The Edexcel C4 (21/06/12 - PM) Revision Thread

aln(b) is equal to ln(b^a). It's one of the rules of logs. Then use the rules of indices to simplify. So -2ln(0.5) = ln(0.5^(-2)) = ln(1/(0.5^2)) = ln(1/(1/4)) = ln(4)

The -2 becomes the power over 0.5. Ie. -2ln(0.5) = ln(0.5)^-2 which is then ln4

http://www.sci2.co.uk/maths/C4/Edexcel/2006%20Jan%20QP.pdf

jan 2006 paper

for question 7b, how do you do it?
I understand that dr.dt = dV/dt / dV/dr but where do I get the dV/dt?? And I don't see where I need to use the chain rule!!

try using the substitution u=tanx and remember what tanx differentiates to. then changing dx into du will leave your sec^2x to cancel out

I think its an error on that paper. Its supposed to say dV/dt = f(t) not dV/dr = f(t)

January 2011 .. what a beautiful paper !!

Do papers from earlier years... Like jan 2008. Those papers are plain nasty.

already done them. not that bad..

it from solomon A and its a trapezium question so i dont think it will take much time

Brilliant thank you! any idea for tan^3(x)?

∫tan³x dx
= ∫tan²x tanx dx
= ∫(sec²x−1)tanx dx
= ∫sec²xtanx dx − ∫tanx dx

Put u=tanx, then du/dx=sec²x, du=sec²xdx

Leading to:

∫u du − ∫tanx dx
=u²/2 − ln|secx| + C
=(tan²x)/2 − ln|secx| + C

http://www.sci2.co.uk/maths/C4/Edexcel/2006%20Jan%20QP.pdf

jan 2006 paper

for question 7b, how do you do it?
I understand that dr.dt = dV/dt / dV/dr but where do I get the dV/dt?? And I don't see where I need to use the chain rule!!

Anyone have general tips to approach rates of change questions?

And also a list of the mensuration formula we may be required to use? Like spheres etc

Anyone have general tips to approach rates of change questions?

And also a list of the mensuration formula we may be required to use? Like spheres etc

I need to write down all the variables and the relationships (none differential equations, ratios, etc) between them at the start. A diagram nearly always helps. It's then usually possible to answer the questions by doing 1 or 2 differentiations and applying the Chain Rule. However, you may need to come up with more than one version of the Chain Rule to answer the question. It's very easiy to correctly apply the Chain Rule, but with a wrong variable and then you're stuffed.

I love these questions. They make you think, but they can take time though if you don't spot something quickly enough.

Did you learn all of the mensuration formulae?

If you're integrating by substitution, what do you do if the substituted higher value is less than the substituted lower value?

For example say the integral says to find between x=4 and x=0, but when you substitute it for t you get t=2 on top and t=3 on the bottom?

I seem to remember that if you switch them round you have to make the integral negative, if that's right can anyone explaint to me why it is? Thank you

I seem to remember that if you switch them round you have to make the integral negative, if that's right can anyone explaint to me why it is? Thank you