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# Edexcel A2 C4 Mathematics June 2015 - Official Thread watch

1. (Original post by lam12)
How would I integrate 2sinycosysiny

So 2siny^2cosy

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(Original post by physicsmaths)
Well cos is the differential of sin. Consider to the higher power (sinx)^3
Or if you prefer substitution to recognition (which some people do), substitute u = sin y. I chose the substitution because siny is effectively the inside of a bracket.
2. (Original post by LukeTownsend123)
I am currently at the stage where I am borderline A*/A in my C4 exams, but I want to get a secure A*. The only marks I am losing are small mistakes or if the question is an abstract one. Anyone have any good techniques to get an A*, securely? (75/75) I've done virtually every past paper and I'm not sure what else I can do!
If you don't have one already. Grab a fx-991 ES plus to check differentiation/integration answers quickly.
Stuff like binomial can be checked easily by subbing a value into the result.
The main thing for me was to find the most efficient way of checking answers - takes all kinds of pressure off you.
http://gyazo.com/cba3450ed278963ff227112c0f8b937a

So when changing the limits the first time round for part C, I expanded the substitution and set it equal to 5, and then to 2 which gave me upper limits of both 6 and 2 and lower limits of 3 and 5.

However, realising I had made a mistake, I just substituted 5 and then 2 into the original substitution and got limits of 6 and 5 only.

Can anyone tell me why I eliminated the 2 and 3 the second time round? I know it has something to do with the positive square root but I don't really understand it. Thanks
4. (Original post by cerlohee)
http://gyazo.com/cba3450ed278963ff227112c0f8b937a

So when changing the limits the first time round for part C, I expanded the substitution and set it equal to 5, and then to 2 which gave me upper limits of both 6 and 2 and lower limits of 3 and 5.

However, realising I had made a mistake, I just substituted 5 and then 2 into the original substitution and got limits of 6 and 5 only.

Can anyone tell me why I eliminated the 2 and 3 the second time round? I know it has something to do with the positive square root but I don't really understand it. Thanks
the first part is moving the x-1 on the left and getting (u-4)^2 makes it easier to solve
5. (Original post by Ripper Phoenix)
the first part is moving the x-1 on the left and getting (u-4)^2 makes it easier to solve
Wait what do you mean sorry? :3
6. (Original post by Ripper Phoenix)
the first part is moving the x-1 on the left and getting (u-4)^2 makes it easier to solve
When you have x=5 u=6 or u=2
When you have x=2 u=5 or u=3

How do you know which limits you should use, in situations like this, would I have to use 6 and 5 because these were calculated using the positive value of the root or 2 and 3 because these were calculated using the negative value of the root?
7. Can someone please explain how/why this works please? It's from the C4 solutionbank, question 3 from Ex 6K:

I'm confused as to where the A comes from? I thought you were supposed to solve for C (which I did and got zero) and then continue, but I don't understand why they put an A in front of the e^t. I've seen this elsewhere as well but never really understood it.

Thank you!
8. (Original post by randlemcmurphy)
When you have x=5 u=6 or u=2
When you have x=2 u=5 or u=3

How do you know which limits you should use, in situations like this, would I have to use 6 and 5 because these were calculated using the positive value of the root or 2 and 3 because these were calculated using the negative value of the root?
How could you tell which is calculated using the positive/negative value? I just got a quadratic!

You would use 6 and 5 because they're caluclated using the positive value though, because the square root symbol conventionally only includes the positive square root!
9. (Original post by themorninglight)
Can someone please explain how/why this works please? It's from the C4 solutionbank, question 3 from Ex 6K:

I'm confused as to where the A comes from? I thought you were supposed to solve for C (which I did and got zero) and then continue, but I don't understand why they put an A in front of the e^t. I've seen this elsewhere as well but never really understood it.

Thank you!
Let e^c=A

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10. (Original post by themorninglight)
Can someone please explain how/why this works please? It's from the C4 solutionbank, question 3 from Ex 6K:

I'm confused as to where the A comes from? I thought you were supposed to solve for C (which I did and got zero) and then continue, but I don't understand why they put an A in front of the e^t. I've seen this elsewhere as well but never really understood it.

Thank you!

Then:

But e is a constant, and b is an arbitrary constant, so e^b is an arbitrary constant. Say e^b = k, and then:

11. (Original post by cerlohee)
http://gyazo.com/cba3450ed278963ff227112c0f8b937a

So when changing the limits the first time round for part C, I expanded the substitution and set it equal to 5, and then to 2 which gave me upper limits of both 6 and 2 and lower limits of 3 and 5.

However, realising I had made a mistake, I just substituted 5 and then 2 into the original substitution and got limits of 6 and 5 only.

Can anyone tell me why I eliminated the 2 and 3 the second time round? I know it has something to do with the positive square root but I don't really understand it. Thanks
There is one simple way to remember it which will never confuse you. When using substitution and you need to square root one of the values of "x" to get "u" don't worry about negative values it's always the positive ones.

That's how I've been getting through this crap
12. (Original post by Krollo)

Then:

But e is a constant, and b is an arbitrary constant, so e^b is an arbitrary constant. Say e^b = k, and then:

Yes it does! Thanks a million
13. (Original post by themorninglight)
Yes it does! Thanks a million
No problem :-)
14. (Original post by cerlohee)
http://gyazo.com/cba3450ed278963ff227112c0f8b937a

So when changing the limits the first time round for part C, I expanded the substitution and set it equal to 5, and then to 2 which gave me upper limits of both 6 and 2 and lower limits of 3 and 5.

However, realising I had made a mistake, I just substituted 5 and then 2 into the original substitution and got limits of 6 and 5 only.

Can anyone tell me why I eliminated the 2 and 3 the second time round? I know it has something to do with the positive square root but I don't really understand it. Thanks

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Attached Images

15. (Original post by cerlohee)
How could you tell which is calculated using the positive/negative value? I just got a quadratic!

You would use 6 and 5 because they're caluclated using the positive value though, because the square root symbol conventionally only includes the positive square root!
Because your are substituting for the function root x-1 and the actual sub they are using is root(x-1)=u-4 hence U>4 since both sides need be positive neglecting the other values!

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16. Ahh thanku!!! Yeah I got the question right I just don't get the limit thing haha As in how can you tell which is the positive and which is the negative limit from the quadratic.

I'll show you what I mean cos idk how to use latex lol

if you can read that... why is half of it wrong?!?!?
Attached Images

17. (Original post by physicsmaths)
Because your are substituting for the function root x-1 and the actual sub they are using is root(x-1)=u-4 hence U>4 since both sides need be positive neglecting the other values!

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Ahh so even expanding it in the first place was wrong! Thanks )) Have given you too many rep but I would rep again if I could haha
18. Our math teacher told us to use the positive rooted value.for instance root 4 would be plus minus 2 but we use plus 2.. i think physicsmaths has already provided a logical explanation though above 😊

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19. And one more question... Sorry!

Where did I go wrong here with part d?
http://gyazo.com/c165ece148120aa86a76fb16ed8973ae
Attached Images

20. (Original post by cerlohee)
And one more question... Sorry!

Where did I go wrong here with part d?
http://gyazo.com/c165ece148120aa86a76fb16ed8973ae
Should be right angle at L2. You can work this out differentiation or completing the square which would work all the time.

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