AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

There must be other ways of showing that y=0, however in this instance I don't think your way works because in the question they have told you that q cannot equal 0 but they haven't given you a restriction for p (correct me if I'm wrong).

So the fact that you've found two equivalent expressions for x in this case is nullified by the fact that they both contain (q/p), as if p=0 then they don't intersect as this would instead be an asymptote, does that make sense?

I see! Argh how I hate some of these questions that contradict everything I thought I knew... 😭


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Well actually I think I'm being a bit overscrupulous here as I doubt the mark scheme will be THAT picky, so you'll probably get the marks anyway, just thought I'd show you a more efficient approach is all...

*walks away*

Ever so sorry, I thought you needed help with 3c and not 3b. Here's my method for both parts.



Is this correct?


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yep that is correct!

but what I don't understand is how finding the dot product of pq and qr will help me find angle q?

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Oh right thanks!

Is this an acceptable way of showing the two tangents meet on the x axis?

(Jan 13 Q4 a ii)




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I would have thought so. But I'm not too sure.

The dot product helps you find the angle between two lines.

Hopefully this clears it up

(https://m.youtube.com/watch?v=5p5gZL6FW28)


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Think I've done it properly

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Here's my workings for the same questions.

I believe our answers agree, even though they're expressed slightly differently




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your way seems much better and more convincing. :-)

how are you revising now for this exam?

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Just purely past papers! How about you?


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Could someone explain where I've gone wrong on June 2010 Q7?



I've set up a quadratic in terms of $\alpha$ like so:

$5 \alpha^{2} - 12 \alpha + 7 = 0$

To which the solutions are:
$\alpha = 1$ or $\alpha = \frac{7}{5}$

This gives co ordinates of C as:
$(3, -1, 3)$ which is correct, and $(\frac{19}{5}, -1, \frac{17}{5})$ which is incorrect.

How can I get to the other possibility of $(-1, -1, 1)$?

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will be starting past papers tomorrow!

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Hey I don't know if anyone gave a solution for this but I think it may be because in your second parallelogram BP isn't parallel to AC due to C being in a different position so the equation is only valid for one solution if that makes sense? Sorry if this is wrong or someone already answered it!

Brilliant! Thank you. No one had answered haha.


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No problem 😊.