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F3 IAL June 2016 Model Answers

solutions to the IAL F3 edexcel.

I thought this was a tricky paper. Much tougher than the joke of a paper the UK lots had.

2b might have caused some problems
(edited 7 years ago)
F3 IAL S16.pdf982.0KB
Reply 1
Avatar for Zacken
Zacken
22
Original post by kprime2
solutions to the IAL F3 edexcel.

I thought this was a tricky paper. Much tougher than the joke of a paper the UK lots had.

2b might have caused some problems


Really? I thought it was about as easy as the UK paper. FWIW, the end of Q3(b) could be simplified by noticing that arsinh was an odd function so you had arshinh(1)=ln⁡(1+2)\text{arshinh}(1) = \ln(1+\sqrt{2}) as your answer (which is equivalent to yours, just neater).

Our matrix question was easier than it usually is, no fancy transformations or anything, so was the vectors, the integration reduction was a joke for 9 marks.

Thanks for the solutions, looks like I have full raw. :smile:
Reply 2
Avatar for kprime2
kprime2
OP
20
Original post by Zacken
Really? I thought it was about as easy as the UK paper. FWIW, the end of Q3(b) could be simplified by noticing that arsinh was an odd function so you had arshinh(1)=ln⁡(1+2)\text{arshinh}(1) = \ln(1+\sqrt{2}) as your answer (which is equivalent to yours, just neater).

Our matrix question was easier than it usually is, no fancy transformations or anything, so was the vectors, the integration reduction was a joke for 9 marks.

Thanks for the solutions, looks like I have full raw. :smile:


I think the UK paper only had one tricky question which I struggled with myself. This was question 7a, reduction formula for sin(nx)/sinx. The IAL still had parts where students could have made mistakes. The integration by substitution was quite tedious.

Well done on the full raw
Reply 3
Avatar for Ayman!
Ayman!
15
Original post by Zacken
arshinh(1)=ln⁡(1+2)\text{arshinh}(1) = \ln(1+\sqrt{2})


Nice - I did this as well!

Original post by kprime2
solutions to the IAL F3 edexcel.

I thought this was a tricky paper. Much tougher than the joke of a paper the UK lots had.

2b might have caused some problems


PRSOM. I can't remember my answers at all, so no predictions. :tongue: Looks like I should have 90 UMS at the very least, which is good enough for me!

A few were saying that it was easy which made me a little concerned about boundaries haha
Reply 4
Avatar for kprime2
kprime2
OP
20
Original post by Zacken
arshinh(1)=ln⁡(1+2)\text{arshinh}(1) = \ln(1+\sqrt{2})


I used the formula in the booklet.
arsinh(x)=ln⁡(x+x2+1){arsinh}(x) = \ln(x+\sqrt{x^2 +1})
(edited 7 years ago)
Do you think 73/75 will be full ums? Also in 2b i used a method with determinants. This method specifically: http://demonstrations.wolfram.com/TheAreaOfATriangleUsingADeterminant/.
Can you use methods that are not taught in the a level syllabus?
(edited 7 years ago)
Reply 6
Avatar for Ayman!
Ayman!
15
Original post by AlexFrangos
Do you think 73/75 will be full ums? Also in 2b i used a method with determinants. This method specifically: http://demonstrations.wolfram.com/TheAreaOfATriangleUsingADeterminant/.
Can you use methods that are not taught in the a level syllabus?


I used that method as well. It's acceptable since many IAL students use it. I've seen F1 mark schemes using it as an alternative.
Original post by Ayman!
I used that method as well. It's acceptable since many IAL students use it. I've seen F1 mark schemes using it as an alternative.


And do you think 73/75 will be full ums?

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