Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

@Zacken Please could you upload a sketch, thanks so much!

Step 1: You know the curves intersect at $\theta = \pi/12, 5\pi/12$ - draw these half-lines.

Step 2: Find the area of the sector bounded between these two half lines and the circle. $\frac{1}{2} \times a^2 \times \frac{\pi}{6}$

Step 3: Find the blue shaded bits using integration of C1 from 0 to pi/12 + integration of C1 from 5pi/12 to pi/2.

Add step 2 + step 3.

Thank you!! Please could you explain a bit of this question to me, I got it right apart from I thought the angle for the sector of the circle would be pi/ 6 as if it's 2pi/3 plus the other area I worked out then wouldn't this be the top bit of the circle rather than the shaded bit? Or is my
Method incorrect? Thanks

How many of your qs has the poor guy helped you with ahaha, he's probs got work of his own to do.

Sorry haha, I got rejected from Cambridge btw if that was why you had animosity towards me 😂

Thank you!! Please could you explain a bit of this question to me, I got it right apart from I thought the angle for the sector of the circle would be pi/ 6 as if it's 2pi/3 plus the other area I worked out then wouldn't this be the top bit of the circle rather than the shaded bit? Or is my
Method incorrect? Thanks

I've got a nice and short question:

If the complex number $\displaystyle z=\left ( 1+i\tan\theta \right )^{3}$,

Show that $\displaystyle 1-3\tan^{2}\theta \equiv \dfrac{\cos3\theta}{\cos^{3} \theta}$.

Side note, you could also just do:



Edit to add: I can't immediately see how to do it the way you intended. I get that $\Re(1 + i\tan \theta)^3 = 1 - 3\tan^2 \theta$, but then what?

$\displaystyle (1 + i\tan \theta)^3$

$\displaystyle \left(\frac{\cos \theta}{\cos \theta} + i\frac{\sin \theta}{\cos \theta}\right)^3$

$\displaystyle \left(\frac{1}{\cos \theta}\right)^3 \: (\cos \theta + i\sin \theta)^3$

then regular de moivre theorem

Hey everyone
Does anyone know which paper is this?or how to get its markscheme? I tried looking it up online couldn't find anything. Thanks


Posted from TSR Mobile

june 2013 withdrawn paper

Hi, could anyone give any advice please on sketching the graph in question 5cii of this paper https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202013%20QP%20-%20FP2%20Edexcel.pdf Thanks

Literally just done this paper. Uhm, you should know that there are minimum points at $(\pm \sqrt{2} ,4)$ and that the $y$-axis is an asymptote (since you have $x^2$ in the denominator) and then the graph tends to positive infinity as $x$ approaches infinity. Then the curve tails off the positive infinity as $x$ tends to positive and negative infinity. Also there is symmetry in the $y$-axis because it's an even function i.e $f(-x) = f(x)$ since all your $x$ terms are squared.

Edit: also your function will always lie above your $x$-axis since it is always positive due to the squared terms.

How did you do? Thanks for that, very helpful

I did okay, it was a nice one, I think. You?