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Original post by RuairiMorrissey
Oh, are there mark schemes available for pre-2007 papers?


No, but I've done all of them so if anyone wants to peer mark feel free.
Original post by Mystery.
What's the way to solve simultaneous equations in three variables including quadratics without using matrices?
like these :


like these? you didnt list anything
Original post by Insecure
No, but I've done all of them so if anyone wants to peer mark feel free.


i have the solutions for 2001
For some reason my other attachments don't seem to be working but for this question I have drawn a graph, determined where the inequality part lies where it satisfies them all but I don't know how to express it in x and y.. I found the POI and I have got the coordinates but stuck on what to do next Capture3.JPG
Original post by DFranklin
You don't, unless there's a trick somewhere.


What's the trick in this one
Capture2.JPG
Original post by Insecure
No, but I've done all of them so if anyone wants to peer mark feel free.


I've been doing them today so shall I pm my answers or what? :smile:

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Original post by Mystery.
For some reason my other attachments don't seem to be working but for this question I have drawn a graph, determined where the inequality part lies where it satisfies them all but I don't know how to express it in x and y.. I found the POI and I have got the coordinates but stuck on what to do next Capture3.JPG


Just an idea- coordinates?

If you've done D1 it's basically linear programming

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Original post by theaverage
Just an idea- coordinates?

If you've done D1 it's basically linear programming

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nope, I didn't do D1.
Original post by Mystery.
nope, I didn't do D1.


Rearrange them into the form y=mx+c, draw the lines, shade the areas indicated by the inequality, then find the area that satisfies all three. Look for integer coordinates by inspection.
Original post by theaverage
I've been doing them today so shall I pm my answers or what? :smile:

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Sure.
Do you guys need to do this for Imperial Computing???
Reply 671
Original post by mrman2000
Do you guys need to do this for Imperial Computing???


Nope, not sure why they don't use it though seeing as they use the MAT anyway.
Original post by Quido
Nope, not sure why they don't use it though seeing as they use the MAT anyway.


Thanks for the response. I've seen a section on the MAT which is for computer science applicants only. I think this is for Oxford's MAT, but I'm not sure. I can't tell whether they share the same test or whether they're different.
Why is a^loga(y)=y?
Original post by Mystery.
What's the trick in this one
Capture2.JPG
OK, firstly, this doesn't "really" involve quadratics (i.e. its linear in the variables you are solving for).

That said, I'd say it's harder than you would be asked for these days; there was usually a question like this on the 1st year Tripos (although to be fair, it was always a little anomalous in terms of being a question-type that didn't really require even A-level knowledge).

So, solve it as you would any other set of linear equations (treating k as a constant), being careful to take account of any "special cases" where you may be dividing by zero (e.g. if you wanted to divide by (k-1), you'd have a special case when k = 1). The solution when "no special cases" occur corresponds to (i). For each special case you need to decide whether (ii) or (iii) occurs and provide a solution.

It's tedious and requires care, but there's no magic or mystery here - as the phrase goes, "Only way to do it is to do it..."
Original post by Mystery.
For some reason my other attachments don't seem to be working but for this question I have drawn a graph, determined where the inequality part lies where it satisfies them all but I don't know how to express it in x and y.. I found the POI and I have got the coordinates but stuck on what to do next Capture3.JPG
I think this predates D1/D2, so I'd say the expected method would be to plot the inequalities. The fact that you're expected to list the integer solutions implies there aren't going to be that many.
Reply 676
Original post by Nonamebzja
Why is a^loga(y)=y?


Pretty much by definition.

a^x = y

and

x = log_a(y)

are equivalent.
Original post by Nonamebzja
Why is a^loga(y)=y?


By definition. a^x=y --> x=loga(y).

For instance, log2(8)=3 since 2^3=8. 2^log2(8)=2^3=8.
Original post by RichE
Pretty much by definition.

a^x = y

and

x = log_a(y)

are equivalent.

Yeah thanks , oh and see like if a function f(x) has 2 turning points then f'(x) has 2 roots right?
Original post by DFranklin
OK, firstly, this doesn't "really" involve quadratics (i.e. its linear in the variables you are solving for).

That said, I'd say it's harder than you would be asked for these days; there was usually a question like this on the 1st year Tripos (although to be fair, it was always a little anomalous in terms of being a question-type that didn't really require even A-level knowledge).

So, solve it as you would any other set of linear equations (treating k as a constant), being careful to take account of any "special cases" where you may be dividing by zero (e.g. if you wanted to divide by (k-1), you'd have a special case when k = 1). The solution when "no special cases" occur corresponds to (i). For each special case you need to decide whether (ii) or (iii) occurs and provide a solution.

It's tedious and requires care, but there's no magic or mystery here - as the phrase goes, "Only way to do it is to do it..."


Thankyou so much.I'll keep that phrase in mind when I get stuck on a question next week.

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