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

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yeah I'd agree! some questions are repetitive though... hoping for some nice integrals

Some are but nothing like FP3 where there was basically only 3 topics in the entire unit. Hope so too

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Reply 81

Jm098

Original post by PiTheta97

Some are but nothing like FP3 where there was basically only 3 topics in the entire unit. Hope so too

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can't wait to see what they throw M2 for the last question

Reply 82

Kedric123

Anyone mind explaining some of the harder questions to me if I post the ones I'm struggling with?

Reply 83

Roxanne18

Hey can some one help me with the proof by induction question on the 2015 paper please ?
I've attached what I've managed to do so far, but don't understand how to simplify

Reply 84

Jm098

Original post by Kedric123

Anyone mind explaining some of the harder questions to me if I post the ones I'm struggling with?

yeah i will

Reply 85

Jm098

Original post by Roxanne18

Hey can some one help me with the proof by induction question on the 2015 paper please ?
I've attached what I've managed to do so far, but don't understand how to simplify

16 is 2^4 so the 2^4k+7 s cancel :smile:

Reply 86

Roxanne18

Original post by Jm098

16 is 2^4 so the 2^4k+7 s cancel :smile:

Oh wow can't believe I didn't see that, thank you !

Reply 87

TheLifelessRobot

Original post by girlruinedme

Someone please explain the factorials in the 2015 paper for

(r+1)! can be expressed as (r+1)r!, it can be expressed as that because for example 4! = 4*3! because 4! = 4*3*2*1 = 4*3!. (r+2)! can be expressed as (r+2)(r+1)r! just like how 4! = 4*3*2!.

Here's some working for the question.

factorials.png71.4KB

Reply 88

Buntar

Could someone kindly explain june 11 q4) aiv), bi) and bii) - I don't understand the method shown in the solutions

Screen Shot 2016-06-23 at 00.01.01.png

Thanks in advance!

Reply 89

C0balt

Where did they get the results in the mark scheme for June 14 4bii?

Attachment not found

image.jpg25.5KB

Reply 90

IrrationalRoot

Original post by C0balt

Where did they get the results in the mark scheme for June 14 4bii?

Attachment not found

Honestly I have no idea. I just used the method shown to the right.

Reply 91

IrrationalRoot

Original post by Buntar

Could someone kindly explain june 11 q4) aiv), bi) and bii) - I don't understand the method shown in the solutions

Screen Shot 2016-06-23 at 00.01.01.png

Thanks in advance!

\alpha

satisfies the equation since it is a root. But so are

\beta

and

\gamma

so they also satisfy the equation.
You want the sum of the cubes of the roots, so clearly you want to add the equations

\alpha^3-2\alpha^2+k=0

\beta^3-2\beta^2+k=0

\gamma^3-2\gamma^2+k=0

to get

\Sigma\alpha^3-2\Sigma\alpha^2+3k=0

and the result follows.

For the next two cases you want the sum of fourth powers and fifth powers, so it makes sense to multiply the equation by

x

and

x^2

respectively and do the same thing.

Reply 92

Hjyu1

Original post by IrrationalRoot

Honestly I have no idea. I just used the method shown to the right.

Do you mind explaining your method

Reply 93

IrrationalRoot

Original post by Hjyu1

Do you mind explaining your method

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.

Reply 94

sam_97

Original post by C0balt

Where did they get the results in the mark scheme for June 14 4bii?

Attachment not found

Original post by IrrationalRoot

Honestly I have no idea. I just used the method shown to the right.

Original post by Hjyu1

Do you mind explaining your method

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

(edited 7 years ago)

Reply 95

Buntar

Original post by IrrationalRoot

\alpha

satisfies the equation since it is a root. But so are

\beta

and

\gamma

so they also satisfy the equation.
You want the sum of the cubes of the roots, so clearly you want to add the equations

\alpha^3-2\alpha^2+k=0

\beta^3-2\beta^2+k=0

\gamma^3-2\gamma^2+k=0

to get

\Sigma\alpha^3-2\Sigma\alpha^2+3k=0

and the result follows.

For the next two cases you want the sum of fourth powers and fifth powers, so it makes sense to multiply the equation by

x

and

x^2

respectively and do the same thing.

Thank you!

Reply 96

C0balt

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

Thanks, I see it but how did you spot going from line 3 to 4 without seeing the ms

Reply 97

sam_97

Original post by C0balt

Thanks, I see it but how did you spot going from line 3 to 4 without seeing the ms

No problem. I haven't done the paper for a while so I can't remember too clearly, I think I just presumed that the terms in line 3 might factorise somehow with all the squared terms knocking around.