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

1. What does anyone think my marks will be, if I got the wrong value for k. But my method for that and the rest of question 10 were correct. (Got the wrong value for k as I STUPIDLY multiplied -3 and 9 to get....-18. Giving k=4) This is why I inquired about ECF marks before.
2. (Original post by a10)
you sure about that bro? I don't think so. It states on the mark schemes the correct method and if its not on there you'll simply lose those method marks but still gain an answer mark. It wouldn't be fair on first year students if people got awarded full marks for using a method not taught in C1.
No, you're wrong. Look over a question which requires you to find the gradient at a point on a circle (possibly the tangent or normal at that point, they're all the same), and then find the mark scheme for it. The mark scheme includes implicit differentiation. If this technique - the tools for which come from C3, and which is not taught directly as far as I know in any of OCR's modules - gets the marks, why would factorisation (which comes from C2) not be allowed?
3. (Original post by Majeue)
Unofficial markscheme in above post.
Cheers
Do you remember the question for either this, or Question 8?
Yeah, you were told point A (2,2) was on the circumference, and had to find the co-ordinates of point B. Line AB was the diameter of the circle which was 2rt.40
5. (Original post by Majeue)
Here's the unofficial markscheme, fairly sure it's all correct:
1)i) 12 rt.5
ii) 4 rt.5
iii) 5 rt.5

2) x^3 = 1/8 and -1 ... therefore x = 1/2 and -1

3) f'(x) = -(12x^-3) +2
f''(x) = 36x^-4

4)i) 3(x+3/2)^2 + 13/4 [Because 3(3/2)^2 = 3(9/4) = (27/4) .. so then take that from the ten to give 13/4]
ii) Vertex is (-3/2, 13/4)
iii) b^2 - 4ac = 9^2 - 120 = -39

5)i) Basically the same graph as 1/x^2... Curves symmetrical and never touch any of the axes
ii) Stretch in Scale Factor 1/2 in y (vertical) direction

6)i) x^2 + (y+4)^2 = 40 Centre is (0, -4) Radius is rt.40 = 2 rt.10
ii) B co-ordinate is (-2, -10)

7)i) x < -1/8
ii) 0 </= x </= 5 ( </= means greater than or equal to! solved this one by drawing out the curve.. )

8) m of perpendicular is 1/3, so m of line is -3.. put into formula to get 3x + y -1/2 = 0 .. x2 to get into integers, 6x + 2y - 1 = 0

9)i) Positive quadratic curve (smiley face, not sad face). Intercepts y at (0,-6) and x at (-3/2,0) and (2,0)
ii) Vertex at x=1/4 so function is decreasing for x < 1/4
iii) Submit into formula for curve, use quadratic formula to solve, find that points P and Q are (-2,4) and (2.5,4) so distance is 4.5

10)i) Solve to find that k = -5
dy/dx = -3x^2 - 6x + 4 - k
0 = -3(-3)^2 - 6(-3) + 4 - k
k = -27 + 18 + 4 = -5

ii) d2y/dx2 = -6x - 6 ... submit -3 in to get 12, 12>0 therefore it's a minimum point
iii) This one was really long winded for 5 marks, basically you had to put the formulas together, solve the quadratic to find the two points on the cubic curve that satisfied a gradient of 9, you then put those two points (0 and -2) into the y=9x-9, and see which works. 0 doesn't work, but -2 does, so A = (-2, -27)
What was question 9iii?
6. How did you do the decreasing function question? I found the minimum point and just said that x<min point ( cannot remember co-ord?

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7. Is it possible if someone could maybe put up some of the actual questions? My mind has gone completely blank to some of them and I'm having trouble trying to work out roughly what I got! thanks
How did you do the decreasing function question? I found the minimum point and just said that x<min point ( cannot remember co-ord?

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That should be correct. X<x co ord of min point = x<1/4
9. Lol i did the entirety of question 3 wrong. went through it three times as well. seems like im always guaranteed to lose a few marks for being so headstrong about a particular answer
10. The only thing I did wrong was to put root40 instead of 2root10... Is that worth -1 mark?
11. (Original post by BankOfPigs)
Do you mean the factorisation?

Perhaps I made a mistake, although it still delivered an answer that seems reasonable.

Mind explaining which part was wrong?

If I recall, both equations could be divided by 1-x since the initial equation already gave you it in such a form and 9x - 9 = -9 (1-x)?
Oh, I see what you've done now, didn't fully understand from your previous posts. That's definitely right! Good job
12. (Original post by Bennouhan)
The only thing I did wrong was to put root40 instead of 2root10... Is that worth -1 mark?
I don't think so because 2√10 = √40. Hope that helps

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13. (Original post by Bennouhan)
The only thing I did wrong was to put root40 instead of 2root10... Is that worth -1 mark?
It depends if it said fully simplify . . .
14. a walk in the park
15. Hi can anyone remember the actual question for 6ii and 3i????
16. (Original post by Bennouhan)
The only thing I did wrong was to put root40 instead of 2root10... Is that worth -1 mark?
Mr M confirmed on his thread that it's fine.

Can anyone tell me what Question 8 and/or 9.iii) were?
Same. I feel like crap. I so so hope you can lose 15 and get an A

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Going my Mr M i seem to have got 55-63, somewhere in between but more likely about 58/59/60- hoping 60 is an A but everyone seems to have found it easy

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18. Hi does anyone have the actual questions for:
7.ii)
4.i)
Thanks
19. What ums does 67 raw marks usually convert to?

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20. Will I lose marks for not not writing the differentiation answers in a more suitable format? For example I wrote 36x^-4 instead of 36 over x^4 and the same with the other differentiation question.

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