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c3 25th January 2013 exam

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Original post by otrivine
That would be so good :biggrin:

The question on x=coty

does my method work i converted the coty in the form of cos/sin and applied the quotient rule and used the sin^2x+cos^2=1 formula/

Also, i used another method where i differentiated the coty and mentioned the 1+cot e.t.c formula



yeah... as long as you've 'show that', you'll be fine.... there are many methods you know.... any will do...
Original post by StrawHat_Shifaz
yeah... as long as you've 'show that', you'll be fine.... there are many methods you know.... any will do...


ok good!

i got 1+sin45


i expanded the bracket and we had to use 2 identites for that question one was
sin2A=2COSASINA and sin^2x + cos^2x =1


the question on rates i got 593.49 then rounded to nearest which was £593.
Original post by bubblegummer
oh gosh... i put the value 1 for cos(theta - 0.927). do you remember what was the mark for this question, plus the part after this question altogether?


i think it was just 4 marks for finding both maximum value of P(@) and @....
Reply 243
For everyone who doesn't why the max value is 2:

Exercise 6B: Question 8 (c) in the C3 book is pretty much the exact question, which You have to apply a similar method.
Original post by otrivine
ok good!

i got 1+sin45


i expanded the bracket and we had to use 2 identites for that question one was
sin2A=2COSASINA and sin^2x + cos^2x =1


the question on rates i got 593.49 then rounded to nearest which was £593.



yep. same! :banana:
Original post by StrawHat_Shifaz
i think it was just 4 marks for finding both maximum value of P(@) and @....


hmm okay :frown: what answer did you get for the last question part b where we have to get the exact value for t?
Original post by StrawHat_Shifaz
yep. same! :banana:


the question on finding t V=9500 bugged me a bit

i did the equation they gave = 9500
then i shifted the 500 on the other side

so got the two e equations = 9000

then mind went on thinking that cause there is a + sign and involving two e equations we can multiply cause e^x + e^y = (e^xy)
Original post by bubblegummer
hmm okay :frown: what answer did you get for the last question part b where we have to get the exact value for t?


i got a wrong answer!!!!! one user posted that it was a factorising question. since 0.5^2 is o.25. it was factorisable! :facepalm::facepalm:
Original post by otrivine
the question on finding t V=9500 bugged me a bit

i did the equation they gave = 9500
then i shifted the 500 on the other side

so got the two e equations = 9000

then mind went on thinking that cause there is a + sign and involving two e equations we can multiply cause e^x + e^y = (e^xy)


i think the answer for this was posted by someone at around page 8... its a factorisable sum man..... who would have thought that...
Original post by StrawHat_Shifaz
i got a wrong answer!!!!! one user posted that it was a factorising question. since 0.5^2 is o.25. it was factorisable! :facepalm::facepalm:

:eek::eek::eek: same here, i didn't think of that!! what a waste!
Original post by otrivine
the question on finding t V=9500 bugged me a bit

i did the equation they gave = 9500
then i shifted the 500 on the other side

so got the two e equations = 9000

then mind went on thinking that cause there is a + sign and involving two e equations we can multiply cause e^x + e^y = (e^xy)


we can't combine both e values together. :s-smilie:
Reply 252
Original post by otrivine
why? cause we had the R value from above calculation and had to just put the 10 in the fraction?

4/12+10 something like that

to get the max value ypu want the denominatorr to be as small as possible. Therefore you have to use -10 because thats the minimumvaluce of 10cos(x-0.927. then you get 4/(12-10). which gives you 2, so 2 is the max value
Reply 253
Original post by Astarplease01
well i was aiming for 100ums but after that its likely ill only get over 85ums if im lucky. 8b) ****ed with me. maximum question ****ed with but i knew exactly how to do it but i thought the wrong of maximums and minimums .


OMG. I thought exactly the same as you in the exam. I thought I'd get 100 at that rate, then I got to 8b. Now I think I messed up other parts too from what people are saying on here.
I really need that A* :frown:
Original post by otrivine
Yes, and the other one did u put cos(x-0.927something)=1 for max of deta


no... it would be cos(x-0.927...)=-1
Original post by justinawe
no... it would be cos(x-0.927...)=-1


what but they asked for maximum not minimum
Original post by otrivine
what but they asked for maximum not minimum


They asked the value of θ at which the maximum for p(θ) occurs. The maximum for p(θ) occurs when cos(θ-0.927)=-1, as others explained earlier.
Reply 257
Original post by Lis slander
I found this one difficult [sin(22.5)+cos(22.5)]^2... but recollected how to do it as soon as I left the xam... :frown:


Me too! :frown:


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Reply 258
Original post by Nia_Beth
OMG. I thought exactly the same as you in the exam. I thought I'd get 100 at that rate, then I got to 8b. Now I think I messed up other parts too from what people are saying on here.
I really need that A* :frown:


Same here, it's soo annoying, personally I thought that paper was so much harder than a lot of the past question papers! The grade boundaries have to be low!
Original post by SuziieB
Lol, look at us insomniacs...

My friend said for 8c that you were supposed to divide £593 by 8? Because it was 8 years? I don't think so though...I left mine as 593.

8b was to form a quadratic. I plugged ky answer in and got 9500 exactly.

__________________________
Save time. See it my way.


I divided by 8 on the last one to get 74


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