AQA A2 MFP2 Further Pure 2 – 24th June [Exam Discussion Thread]

Original post by MrMagic12345

No

And for that question from June 2014 I would just try to memorise that Sum(a^2b)=sum(alpha) x sum(alphabeta)-3alphabetagamma

Reply 102

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-QP-JUN10.PDF

On question 4(c)(i), how are you supposed to know that alpha = -2???
The mark-scheme just suddenly assumes you already know it. Is it just trial and error or something?

Reply 103

-jordan-

5

Original post by 2014_GCSE

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-QP-JUN10.PDF

On question 4(c)(i), how are you supposed to know that alpha = -2???
The mark-scheme just suddenly assumes you already know it. Is it just trial and error or something?

Yeah just trial and error

Reply 104

Original post by 2014_GCSE

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-QP-JUN10.PDF

On question 4(c)(i), how are you supposed to know that alpha = -2???
The mark-scheme just suddenly assumes you already know it. Is it just trial and error or something?

You've got a quadratic in alpha.. (You just need to put the p value you found in it)

Reply 105

Original post by -jordan-

Yeah just trial and error

Thanks man!

Also, does multiplying sinht by sinht equal -sinh^2t?

I'm confused why the equations have you switch the sign on sinh^2(t).

Reply 106

Original post by MrMagic12345

You've got a quadratic in alpha.. (You just need to put the p value you found in it)

No, you have a cubic with alpha.

Reply 107

-jordan-

5

Original post by 2014_GCSE

Thanks man!

Also, does multiplying sinht by sinht equal -sinh^2t?

I'm confused why the equations have you switch the sign on sinh^2(t).

No that's Osborne's rule, because of the way the hyperbolics in expoential form are, it allows you to convert any standard trigonometric identity into the hyperbolic equivalent. When you know sinht has been multiplied by sinht then you put a negative in front of that term in the expression. If you look at the normal identities and then their equivalent hyperbolic counterparts you'll see what I mean.

Reply 108

Original post by 2014_GCSE

No, you have a cubic with alpha.

Yeah my bad

Reply 109

Original post by 2014_GCSE

Thanks man!

Also, does multiplying sinht by sinht equal -sinh^2t?

I'm confused why the equations have you switch the sign on sinh^2(t).

You only use osborne's rule when trying to 'convert' a trig formula describing non-hyperbolic function's into a formula describing hyperbolics

Reply 110

Jm098

can someone show me their working to this question please? i know its probably really simple but the mark scheme is vague

Reply 111

Original post by IrrationalRoot

There's no method to explain, the whole method is shown in two lines to the right on the mark scheme.
If there's something specific you don't understand within it, let me know and I'll explain it.

I don't get where the -2 came from

Reply 112

Original post by Jm098

can someone show me their working to this question please? i know its probably really simple but the mark scheme is vague

Then u can do X coordinate + X coordinate /2 to get midpoint etc

Reply 113

Jm098

Original post by Hjyu1

Then u can do X coordinate + X coordinate /2 to get midpoint etc

thanks! where is the first line from though? have you just set the both mods equal?

Reply 114

IrrationalRoot

20

Original post by sam_97

This is how I did it:

(\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)[br]= (\alpha + \beta)(\alpha\beta + \beta\gamma + \gamma\alpha +\gamma^2)[br]= \alpha^2\beta + \gamma\alpha^2 + \alpha\beta^2 + \beta^2\gamma + \beta\gamma^2 + \gamma^2\alpha + 2\alpha\beta\gamma [br]= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - 3\alpha\beta\gamma + 2\alpha\beta\gamma[br]= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma[br]= \Sigma\alpha\Sigma\alpha\beta - \alpha\beta\gamma

Expanding the brackets on line 4 gives us an extra

3\alpha\beta\gamma

which we don't want, hence the

-3\alpha\beta\gamma

.

I mean, yeah it's very easy to show the equality, but we were wondering how one would recognise the equality in the first place; the method shown to the right in the mark scheme is what I used and makes a lot more intuitive sense.

Reply 115

IrrationalRoot

20

Original post by Hjyu1

I don't get where the -2 came from

-2 sum of roots.

Reply 116

Original post by Jm098

thanks! where is the first line from though? have you just set the both mods equal?

Yeah just converted to cerstian and which line

Reply 117

Original post by IrrationalRoot

-2 sum of roots.

Ohhhhhhhhhh **** I get it now thanks

Reply 118

Jm098

Original post by Hjyu1

Yeah just converted to cerstian and which line

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

Reply 119