrt of 1 which is -80. I got confused what sqrt of 1 was
15. (Original post by Ishan_2000)
Differentiate first to get: x^2 - 2x - 3 (it was 3x not 5x, I think)

Then, make this equal to 0, as it was a minimum point, so has a gradient of 0, so:
x^2 - 2x - 3 = 0
Factorise: (x - 3)(x + 1) = 0
Solve: x = 3, -1.

Sub in x = 3 into the equation, as that was where y was, and you should end up with k - 9 = 0 (y = 0, when x = 3, as the x-axis was a tangent to the minimum point.)

k = 9 (4 marks)
Thank you! Yes sorry thats a typo it should have been 3 not 5
16. For the perimeter of the trapezium question, to clear any confusion:

BA + BC + CD + AE + DE = Perimeter,
BA = 3
BC = 3
CD = 3 (square has equal sides, all were 3)
AE = Right angled triangle, angle of 60 degrees was given, side of 3 cm was the opposite. AE was the hypotenuse.
So, sin(60) = 3 / AE
AE = 3 / sin(60), so it is 3 / (root3 / 2), which equals 6 / root3

DE = Right angled triangle, angle of 60 degrees was given, side of 3 cm was the opposite. DE was the adjacent.
So, tan(60) = 3 / DE
DE = 3 / tan(60), so it is 3 / root3

Add them all up, to get 9 + 3 / root3 + 6 / root3
Common denominator of root3, and rationalise to get 9 + 3 root3, which factorises to 3 ( 3 + root3) cm
17. (Original post by Ishan_2000)
For the perimeter of the trapezium question, to clear any confusion:

BA + BC + CD + AE + DE = Perimeter,
BA = 3
BC = 3
CD = 3 (square has equal sides, all were 3)
AE = Right angled triangle, angle of 60 degrees was given, side of 3 cm was the opposite. AE was the hypotenuse.
So, sin(60) = 3 / AE
AE = 3 / sin(60), so it is 3 / (root3 / 2), which equals 6 / root3

DE = Right angled triangle, angle of 60 degrees was given, side of 3 cm was the opposite. DE was the adjacent.
So, tan(60) = 3 / DE
DE = 3 / tan(60), so it is 3 / root3

Add them all up, to get 9 + 3 / root3 + 6 / root3
Common denominator of root3, and rationalise to get 9 + 3 root3, which factorises to 3 ( 3 + root3) cm
Thats what I did but for both AE and DE I rationalised them to make it easier for myself
18. (Original post by Ishan_2000)
For the perimeter of the trapezium question, to clear any confusion:

BA + BC + CD + AE + DE = Perimeter,
BA = 3
BC = 3
CD = 3 (square has equal sides, all were 3)
AE = Right angled triangle, angle of 60 degrees was given, side of 3 cm was the opposite. AE was the hypotenuse.
So, sin(60) = 3 / AE
AE = 3 / sin(60), so it is 3 / (root3 / 2), which equals 6 / root3

DE = Right angled triangle, angle of 60 degrees was given, side of 3 cm was the opposite. DE was the adjacent.
So, tan(60) = 3 / DE
DE = 3 / tan(60), so it is 3 / root3

Add them all up, to get 9 + 3 / root3 + 6 / root3
Common denominator of root3, and rationalise to get 9 + 3 root3, which factorises to 3 ( 3 + root3) cm
You seem to have done very well on the exam, well done
19. (Original post by SmithAmelia)
I completely forgot about the obtuse angle bit
nothing to do with it being an obtuse angle it just said y <0
20. (Original post by Dorisdolphine)
What did u get for x/y
I got -80 for that because it was 16/-0.4 which is the same as 16 x -5 which was -80

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