PAT Solutions.

I have no idea what you mean feel free to copy them and post them in this thread so that you have a wider catalogue

I thought I'd contribute since I'm bored.

Q7

Which paper?

Q3 2010



Edit: Doh! It seems this too has been answered already! wcp100, could you make it clear in the original post which questions have actually been answered?

When you divided through in part (i) you lost some solutions - also, the given range implies that they want the answer in radians... part 2 is fine, though.

I'm scratching my head over this one, because he didn't lose any solutions. Had he found the solutions over the range specified by the question, he would have got 20, 80 and 140 which are all the solutions you got using your method.

If you factor out cos(3x) however, you get incorrect solutions.

$sin(3x)-\sqrt{3}cos(3x)=0[br]cos(3x)(tan(3x)-\sqrt{3})=0[br]cos(3x)=0, 3x=90, 270, 450, x=30, 90, 150[br]tan(3x)=\sqrt{3}, 3x=60, 240, 420, x=20, 80, 140$

All of the answers for cos(x) are wrong though, since when cos(x) is zero, sin(x) is non-zero. I must be overlooking something...

What I meant by losing solutions is that he got only 20 with his method and not 80 and 140, not that the solutions were wrong...

That's very intruiging. The only reason I can think of to explain it is that both factors are not necessarily zero - only one of them needs to be, really. Not sure if that is valid, though...

He just forgot to find the other solutions in the range $0\leq3x\leq3\pi$, which is not a case of losing solutions by division of zero.

But then if you look at the unfactorised expression, this cannot be the case. I think it has sometimes to do with dividing sin(x) by cos(x), but there must be some simple explanation that we're missing!

Ah OK.

No idea. Hopefully someone will come along and explain it to us.

2010 - Q1(i)

Couldn't you divide by zero. Then find all solutions for 3x between 0 and 3pi?

Divide by zero?!

I fixed it