Edexcel C3,C4 June 2013 Thread

Yes it should. My mistake. Fixed it now though.

For $\frac{x+2}{x-2}$, you want to write $x+2$ into something that has $x-2$. You can't do anything with the $x$, but you can rewrite the $+2$ differently. Think of two numbers that make $+2$. From the bottom of the fraction you know one number is $-2$. What is the other?

edit: Perhaps a better way to think of it is like this: The aim to to eventually cancel out the $x-2$ on the bottom with $x-2$ on the top to get $1$. So you know the top has to be written as: $(x-2 + \mathrm{something})$, where the something has to be set so that the top is still $x+2$, but written differently.

The last one follows similarly.

Give this a go too:
$\dfrac{2x-1}{x-1}$

thank youuuu !!!! i think im getting it nowww ... so in what circumstances should i KNOW when to use this method?

thank youuuu !!!! i think im getting it nowww ... so in what circumstances should i KNOW when to use this method?

Give this a go too:
$\dfrac{2x-1}{x-1}$

draw the lines intersecting and you can see which angle you have just worked out of 111 between the lines, then you can see there is another angle which is also between the lines but acute

just did the c4 2012 replacement paper don't understand qu 7a it's basically integrate cosecx and the answer is -ln(cosecx + cotx) - how do you get this? X

is there usually a rule with this method? (making the fractions equal)
i usually just divide the top by the bottom

just did the c4 2012 replacement paper don't understand qu 7a it's basically integrate cosecx and the answer is -ln(cosecx + cotx) - how do you get this? X

just did the c4 2012 replacement paper don't understand qu 7a it's basically integrate cosecx and the answer is -ln(cosecx + cotx) - how do you get this? X

could there be occassions where you have to subtract my 360?

where is everyone in terms of revision??

For the last part to this question, I understand that the gradient is undefinable. The mark scheme then states that therefore



What is the singnificance of the x=1?

thank you

it asks for the equation of the normal when t= Pi/4

if the gradient is 0, the tangent is parallel to the x axis. as they ask for the normal, the line bisects the point of (1,1) and is parallel to the Y AXIS. therefore the equation of the line is x=1

Thank you for your reply.

I now understand this. Just one question, bit of a silly questions, but just to confirm...as we know that the equation of the tangent is y=1 and the normal goes through the point (1,1) are just using the fact that tangent and normal are pependicular to conclude that the equation of the normal is x=1?

yes in short