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Further Maths GCSE AQA Paper 1 (Unofficial mark scheme)

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Original post by NiamhM1801
x


Lol, I don't even know what trig identities are. RIP.
[QUOTE="Ano123;65980073"]
Original post by Joelo00


Given that d/dx (sin2x+cos2x)=k.
It is given further that d/dx (sin2x)= 2sinxcosx,
Find the value of k and hence find the derivative of cos2x.


Too hard?
(You do not have to know how to differentiate trig functions to do this).
Original post by mattedwards8972
I hope there's nothing to do with the cosine rule with algebra because, even though I can pick up marks on doing the basic part of the questions, I can never seem to pick up all the marks on the question. I hope something like indices comes up on the exam because these are easy marks, more differentiation stuff would be nice too, anything with the second derivative I wouldn't mind. Also maybe a trigonometric proof would be nice as well, if it's a nice easy one. What topics do you hope come up?


Hmm...I don't know I don't mind them :tongue:
Although topics I'd like would be:
Trig identities
Differentiation/nature of stationary points/increasing and decreasing functions
Something on proof, even though there was one last time
Circles on graphs
Matrices (including combinations of transformations)
Quadratic nth term (please!!! :biggrin:)
F(x) graphs where they give you domains and you have to plot the ranges - absolutely love them!

Things I really don't want:
Angle between side and plane/angle between two planes
Those things that give you something like 5sin²x = 125, find all values of X. Do you know the ones I'm on about? (Obviously you wouldn't get that one as it wouldn't work, was for explanation purposes only)
Can't really think of anything else I hate...
Original post by Chittesh14
Lol, I don't even know what trig identities are. RIP.


You're given the only ones we need to know on the formula page
Original post by NiamhM1801
You're given the only ones we need to know on the formula page


No, as in I don't even know what they are - so I wouldn't be able to recognise them.
Original post by NiamhM1801
Hmm...I don't know I don't mind them :tongue:
Although topics I'd like would be:
Trig identities
Differentiation/nature of stationary points/increasing and decreasing functions
Something on proof, even though there was one last time
Circles on graphs
Matrices (including combinations of transformations)
Quadratic nth term (please!!! :biggrin:)
F(x) graphs where they give you domains and you have to plot the ranges - absolutely love them!

Things I really don't want:
Angle between side and plane/angle between two planes
Those things that give you something like 5sin²x = 125, find all values of X. Do you know the ones I'm on about? (Obviously you wouldn't get that one as it wouldn't work, was for explanation purposes only)
Can't really think of anything else I hate...


Yeah, I know what you mean, I don't like the angles between a side and a plane either because I can never picture what the angle would be. I'd love quadratic nth terms to come up, and circle on a graphs (maybe with a straight line intersecting the circle twice and in other words using simultaneous equations to work out the values of x and y, I dont even know if thats possible but it seems like an interesting and juicy question I'd like)
Original post by Chittesh14
No, as in I don't even know what they are - so I wouldn't be able to recognise them.


Oh...I think questions on them are normally worth 3 marks.
If you have a look in some past papers they're normally near the back - you'll just see a lot of tan, sin, cos and 1s normally in a fraction but not always and be asked to prove something.
Reply 147
Does anyone have a link to the paper in pdf form that I can view and maybe have a look over for myself.
Original post by mattedwards8972
Yeah, I know what you mean, I don't like the angles between a side and a plane either because I can never picture what the angle would be. I'd love quadratic nth terms to come up, and circle on a graphs (maybe with a straight line intersecting the circle twice and in other words using simultaneous equations to work out the values of x and y, I dont even know if thats possible but it seems like an interesting and juicy question I'd like)


Yeah - that's the hardest part for me too! Normally I actually draw out the little triangle of sides to try and find the angle, usually it's just a bit of basic trig after that.
And yeah I've seen that a few times actually, that'd be pretty good too!
Original post by jay1212
Does anyone have a link to the paper in pdf form that I can view and maybe have a look over for myself.


Which paper?
Original post by mattedwards8972
Yeah, I know what you mean, I don't like the angles between a side and a plane either because I can never picture what the angle would be. I'd love quadratic nth terms to come up, and circle on a graphs (maybe with a straight line intersecting the circle twice and in other words using simultaneous equations to work out the values of x and y, I dont even know if thats possible but it seems like an interesting and juicy question I'd like)


Find the points of intersecgtion of the ellipses with equations
(x2)2+5y2=4 (x-2)^2 +5y^2 =4
and
2(x1)2+y2=1 2(x-1)^2 +y^2 =1 .

[Give the each of the coordinates as exact answers.]

You said 'juicy' simultaneous equations.
(edited 7 years ago)
Original post by NiamhM1801
Oh...I think questions on them are normally worth 3 marks.
If you have a look in some past papers they're normally near the back - you'll just see a lot of tan, sin, cos and 1s normally in a fraction but not always and be asked to prove something.


It's fine, I'll just learn them and that's it lol tbh.
Original post by Ano123
Find the points of intersecgtion of the ellipses with equations
(x2)2+5y2=4 (x-2)^2 +5y^2 =4
and
2(x1)2+y2=1 2(x-1)^2 +y^2 =1 .

[Give the each of the coordinates as exact answers.]

You said 'juicy' simultaneous equations.


x^2 - 4x + 4 + 5y^2 = 4
x^2 - 4x + 5y^2 = 0

2(x^2 - 2x + 1) + y^2 = 1
2x^2 - 4x + 2 + y^2 = 1
2x^2 - 4x + 1 + y^2 = 0

x^2 - 4x + 5y^2 = 2x^2 - 4x + 1 + y^2
0 = x^2 - 4y^2 + 1
y^2 = -2x^2 + 4x - 1
0 = x^2 - 4(-2x^2 + 4x - 1) + 1
0 = x^2 + 8x^2 - 16x + 4 + 1
0 = 9x^2 - 16x + 5

Using the quadratic formula:

x=±113x = \pm \frac{\sqrt 11}{3}

5y^2 = -x^2 + 4x
5y^2 = - -(\frac{\sqrt 11}{3}) - 4(\frac{\sqrt 11}{3})
5y^2 = - -(11/9) - 4 \sqrt 11 + 12
5y^2 = 11/9 + 4 \sqrt 11 + 12
5y^2 = - 0.04
leave it, i give up on that one.

or otherwise

5y^2 = - (\frac{\sqrt 11}{3}) + 4(\frac{\sqrt 11}{3})
5y^2 = -11/9 + 4 \sqrt 11 + 12
5y^2 = 24.04
y^2 = 4.8
y = 2.19
Original post by Ano123
x


Sorry, are these answers:

x = 0.40, y = -0.54
x = 0.40, y = 0.54
x = 1.37, y = -0.85
y = 1.37, y = 0.85
(edited 7 years ago)
Original post by NiamhM1801
Yeah - that's the hardest part for me too! Normally I actually draw out the little triangle of sides to try and find the angle, usually it's just a bit of basic trig after that.
And yeah I've seen that a few times actually, that'd be pretty good too!


I think this exam will go well though, I really hope there is something to do with simultaneous equations though because I really did revise that
Original post by Chittesh14
x^2 - 4x + 4 + 5y^2 = 4
x^2 - 4x + 5y^2 = 0

2(x^2 - 2x + 1) + y^2 = 1
2x^2 - 4x + 2 + y^2 = 1
2x^2 - 4x + 1 + y^2 = 0

x^2 - 4x + 5y^2 = 2x^2 - 4x + 1 + y^2
0 = x^2 - 4y^2 + 1
y^2 = -2x^2 + 4x - 1
0 = x^2 - 4(-2x^2 + 4x - 1) + 1
0 = x^2 + 8x^2 - 16x + 4 + 1
0 = 9x^2 - 16x + 5

Using the quadratic formula:

x=±113x = \pm \frac{\sqrt 11}{3}

5y^2 = -x^2 + 4x
5y^2 = - -(\frac{\sqrt 11}{3}) - 4(\frac{\sqrt 11}{3})
5y^2 = - -(11/9) - 4 \sqrt 11 + 12
5y^2 = 11/9 + 4 \sqrt 11 + 12
5y^2 = - 0.04
leave it, i give up on that one.

or otherwise

5y^2 = - (\frac{\sqrt 11}{3}) + 4(\frac{\sqrt 11}{3})
5y^2 = -11/9 + 4 \sqrt 11 + 12
5y^2 = 24.04
y^2 = 4.8
y = 2.19


I doubt this will come up in the exam??
Original post by guy321
I doubt this will come up in the exam??


Obviously not :smile:, I'm probably wrong anyway lool.
Original post by guy321
I doubt this will come up in the exam??


Something similar may come up though with a quadratic and a linear straight line where it intersects the quadratic at 2 points and it might ask you to work out the co-ordinates of where both points intersect. That is a much more likely question.
Original post by Chittesh14
x^2 - 4x + 4 + 5y^2 = 4
x^2 - 4x + 5y^2 = 0

2(x^2 - 2x + 1) + y^2 = 1
2x^2 - 4x + 2 + y^2 = 1
2x^2 - 4x + 1 + y^2 = 0

x^2 - 4x + 5y^2 = 2x^2 - 4x + 1 + y^2
0 = x^2 - 4y^2 + 1
y^2 = -2x^2 + 4x - 1
0 = x^2 - 4(-2x^2 + 4x - 1) + 1
0 = x^2 + 8x^2 - 16x + 4 + 1
0 = 9x^2 - 16x + 5

Using the quadratic formula:

x=±113x = \pm \frac{\sqrt 11}{3}

5y^2 = -x^2 + 4x
5y^2 = - -(\frac{\sqrt 11}{3}) - 4(\frac{\sqrt 11}{3})
5y^2 = - -(11/9) - 4 \sqrt 11 + 12
5y^2 = 11/9 + 4 \sqrt 11 + 12
5y^2 = - 0.04
leave it, i give up on that one.

or otherwise

5y^2 = - (\frac{\sqrt 11}{3}) + 4(\frac{\sqrt 11}{3})
5y^2 = -11/9 + 4 \sqrt 11 + 12
5y^2 = 24.04
y^2 = 4.8
y = 2.19


What I would of done is make both equations equal to -y^2 so then you could make them equal to eachother, rearrange to make it equal to 0, factorise (either normally or through the quadratic formula) and then substitute it back into the original equation like you did afterwards. But I could be completely wrong on that, if I am, lets blame it on being a long day ay :smile:
Original post by mattedwards8972
What I would of done is make both equations equal to -y^2 so then you could make them equal to eachother, rearrange to make it equal to 0, factorise (either normally or through the quadratic formula) and then substitute it back into the original equation like you did afterwards. But I could be completely wrong on that, if I am, lets blame it on being a long day ay :smile:


I don't even know what I done.
I done it instead of revising for my Physics exam -__

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