Edexcel FP3 - 27th June, 2016

you're looking for the midpoint right? that's the sum of the first x coordinate and the second x coordinate, i.e (x1+x2)/2

Don't really understand part d. Could someone explain it to me please?

When you differentiate (-1/3)(16-x^2)6(3/2) you get x(16-x^2)^(1/2) so you can see that there is an x on the outside. This x is needed so that you can integrate the expression exactly into what I just differentiated.

@Zacken, Tom, could you guys help with 8(c) here? I can see how implicit differentiation works, but I'm not sure I understand the main mark scheme method of using summation of roots (which is supposedly not even in the UK spec?)

Sweet man, what do you think of the department? I'm really excited for next year tbh

The square root part cancels when you add them. See attached.

I get that, but why can't you just integrate the initial expression, using u=x^n and dv=(16-x^2)^(1/2)? this also forms an expression in terms of In and I(n-2) but it's not what they are looking for.

Have you already done 3 A2 units?

This was a weird one!

I know how to do the algebra, I just wasn't sure why we wanted (a + b)/2. Cheers though!

More, I think. I've already done C1-4, M1-3, S1-3, FP1-2 - just FP3 left.

I'm still not really sure I understand.

And how would you integrate the dv part nicely? If you had an x in front it would make life so much easier because there is an -x^2 term inside the root which differentiates to -2x outside the root by the chain rule. You would end up doing a lot more work than needed.

**** I'm a dumbass, of course, dv would be a ***** to work out. I thought it would integrate to -1/3x*(16-x^2)^(3/2), gg

Find the locus as you would normally do, by eliminating your parameter $\theta$.

Look at the coordinates of the point R. Note that the y coordinate must always be zero so the locus must lie on the real axis. Your x-coordinate is $\frac{a}{cos\theta}$ where $cos\theta$ is always less or equal to than one. Hence the x coordinate must be greater than or equal to a.

Apply the same logic to the other side of the ellipse.

Don't really understand part d. Could someone explain it to me please?

Think you're complicating it more than it needs to be.

You know that $R$ is the point $\left(\frac{a}{\cos \theta} , 0\right)$.

Since $-1 \leq \cos \theta \leq 1$ then that means that $\frac{1}{\cos \theta} \geq 1$ or $\frac{1}{\cos \theta} \leq -1$.

So that means that $\frac{a}{\cos \theta} \geq a$ or $\frac{a}{\cos \theta} \leq -a$.

But that is precisely $x \geq a$ or $x \leq -a$.

Does anyone have a copy of the FP3 June 13 withdrawn paper ?