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# OCR (Not MEI) Core 2 Discussion Thread 17th May 2013 + Jan 13 Paper and MS Watch

1. I tried to do the tangent area under the graph question 2 ways...

First I got 37/30 by subtracting the area of the right angled triangle.

Then, to check, i tried integrating. (x^3/2 - 1) - (3x-5) and got a completely different answer of 1.9

Surely both methods are valid and should both produce the exact same answer??

In the end I stuck with the original 37/30, the rest of the paper was pretty fair.

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2. (Original post by purplemind)
I got -11.
Me too.
3. (Original post by mia_hilton)
but how did you know the coordinate of the tangent when it hit the x axis?
The equation of the tangent was y=3x-5, so when y=0, 3x=5, x=5/3 so coordinates are (5/3, 0)

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4. My answers here. Hope it helps http://www.thestudentroom.co.uk/show...9#post42661909
5. (Original post by _JC95)
I tried to do the tangent area under the graph question 2 ways...

First I got 37/30 by subtracting the area of the right angled triangle.

Then, to check, i tried integrating. (x^3/2 - 1) - (3x-5) and got a completely different answer of 1.9

Surely both methods are valid and should both produce the exact same answer??

In the end I stuck with the original 37/30, the rest of the paper was pretty fair.

Posted from TSR Mobile
Yeah 37/30 is the right answer!
If you integrated then you had to use the coord where the tangent crosses the x axis which was 5/3
Therefore you integrate between 5/3 and 1
You used 4 and 1 which is why you got 1.9 because I did that too but realised..
6. (Original post by _JC95)
The equation of the tangent was y=3x-5, so when y=0, 3x=5, x=5/3 so coordinates are (5/3, 0)

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thanks it makes so much sense........wishing i could turn back time right now
7. anyone want to hazard a guess at grade boundaries?
8. Could you just quote a value of a and b because that is what it said?
9. That log question was horrifichorrific
10. Okay that was horrible! At least most people found it hard so hopefully low grade boundaries! What was going on with the big log question!? Had no clue!
11. (Original post by stefanconstant)
Could you just quote a value of a and b because that is what it said?
Yeah it asked for possible values for a and b, so as long as your a was >1 and 0<b<1
12. Oh nooooo. I put 1.9. Are you sure it's not 1.9???

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13. (Original post by niaphonic)
Yeah it asked for possible values for a and b, so as long as your a was >1 and 0<b<1
Sweet I said a=2 and b= 1/2
14. (Original post by stefanconstant)
Sweet I said a=2 and b= 1/2
Should be fine, I put a=1.5 b=0.5 too. I just tried out random values that looked roughly right, brain wasn't working... nearly wrote b= -1
15. (Original post by stefanconstant)
Yeah 37/30 is the right answer!
If you integrated then you had to use the coord where the tangent crosses the x axis which was 5/3
Therefore you integrate between 5/3 and 1
You used 4 and 1 which is why you got 1.9 because I did that too but realised..
Yeah but in dying minutes i also did it with 1 and 5/3 and did not get the same answer either...

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16. (Original post by Cowcat)
Okay that was horrible! At least most people found it hard so hopefully low grade boundaries! What was going on with the big log question!? Had no clue!
It threw me the first time so I left it to the end and tried again, managed it the second time! The question showed you had to use log to base 2, and then had to use ab=2 to simplify the expression at the end (ie. replace b with 2/a). I forgot both of those the first time I tried to do it..
17. (Original post by niaphonic)
Should be fine, I put a=1.5 b=0.5 too. I just tried out random values that looked roughly right, brain wasn't working... nearly wrote b= -1
Haha yeah, thanks!
18. (Original post by _JC95)
Yeah but in dying minutes i also did it with 1 and 5/3 and did not get the same answer either...

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Well 37/30
is the right answer so you're fine..
Did you integrate it right?
19. (Original post by _JC95)
I tried to do the tangent area under the graph question 2 ways...

First I got 37/30 by subtracting the area of the right angled triangle.

Then, to check, i tried integrating. (x^3/2 - 1) - (3x-5) and got a completely different answer of 1.9

Surely both methods are valid and should both produce the exact same answer??

In the end I stuck with the original 37/30, the rest of the paper was pretty fair.

Posted from TSR Mobile

I'm on the same boat as you,

the rest of the paper was fine,
except when i wanted to make sure my answer was right for that question,
20. How many marks would I get for putting 1.9 for that graph question?

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