You are Here: Home >< Maths

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

Announcements Posted on
Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. (Original post by marioman)
I've just updated my post with the link to the 2015 paper.
Thanks
2. Is it me or is question 4 in the 2015 paper worth way too many marks for what it is?
3. Can anyone think of any small things which i might not know i should know for tomorrow?
4. (Original post by ughexams)
the answer is 1/36 pi.... (for part iii)
I know
5. Hi! How do you find the range and domain from the equation? I always forget!
6. (Original post by r34_786)
Hi! How do you find the range and domain from the equation? I always forget!
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
7. How do I show that:

1 – ln(2) = ln(1/2e)
8. (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)
9. Thank you so much!
(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
10. I think proof in vectors will come up. Although I've never seen one in a paper, I've seen it in my textbook 😬
11. (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.
12. 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.
Attached Images
13. Specimen QP - C3 OCR.pdf (42.6 KB, 46 views)
14. (Original post by 16characterlimit)
Vectors is C4.
Oh yeah. My mind is just mental atm.
15. 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
16. (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?
17. (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.
18. (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!
So at a turning point the gradient of a curve is the maximum. I always thought the gradient was 0 at turning points.
19. (Original post by SI 1)
Thank you so much!
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

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!
20. (Original post by AlfieH)
The gradient is always 0 at turning points, be it maximum or minimum

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.
21. (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?

Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: August 21, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

University open days

Is it worth going? Find out here

How to cope with a heavy workload

Poll
Useful resources
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.