Uhm its effectively the same graph yes, because it has the same roots. But if you think of it in terms of transformations in C2, the original factorised form is f(x).(Original post by Stephybob)
Hmm you're probably right but seeing as both are f(x) then it shouldn't matter which one you use if they equal the same thing? As both should give the same y intercept? Unless they wanted us to find g(x) or something which has the same roots as f(x) but a different intercept?
What we've then factorised it into is 2(fx) i.e. a stretch s.f. 2 parallel to the yaxis (basically a vertical stretch scale factor 2). This means where it crosses the xaxis will remain the same (only a vertical stretch) but where it crosses the yaxis, that value will double.
I don't really know if I've explained this clearly, but that's my interpretation of the difference..
C1, OCR MEI  16/05/12 (Answers)
Announcements  Posted on  

Live webchat: Student Finance explained  on TSR from 2  3pm  17092014  
Got a question about Student Finance? Ask the experts this week on TSR!  14092014 


(Original post by thesaintx)
Uhm its effectively the same graph yes, because it has the same roots. But if you think of it in terms of transformations in C2, the original factorised form is f(x).
What we've then factorised it into is 2(fx) i.e. a stretch s.f. 2 parallel to the yaxis (basically a vertical stretch scale factor 2). This means where it crosses the xaxis will remain the same (only a vertical stretch) but where it crosses the yaxis, that value will double.
I don't really know if I've explained this clearly, but that's my interpretation of the difference..
I must admit I wasn't sure so on the last two parts of 11 I wrote the alternative and put a neat line through it, just in case they want to give me the benefit of the doubt... 
I'm sure you'll do well its only 1 question, and C1s kinda tricky. I think this paper was quite a hard one imo (in comparison), so I'm sure you'll be ok

I think in question 9, if (n+3)^2n^2 was divisible by 9, n was a multiple of 3, not a factor of 3, i believe

(Original post by Jack:))
Judging on previous papers my geuss would be 63 Max and 58 Min but personally im thinking around 61. Of course it is hard to tell
for A im guesing 5758 tbh
:/
havent seen a paper wiv 60 for an A o.o
i found this was a medium difficulty paper :P
last year's june 11 was 55 
https://docs.google.com/viewer?a=v&p...Q8JRE3A&pli=1
Here is the question paper. 
for question 2, can you right it as SQRT(3/2a)/c?
or does it have to be written as SQRT(3a)/(2c)?
are they the same thing or will i lose marks for writing it in the first form?
Any help appreciated. 
(Original post by cloudhang)
i doubt it'll be that high ...i did an really easy paper a few days ago and it was 58 for an A..
for A im guesing 5758 tbh
:/
havent seen a paper wiv 60 for an A o.o
i found this was a medium difficulty paper :P
last year's june 11 was 55 
(Original post by kimmar)
I think in question 9, if (n+3)^2n^2 was divisible by 9, n was a multiple of 3, not a factor of 3, i believe 
(Original post by Jack:))
Yes i see, but i found this paper really easy. I guess it is hard too tell. there has been a 60 and a few 58s. i would love it to be 55, fingers crossed!
I looked through all grade boundaries from 08 to jan 12. The lowest was 52 for an A and highest 62 for an A. I think it will be around 59 like Jan 
Issue with question 7
"SQRT(5) < K < SQRT(5)"
Shouldn't it be two separate equalities: K > Sqrt5 and K<Sqrt5 
whats the paper out of, like how many marks?

(Original post by Dillanp1)
Issue with question 7
"SQRT(5) < K < SQRT(5)"
Shouldn't it be two separate equalities: K > Sqrt5 and K<Sqrt5 
(Original post by Jack:))
I dont think so, i believe the question was for when there was no intersection which would mean SQRT(5) < K < SQRT(5) although if you mean shoud it just be written as 2 equalities then i believe it does not matter. 
(Original post by Dillanp1)
I'm pretty sure if K was a value between sqrt5 and sqrt5 then it would intercept the axis, however if K was either bigger then sqrt5 or smaller then sqrt5 then there will be no roots.
I will double check my working tomoro and hopefully revise the answers! 
(Original post by Stephybob)
I have an issue with your question 11, if you take the original factorised form of (x+1/2)(x+2)(x5) then the y intercept would be 5 not 10. I put the equations 2x^3 5x^2 23x 10 and the expanded form of the factorised equation into autograph and they aren't the same. So to get from the factorised equation to one with a coefficient of 2, you can't just double everything or I believe you get a false y intercept (but no idea how you actually get there).
So if I'm correct, the answer for the y intercept of f(x) 8 is (0, 13) and the answer to f(x3) is 20?
Sorry if someone has already pointed this out
To get it in the form of 2x^3 + bx^2 + cx + d as required by the question, you need to have one of your factors as (2x + h) where h is a constant.
(x + .5) => (2x + 1)
Otherwise you don't get 2x^3, you only get x^3 which wasn't asked for in the question. 
(Original post by Dillanp1)
I'm pretty sure if K was a value between sqrt5 and sqrt5 then it would intercept the axis, however if K was either bigger then sqrt5 or smaller then sqrt5 then there will be no roots.
Take k=1,
Therefore, b^24ac:
=2^2(4*1*5)
=420
=16
The discriminant is less than 0, so no real roots so no intersections with the xaxis. 
(Original post by Jack:))
Yes i see, but i found this paper really easy. I guess it is hard too tell. there has been a 60 and a few 58s. i would love it to be 55, fingers crossed! 
(Original post by kimmar)
The original equation was x^2+2kx+5 and the answer was sqrt5<k<sqrt5, *because if you take 3 as a value of k, for example, which is bigger than sqrt5, the discriminant of the equation becomes to be bigger than 0 which means that there are 2 real roots. also if you take 3 as a value of k, which is smaller than sqrt5, the discriminant also gives a value greater than 0. The question asks for range of values of k where the equation doesn't intersect the xaxis therefore it has to be sqrt5<k<sqrt5 and you can try any values in between and the discriminant will always be less than 0.*
Take k=1,
Therefore, b^24ac:
=2^2(4*1*5)
=420
=16
The discriminant is less than 0, so no real roots so no intersections with the xaxis. 
For question 9, n is a factor of 1.5 also works.
And for question 8, I got something completely different
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: