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Edexcel A2 C4 Mathematics June 2015 - Official Thread

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Original post by Dante991
Sounds like good practice, what website is this mate?

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http://www.thestudentroom.co.uk/showthread.php?t=3361905
Original post by Gome44
Haven't seen this on a c4 paper and there is a very slim chance it will come up but if you want a challenge:

∫ cos(x)e^x dx

It's pretty easy once you spot the trick :smile:


What's the answer ?


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Original post by Tazmain
When I started reading I thought it'd be this kinda one, thanks for sharing :smile:
I don't think its in our spec, but you never know with edexcel..

Is it...?

Spoiler



Yep, but don't forget your +c :tongue:
Can someone help me with 4b, the mark scheme isn't helpful.

https://a086a5a2f39bda93734c56a63fab1d7be0a9ba38.googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/June%202013%20%28R%29%20QP%20-%20C4%20Edexcel.pdf
Original post by Medicjug
What's the answer ?


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Have a look at Tazmain's post :smile:
Original post by jubilant_joej
Yes the larger limit is always on top, unless you're integrating using substitution, then you put the largest x value on top, but say when you change the limits to u, the equivalent numbers stay in the original position of the x values, so a smaller u value could end up being on top


Original post by Genesis2703
I believe so, when finding an area under a curve in the region a<x<b you integrate with respect to x with lower limit a and upper limit b as you have done for a few units now.

When we have parametric equations, we still want to integrate over that x range, but we have to do it in terms of t, or theta (or whatever parameter is being used) as it makes the integral easier. But SIMILARLY (but not the same) to integration by substitution, you need to change the limits and put them with respect to t.

So if your larger x-value (b, from above) occurs at time t2, and the smaller one (a) occurs at time t1, then your integral is with upper limit t2 and lower limit t1, regardless of whether t2>t1 or not.

Hope I cleared things up!


Thanks, that is a lot of help guys!
Original post by Tazmain
When I started reading I thought it'd be this kinda one, thanks for sharing :smile:
I don't think its in our spec, but you never know with edexcel..


Dunno if anyone's still interested in solving this so I'll put spoilers

Spoiler

I'm feeling a trickier implicit differentiation question......

How would one implicitly differentiate 2^y?

would it be: dy/dx 2^y ln2?
Original post by nayilgervinho


You have to get a value for X which is within the ranges of the expansion 􀀁|x| <8/9 , the only way to get that is by going through that method to find X.

By factoring 1000 out and cubing it to give 10, you can 10sqroot7.1. By finding a value of X from
8-9x = 7.1, you get 0.1.
That x value is within the range of 􀀁|x| <8/9, and can therefore be substituited to get an approximation for sqroot7.1. . You then need to multiply the approximation by 10, to get an approximation for 10sqroot7.1.

Hope that makes sense :smile:
Original post by TheAnnabelle
I'm feeling a trickier implicit differentiation question......

How would one implicitly differentiate 2^y?

would it be: dy/dx 2^y ln2?


I'm pretty sure it would be
Hey guys,

Maybe this has already been said but when doing IBP the best way to decide what to make the first "u" is to use the L.A.T.E acronym

(ln, then any algebra, then any trig, then any e)

(has never not worked for me)
Original post by 1 8 13 20 42
Dunno if anyone's still interested in solving this so I'll put spoilers

Spoiler



True, thanks for the heads up.
Original post by Tazmain
You have to get a value for X which is within the ranges of the expansion 􀀁|x| <8/9 , the only way to get that is by going through that method to find X.

By factoring 1000 out and cubing it to give 10, you can 10sqroot7.1. By finding a value of X from
8-9x = 7.1, you get 0.1.
That x value is within the range of 􀀁|x| <8/9, and can therefore be substituited to get an approximation for sqroot7.1. . You then need to multiply the approximation by 10, to get an approximation for 10sqroot7.1.

Hope that makes sense :smile:


Thanks, I was confused as the mark scheme said 71 instead of 7.1, maybe the dot was hard to see.
Original post by TheAnnabelle
I'm feeling a trickier implicit differentiation question......


I have no idea if this is mathematically possible, but can you implicitly differentiate 3^xy
Original post by nayilgervinho
Thanks, I was confused as the mark scheme said 71 instead of 7.1, maybe the dot was hard to see.


No, they genuinely made a mistake, I know cos I did the same question in the morning XD.
Original post by veggienator
I have no idea if this is mathematically possible, but can you implicitly differentiate 3^xy


Hmm....I'm not sure...I'll see if I can ask a teacher tomorrow and get back to you guys
Original post by TheAnnabelle
I'm feeling a trickier implicit differentiation question......

How would one implicitly differentiate 2^y?

would it be: dy/dx 2^y ln2?

Think so. Irrelevant of what it is, I'm pretty sure the approach of differentiating it as per normal and sticking a dydx\frac{dy}{dx} at the end holds.
Original post by TheAnnabelle
Hmm....I'm not sure...I'll see if I can ask a teacher tomorrow and get back to you guys


That'll be great :yy:
Original post by Oxyfrost
Hello, can someone help me? Either I'm very confused, or Solomon papers have got the mark scheme wrong. Paper A, Question 5. Trapezium rule. Equation is 4(x^0.5)(e^-x).
My value for x=2 is 0.766 (3sf), but on the mark scheme it says 1.083. Can someone please help me to understand this discrepancy? All my other values are correct.

Question and mark scheme attached.


Please can anyone help me with this question?
Original post by 1 8 13 20 42
Dunno if anyone's still interested in solving this so I'll put spoilers

Spoiler



I did the integration by parts twice, but I end up needing to integrate the original problem, what's the trick?

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