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Original post by Non-Euclidean
Hi, can anyone shed any light on 2013 1J multi choice, i can't make sense of some of their justifications in the mark scheme thank you


All I did was draw a graph, which you can then see the area easily for n=1 and n=2.

Sub n=1 and n=2 into the four choices and you are left with one choice.
Can someone tell me if my answer to question 4 part iii 2013 would still get full marks?
The questions asks to show Sinθ+Cosθ0.5\displaystyle Sin\theta+Cos\theta \leq 0.5

The mark scheme did it by factorising but I did it a lot different. The question says the only thing you are allowed to assume Sin2θ+Cos2θ=1.\displaystyle Sin^2\theta+Cos^2\theta=1. Have I made another assumption by showing you can construct a square or is that okay?

Spoiler

Can anyone help me with marking? Do you get full marks if you get the right answer or do you have to show ALL the working? Especially in the multiple choice ones.
Original post by Hiling99
Can anyone help me with marking? Do you get full marks if you get the right answer or do you have to show ALL the working? Especially in the multiple choice ones.


multiple choice is you only get the 4 marks if you have ticked the correct box, anything else is zero (you can guess and get full marks). Im unsure about questions 2-7
Original post by KloppOClock
multiple choice is you only get the 4 marks if you have ticked the correct box, anything else is zero (you can guess and get full marks). Im unsure about questions 2-7

Thx, I usually show all working for all the others, it's only question one where my working makes no sense.
Original post by KloppOClock
All I did was draw a graph, which you can then see the area easily for n=1 and n=2.

Sub n=1 and n=2 into the four choices and you are left with one choice.


hmm..that helped eliminate a and c as the are of n=1 is easily calculated as 1 but for n=2 i not sure the area is so simple since for what n does the graph instantaneously become 3 as opposed to 2, it clearly becomes 4 for x=2 but the area is harder to calculate as the step to 3 isnt obvious (would be log2(3) which i cant work out in my head :P? Did your graph look like a set of stairs kind of thing?
Original post by Hiling99
Thx, I usually show all working for all the others, it's only question one where my working makes no sense.
As long as you make a logical, correct argument without making assumptions, that's all the working you need for 2-5 the mark scheme is only modelled it shouldnt be exactly what you put :smile:
Original post by Non-Euclidean
hmm..that helped eliminate a and c as the are of n=1 is easily calculated as 1 but for n=2 i not sure the area is so simple since for what n does the graph instantaneously become 3 as opposed to 2, it clearly becomes 4 for x=2 but the area is harder to calculate as the step to 3 isnt obvious (would be log2(3) which i cant work out in my head :P? Did your graph look like a set of stairs kind of thing?


Spoiler



if u still dont get it i can explain it over skype
(edited 7 years ago)
Original post by KloppOClock
Can someone tell me if my answer to question 4 part iii 2013 would still get full marks?
The questions asks to show Sinθ+Cosθ0.5\displaystyle Sin\theta+Cos\theta \leq 0.5

The mark scheme did it by factorising but I did it a lot different. The question says the only thing you are allowed to assume Sin2θ+Cos2θ=1.\displaystyle Sin^2\theta+Cos^2\theta=1. Have I made another assumption by showing you can construct a square or is that okay?

Spoiler


bump
Original post by KloppOClock

Spoiler


if u still dont get it i can explain it over skype
Yeah thanks a lot that makes sense, still want to be able to derive it though haha but process of elimination wins this one!
Original post by KloppOClock
bump


hm.. i think you'd probably get all the marks, maybe just for a=b r=0 by saying sinx=cosx for x=45, but that's a pretty cool way of thinking about it :smile:
Original post by Non-Euclidean
hm.. i think you'd probably get all the marks, maybe just for a=b r=0 by saying sinx=cosx for x=45, but that's a pretty cool way of thinking about it :smile:


thanks, ill add that extra bit to my solution
Man, there is not long till the exam at all! My revision has been totally messed up as I needed to have surgery today and these last 3 weeks I have not been able to focus at all. However, I realise not everything goes the way you want it to. I appreciate the posts in this thread and I really need to get my head down.
Original post by alfmeister
Man, there is not long till the exam at all! My revision has been totally messed up as I needed to have surgery today and these last 3 weeks I have not been able to focus at all. However, I realise not everything goes the way you want it to. I appreciate the posts in this thread and I really need to get my head down.


Hope you're okay, still plenty of time to get prepared for it.
Original post by KloppOClock
Hope you're okay, still plenty of time to get prepared for it.


Thanks for the concern, I'm doing alright just a bit drowsy from the anaesthetic but will be fine. didn't feel good for the three weeks prior to today. That is true ,Ill just have to get my head down over the next few days and try forget about the pain.
What did everyone do for part iii question 5 2012?

I don't see how they used the hint?

Posted from TSR Mobile
For 2011 Q5 ii can someone tell me if n^2-3n+4 would be an acceptable answeer because its not the answer they have given but I provided working and it seems to work ?
Reply 317
Original post by theaverage
What did everyone do for part iii question 5 2012?

I don't see how they used the hint?

Posted from TSR Mobile


The recurrence is a lot easier to understand in terms of m_n = l_n + 2.

Have you looked at the solution?
Reply 318
Original post by Mystery.
For 2011 Q5 ii can someone tell me if n^2-3n+4 would be an acceptable answeer because its not the answer they have given but I provided working and it seems to work ?


Do you mean 2011 Q5 (ii)? That is not asking for a specific answer. Part (iii) is, but the given answer of 2^(n-1) is very different from yours.
Original post by theaverage
What did everyone do for part iii question 5 2012?

I don't see how they used the hint?

Posted from TSR Mobile


I didn't get the same answer as them but I got a different answer that works, I don't get how they used the hint either.

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