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# OCR (not MEI) C1 13/05/13 Watch

1. (Original post by a10)
yup k is -5
Thank you kind sir
2. (Original post by jt663)
What did people get for the length PQ? on y=4 on the x^2 graph? I got 11/2, I heard it might be 9/2 though
I got 9/2 , but I can't say for certain that i am right.
3. (Original post by a10)
wat wer the coordinates of B for 6ii

(I got -2,-10) I think cant remember if both wer negative...
this was my answer as well, through a very awkward working out procedure i might add
4. (Original post by a10)
wat wer the coordinates of B for 6ii

(I got -2,-10) I think cant remember if both wer negative...
I got that too.
5. What did you all get for the minimum value? for 3(x+1.5)^2+3.25 ???
6. Hi guys,
I accidentally put k=5 for 10i even though I did the right workings (can't add up). I also then used this for part iii with the tangent 9x-9 but It wouldn't factories as I got the wrong quadratic. How many marks will I lose?
thanks
7. (Original post by StephHowarth)
I got 9/2 , but I can't say for certain that i am right.
I got 9/2.
8. What did you all get for the minimum value? for 3(x+1.5)^2+3.25 ???
9. A lot of my smart friends got K=-5

I lost at least 12 marks. and say 3 marks for mistakes

if i lost 15 marks what grade would that be?

I had trouble with 10ii and 9iii and the K = question
10. (Original post by styles4777)
What did you all get for the minimum value? for 3(x+1.5)^2+3.25 ???
(-3/2, 13/4), something like that.

Got a horrible feeling right now, I did -18-9= -9 for some reason, if i did that for the last question and then put (-2, -9) as co-ordinates how many marks will i lose?
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11. douse any one have the paper or is any one working on an unofficial mark scheme i would really like to know what i got ruffly.
12. (Original post by eggfriedrice)
Remind me what was question two and I might remember my answer.
It was to solve a quadratic, with x^6 + x^3 + (..) or something and a typical method would be to substitute a = x^3 or something, and rewrite it as a^2 + b + c

I got x = + - 1/2

and my other x value was cuberoot of -1 ? But i mistook that for a square root though X(
13. (Original post by styles4777)
What did you all get for the minimum value? for 3(x+1.5)^2+3.25 ???
-3/2,13/4

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14. (Original post by eggfriedrice)
I got that too.
today is getting better xD so far I reckon iv got 100ums if im lucky with the grade boundaries.

wat did u get for the question that said wat values of x is the curve decreasing?
15. 3 - 8x > 4

x should be less than -1/8, e.g. if you choose -1/10 as x.

3+ (8/10) > 4 which is not true, therefor x must less than -1/4
16. (Original post by StickySteve)
It was to solve a quadratic, with x^6 + x^3 + (..) or something and a typical method would be to substitute a = x^3 or something, and rewrite it as a^2 + b + c

I got x = + - 1/2

and my other x value was cuberoot of -1 ? But i mistook that for a square root though X(
It's only +1/2

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17. (Original post by eggfriedrice)
No :s last year was 61 for an A I think.

Aha, I did that and realised root a square would just cancel out. Hmm I'm not sure, at most you'll lose one mark.
FUUUUUU. So annoyed with myself, i reckon i've got around 60/72, will that be enough for an A?

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18. For the decreasing function, -3/2<x<1?
19. (Original post by a10)
wat wer the coordinates of B for 6ii

(I got -2,-10) I think cant remember if both wer negative...
I got that!
20. Question 10iii) For the coordinates of A, I set the line equal to the curve, and then equated it to 0, then solved it using the remainder theorem, to obtain a quadratic, solved the quadratic and got X to be -2, subbed X value into the curve and got Y to be -27. So the point = (-2,-27) Would I get 5/5 marks for this method?

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