Edexcel A2 C3 Mathematics 12th June 2015 Watch

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sj97
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#1321
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#1321
(Original post by rache11ouise)
f(x) = 2[3-x]+5
at the root x=0 therefore f(0) ---> f(x) = 11
so k<= 11
since ---> f(x)=k ---> 2[3-x]+5=k
2[3-x]=k-5 so k>5
so 5<k<=11
Could you please explain how to do 4a)?I don't understand the expansion?
https://b3755649dbd1afe3db91a899c3b9...%20Edexcel.pdf
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1122ag
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#1322
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(Original post by adorablegirl1202)
Ah thanks, i get it! I know that we would find the point it intersects the x axis by x = 0, how about the minimum point?
I just think about it logically:
f(x)=2|3-x|+5

in order for f(x) to be the smallest value possible the 2|3-x| part should equal 0, this is because its a modulus so it cant equal a negative number.

So the minimum value of f(x)=5.

That's how i'd do it but there's probably a different way... otherwise i'm not sure.

Hope this helps..
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gregsey28
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(Original post by Kingnig)
Definitely June 13 and 14, probs the hardest of the normal papers
Thanks I've done June 14 in a class test, but I'll give June 13 a go.
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gregsey28
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(Original post by mat1996)
try june 2013 btw where are the elmwood paper?
Thanks, I think I got them from physicsandmathstutor or some website of a similar name.
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phawkins96
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#1325
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#1325
how do you prove the differential of a^x as being a^x lna?
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rache11ouise
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#1326
tanx=sinx/cosx

differentiating sinx ---> cosx
differentiation cosx ---> -sinx
using quotient rule sinx/cosx (u/v)
u=sinx v=cos u'=cos v'=-sinx
(cosx)(cosx)-(-sinx)(sinx)/(cosx)^2
(cos^2x+sin^2x)/cos^2x ---> sin^2x+cos^2x=1
the numerator now equals 1
1/cos^2x=sec^2x
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gregsey28
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#1327
(Original post by anonwinner)
June 2013 is by far the hardest, you only needed 57/75 for 90 UMS
That sounds very low, I'll do that one next!
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rache11ouise
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#1328
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#1328
360-PV using CAST and going clockwise
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mat1996
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(Original post by gregsey28)
Thanks, I think I got them from physicsandmathstutor or some website of a similar name.
Hi thanks which one did you find hardest out of the elmwood papers
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adorablegirl1202
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(Original post by 1122ag)
I just think about it logically:
f(x)=2|3-x|+5

in order for f(x) to be the smallest value possible the 2|3-x| part should equal 0, this is because its a modulus so it cant equal a negative number.

So the minimum value of f(x)=5.

That's how i'd do it but there's probably a different way... otherwise i'm not sure.

Hope this helps..
Thanks so much! Another thing, which graps should we able to sketch???
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suyoof123
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#1331
https://www.youtube.com/watch?t=368&v=nYOh4QbULsM

For this question, do you have to revert to dy/dx , cant we just keep it dx/dy and then sub y into the dx/dy to find the gradient?
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rache11ouise
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#1332
g(x)=ln2(x)
g(x^2)=ln2(x^2) ---> ln2x^2
g(x^3)=ln2(x^3) ---> ln2x^3
Your not squaring lnx, you're only squaring x
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gcsestuff
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#1333
If I do all the iygb papers ( the none extreme ones) will that be enough revisin?


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rache11ouise
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#1334
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you can sub y into dx/dy but because you want the gradient of a normal which would normally be the negative reciprocal putting y into dx/dy means you have the reciprocal of dy/dx but you would still need to multiply by -1
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TeeEm
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#1335
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#1335
just did this question with one of my students ... (he got eaten badly)
Any takers

solution in post 1741
Attached files
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becca9621
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#1336
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#1336
Is there a simpler way to find the range and domain of functions without drawing the graph?
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suyoof123
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#1337
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(Original post by rache11ouise)
you can sub y into dx/dy but because you want the gradient of a normal which would normally be the negative reciprocal putting y into dx/dy means you have the reciprocal of dy/dx but you would still need to multiply by -1
You lost me there..

I know normally the gradient of normal is the tangent's gradient times by -1 and switching if its top or bottom.

So whats the difference of putting y into dx/dy and then multiplying that by -1 and switching its top or bottom and

finding dy/dx by dividing 1 by dx/dy and then shoving the x he found from using the original equation into that for the gradient and then doing the things you do to find gradient of normal from gradient of tangent.
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suyoof123
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(Original post by TeeEm)
just did this question with one of my students ... (he got eaten badly)
Any takers
will you have a worked solution put up?

i'll also attempt to not get eaten alive
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Gilo98
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(Original post by TeeEm)
just did this question with one of my students ... (he got eaten badly)
Any takers
does p = purchase price or is it just so constant?
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fm81
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#1340
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#1340
Yes! If the question gives you a limit as to what x values you can put in, for example x>0 , sub that in and see what it gives you. If it's a function like e^x or X^2 you know it must be greater than zero. Finally, stick in a really large x value to see what it tends to. Its still a lot easier to play it safe and draw a quick sketch though!

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