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OCR C3 (not MEI) Official Thread - Tuesday 21st June 2016

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Reply 40
Avatar for msulli
msulli
2
Original post by marioman
I've just updated my post with the link to the 2015 paper.


Thanks :smile:
Is it me or is question 4 in the 2015 paper worth way too many marks for what it is?
Can anyone think of any small things which i might not know i should know for tomorrow?
Reply 43
Avatar for S.G.
S.G.
22
Original post by ughexams
the answer is 1/36 pi.... (for part iii)


I know
Reply 44
Avatar for r34_786
r34_786
Hi! How do you find the range and domain from the equation? I always forget!:s-smilie:
Original post by r34_786
Hi! How do you find the range and domain from the equation? I always forget!:s-smilie:


not sure if there is a perfect way of doing it, but look for important things. If you have a fraction the denominator can not be 0, a square root can not be negative etc, and for range maybe find the stationary point
Reply 46
Avatar for micycle
micycle
3
How do I show that:

1 – ln(2) = ln(1/2e)
Original post by micycle
How do I show that:

1 – ln(2) = ln(1/2e)


1 = ln(e)

so 1 - ln(2) = ln(e) - ln(2) = ln(e/2) = ln(0.5 e)
Reply 48
Avatar for r34_786
r34_786
Thank you so much! :biggrin:
Original post by LukeWeatherstone
not sure if there is a perfect way of doing it, but look for important things. If you have a fraction the denominator can not be 0, a square root can not be negative etc, and for range maybe find the stationary point
I think proof in vectors will come up. Although I've never seen one in a paper, I've seen it in my textbook 😬
Original post by ArbaazMalik
I think proof in vectors will come up. Although I've never seen one in a paper, I've seen it in my textbook 😬


Vectors is C4.
Reply 51
Avatar for SI 1
SI 1
1
Could anyone explain how to do question 8 ii on this specimen paper??
the mark scheme makes the second derivative equal to 0 but I don't know why.
Specimen QP - C3 OCR.pdf43.6KB
Original post by 16characterlimit
Vectors is C4.


Oh yeah. My mind is just mental atm.
can someone help me with this question from jan 2006

3sin6Bcosec2B=4 where beta lies between zero and 90 degrees

any help would be appreciated
Original post by LukeWeatherstone
Can anyone think of any small things which i might not know i should know for tomorrow?


perhaps the graphs of cosec sec cot etc?
Reply 55
Avatar for duncant
duncant
2
Original post by SI 1
Could anyone explain how to do question 8 ii on this specimen paper??
the mark scheme makes the second derivative equal to 0 but I don't know why.

the second derivative is the rate of change of gradient. so when the second rerivative = 0, the rate of change or gradient is at a turning point, thus it is maximum.
Reply 56
Avatar for SI 1
SI 1
1
Original post by duncant
the second derivative is the rate of change of gradient. so when the second rerivative = 0, the rate of change or gradient is at a turning point, thus it is maximum.

Thank you so much! :smile:
So at a turning point the gradient of a curve is the maximum. I always thought the gradient was 0 at turning points.
Reply 57
Avatar for AlfieH
AlfieH
2
Original post by SI 1
Thank you so much! :smile:
So at a turning point the gradient of a curve is the maximum. I always thought the gradient was 0 at turning points.


The gradient is always 0 at turning points, be it maximum or minimum :smile:

The turning point could be either a minimum or a maximum, hence why you differentiate again to prove. If the value of x substituted into D^2y/Dx^2 is greater than 0, it's a minimum point and if it's less then it's a maximum point!
Reply 58
Avatar for SI 1
SI 1
1
Original post by AlfieH
The gradient is always 0 at turning points, be it maximum or minimum :smile:

The turning point could be either a minimum or a maximum, hence why you differentiate again to prove. If the value of x substituted into D^2y/Dx^2 is greater than 0, it's a minimum point and if it's less then it's a maximum point!


Sorry I probably sound really stupid but how would you know that the maximum value of the gradient is 0? Could the max gradient not be more than 0? I would think that there'll be a point on the curve where the gradient is more than 0.
Reply 59
Avatar for AlfieH
AlfieH
2
Original post by SI 1
Sorry I probably sound really stupid but how would you know that the maximum value of the gradient is 0? Could the max gradient not be more than 0? I would think that there'll be a point on the curve where the gradient is more than 0.


It's not the point at which the gradient is the maximum - so you may be getting confused there.

Nor is it the point where the gradient is the lowest.

Instead the gradient is always 0 at the turning points/ min or max points. It is the maximum/ minimum as in the highest/lowest POINT of the curve not gradient.

If that makes sense?
(edited 7 years ago)

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