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# Maths Ocr 2016 C1

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1. (Original post by Buymoria)
Just did the June 2007 FP1 paper (grade boundary for an A was 54) and was far far easier
Lets just hope we get really low grade boundaries XD
2. how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?
3. What was the question number 10? about the second tangent?
4. Question 1?! Please someone help
5. How many marks was question 8 worth? It was the complete the square question.
6. By getting an A in this module, do people mean above 80 ums?
7. (Original post by student199919)
how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?
Expand the brackets on the right to get x^2 +2x +11= 2kx-k

Move to one side to get = 0, so x^2 +(2- 2k)x+11+k= 0

Sub the values into b^2-4ac > 0
a=1, b=2-2k, c=11+k

Then solve to get k^2-3k-10 = 0, factorise to get (k-5)(k+2)>0,

Draw a quadratic graph that intercepts x axis at k=5 + k=-2, so k<-2 and k>5

Hope this helps
8. (Original post by DanMcGrathKB)
Question 1?! Please someone help
I think it was 2x^2-9
9. Anyone got kik? NEED TO DISCUSS ANSWERS
10. hoping c2 is better 🙌🏽
11. (Original post by liziepie)
Expand the brackets on the right to get x^2 +2x +11= 2kx-k

Move to one side to get = 0, so x^2 +(2- 2k)x+11+k= 0

Sub the values into b^2-4ac > 0
a=1, b=2-2k, c=11+k

Then solve to get k^2-3k-10 = 0, factorise to get (k-5)(k+2)>0,

Draw a quadratic graph that intercepts x axis at k=5 + k=-2, so k<-2 and k>5

Hope this helps

Thank you! i followed all those steps but i got 2 surds as my answers rather than -2 and 5
12. (Original post by student199919)
how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?
That was one of the questions I didn't think was too bad but that I most likely got wrong anyway. Basically put everything on one side so that would be:
X^2 +2X -2kx +11 +K=0 (think I went wrong here put +2KX and -K aka simple adding and subtraction error :/)
You then you the discriminant and I believe the question specified that there would be 2 real roots so b^2-4ac>0.
Substitute in the values of a, b and c:
(2+2K)^2 -4x1x(11+K)>0
4K^2 +8K +4 -44-4K>0
4K^2+4K-44>0
Now what I did here was divide by 4 which should be correct.
K^2 +K -11>0
So as this is a positive K squared graph, I said there would be 2 separate inequalities as the graph would dip below the x-axis therefore 2 distinct areas where K is greater than 0.
I also used completing the square to solve this which may not have been a good idea...anyway I got:
(k+0.5)^2 -0.25-11>0
K>-0.5+11.25^0.5
K<-0.5-11.25^0.5
(^0.5 slightly award to write compared to square root but I don't know how to express that in the forum's text).
So yeah don't think this is right and I'm pretty sure I made a stupid error in putting 10 instead of 11 or something so I would have lost the one of the method marks I may have got for that last step....
13. Thought was harder than usual but alright overall, managed to get the last question 👍
14. what are the predictions for grade boundaries?
15. How many marks do you think I will get for the last question I got 18root27/12 which is equivalent to 27 but will I lose marks for not having it as a whole number? Thanks
16. One the last question I got as far as a=8x^3 but after that I am sure my method was the incorrect way of getting to the answer, hence my answer was definitely wrong, being a=216 XD.
17. (Original post by black1blade)
That was one of the questions I didn't think was too bad but that I most likely got wrong anyway. Basically put everything on one side so that would be:
X^2 +2X -2kx +11 +K=0 (think I went wrong here put +2KX and -K aka simple adding and subtraction error :/)
You then you the discriminant and I believe the question specified that there would be 2 real roots so b^2-4ac>0.
Substitute in the values of a, b and c:
(2+2K)^2 -4x1x(11+K)>0
4K^2 +8K +4 -44-4K>0
4K^2+4K-44>0
Now what I did here was divide by 4 which should be correct.
K^2 +K -11>0
So as this is a positive K squared graph, I said there would be 2 separate inequalities as the graph would dip below the x-axis therefore 2 distinct areas where K is greater than 0.
I also used completing the square to solve this which may not have been a good idea...anyway I got:
(k+0.5)^2 -0.25-11>0
K>-0.5+11.25^0.5
K<-0.5-11.25^0.5
(^0.5 slightly award to write compared to square root but I don't know how to express that in the forum's text).
So yeah don't think this is right and I'm pretty sure I made a stupid error in putting 10 instead of 11 or something so I would have lost the one of the method marks I may have got for that last step....
Thank you. I was going on the right lines but I simplified -8k - -4k as -12k rather than -4k which is annoying
18. (Original post by Bobby21231)
How many marks do you think I will get for the last question I got 18root27/12 which is equivalent to 27 but will I lose marks for not having it as a whole number? Thanks
did it ask to give a as an integer?
19. Any mark-schemes yet?
20. No the question told us that "a is a positive constant"

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