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# AEA June 2011 watch

1. (Original post by superkinetic)
Referring to the attached diagram Qn 6(d) the graph y=f(x-2)+2, the red line is a reflection of the green line on line y=x, both lines have equal gradient, 4, as stated in the question. On the graph y=f(x-2)+2, if the red line has equation y=4x-h, then the green line has equation y=4x+h, as both lines are reflections of each other. This means you have to find the equation of the red line on the graph y=f(x-2)+2 to find the equation of the green line on graph y=f(x-2)+2. Then, you work back to find the equation of the green line on graph y=f(x).
Thank you! I get it now. but I still hope a question like it doesn't come up tomorrow.

Has anyone else noticed them mistake on the 07 paper it says half of x instead of half of pi.
2. Can someone explain me whole 2006 question 7(about circles)
and q1 part b) about convergent thing..
3. (Original post by cazzy-joe)
Can someone explain me whole 2006 question 7(about circles)
and q1 part b) about convergent thing..
Which part of Qn 7 do you not understand? What do you wish to ask regarding Qn 1?
4. (Original post by superkinetic)
Which part of Qn 7 do you not understand? What do you wish to ask regarding Qn 1?
Question 7 actually I dont understand part b) onwards...

Qn1 part c) set of values of x in which series in convergent.
5. (Original post by cazzy-joe)
Question 7 actually I dont understand part b) onwards...

Qn1 part c) set of values of x in which series in convergent.
Have you looked through the mark scheme? I think it will provide a clearer explanation.

But anyway for Qn 7, the radius of the circles form a geometric series, as mentioned in (a). Thus, the sum of all circles is the sum of a geometric series. For (c), you are required to find S = Area of PAOB - Value in (b). For (d), you are to differentiate S. For the last part, the question wants the least value of S for the given interval. So you have to look at (d) to deduce if the derivative is negative for the given interval, meaning the function is decreasing.

For Qn 1, from (b), , which means , if you recall the condition for a binomial series to be convergent, which is for , where n<0.
6. (Original post by Ultimate1)
Erm jun 2010 - 81 - disintction, 61 - merit
Woah! if thats accurate, then its gone up by ~10% as for the last few years before 2009 Distinction has been ~70%!! Was the June 2010 paper THAT much easier?
7. (Original post by saajan_92)
Woah! if thats accurate, then its gone up by ~10% as for the last few years before 2009 Distinction has been ~70%!! Was the June 2010 paper THAT much easier?
Yeah it was quite easy. I only really got stuck on 1 Q.
8. Can someone explain Q3c on the June 2006 paper?
9. (Original post by jp241)
Can someone explain Q3c on the June 2006 paper?
The first equation rearranges to y=x^-1 as deduced from the earlier part. Then for the second equation, let logx(x-y) = z and let logy(x+y) = z

Then try and 'cancel' the z's after rearranging and sub in y=x^-1.
10. (Original post by jp241)
Can someone explain Q3c on the June 2006 paper?
Well the first equation is

11. How likely is it that a question will come up that is similar to one that has already come up? I don't mean the chapter/topic but rather the method of working out? not sure if i'm revising right for this, i'm just going through a few past papers, and remembering how to do questions I can't answer (and sometimes quickly look through the chapter in my books)

Also i'm a bit confused on question 7 of 2008. Part b, the equation of the bisector, the mark scheme makes sense but why can't the direction be 0.5(BC + BA) ? If it bisects ABC then it bisects the line AC?
In part D, for their other line, L1, they seemed to have just used 0.5 (AB + AC) so why doesn't this work for the bisector line in part b?
12. It's a shame this exam is right at the end of all my other exams as I haven't really done much work towards it as I've been busy with other exams. But I'm still looking forward to it
13. (Original post by atofu)
Heya, does anyone have the 2010 paper and mark scheme? It's secure/locked on the edexcel website

Also, what things will I need to know how to do without a calculator? xD All I can think of so far are:
-Trig functions for 30, 45, 60, 90 etc.
- Working with surds
Could you work out for example tan(pi/8) ?
14. (Original post by ian.slater)
Could you work out for example tan(pi/8) ?
Use tan(pi/8+pi/8) formulae and then rearrange to get a quadratic in tan (pi/8)?
15. (Original post by ian.slater)
Could you work out for example tan(pi/8) ?
Hmmm would it have something to do with tan (A-B)? can't think of what angles would work though :/
16. (Original post by soczek322)
Use tan(pi/8+pi/8) formulae and then rearrange to get a quadratic in tan (pi/8)?
Looks like the best option. Although you might look at tan(pi/8 - pi/4) and so on.

Your quadratic gives you two roots. Which do you choose and what does the other one represent?
18. So who's ready for this? As long as no dodgy vectors come up and series (God I hate them ) I think I'm good to go.
19. (Original post by ian.slater)
Looks like the best option. Although you might look at tan(pi/8 - pi/4) and so on.

Your quadratic gives you two roots. Which do you choose and what does the other one represent?
I got to tan(pi/8) = +/- root(2) -1 and because tan(pi/8)>0 I got for the plus sign but I'm not sure what the other solution represents?

Thanks for the help!
20. I think I heard someone say that if you can get an A* in regular maths you should be able to get a merit on AEA. Let's just hope that's the case.
Does anyone else think the 2007 paper is ridiculously more difficult than the other papers? I start crying each time I look at it.

Also I hope the vectors question isn't like the kite one. If I knew that the sides of the kite were so that there were two pairs of similar sides then the question becomes very easy, but if you don't know that you lose almost all of the marks . Quite scary, hope nothing like that comes up tomorrow.

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