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

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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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Reply 61

Music99

Original post by purplemind

I got -11.

Me too.

Reply 62

_JC95

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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Reply 63

Good@Math

My answers here. Hope it helps http://www.thestudentroom.co.uk/showthread.php?t=2354287&p=42661909#post42661909

Reply 64

stefanconstant

Original post by _JC95

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..

Reply 65

mia_hilton

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 :frown:

Reply 66

mia_hilton

anyone want to hazard a guess at grade boundaries?

Reply 67

stefanconstant

Original post by Good@Math

My answers here. Hope it helps http://www.thestudentroom.co.uk/showthread.php?t=2354287&p=42661909#post42661909

Could you just quote a value of a and b because that is what it said?

Reply 68

Jtw95

That log question was horrifichorrific

Reply 69

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!

Reply 70

niaphonic

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

Reply 71

DoeADeer

Oh nooooo. I put 1.9. Are you sure it's not 1.9???

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Reply 72

stefanconstant

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

Reply 73

niaphonic

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

Reply 74

_JC95

Original post by stefanconstant

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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Reply 75

niaphonic

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..

Reply 76

stefanconstant

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!

Reply 77

stefanconstant

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?

Reply 78

flyhigh99

Original post by _JC95

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,
i got different answers :/