BOS MX2 Trials discussion thread. (3 Viewers)

[barbernator]

You have something like -|x - y| and when x > y, it becomes -(x - y)

Oh whoops, I was considering Bk and Bk-1 as individual terms, but the x is to the same power so its the one term. sweet thanks

 

[deswa1]

It's not that much effort lol. Quickest way to find two remaining roots is to use the relationship between roots and coefficients.

Yeah I know- I didn't think of that in the moment though Though were you there when I outlined my solution to finding A(x) and B(x) for Q11? I think that is the most unco solution anyone did in the whole paper. Though I made up for it by getting the whole of Q14 pretty quickly which was good

 

[JaySimmo]

where are the solutions?

 

[alexandred]

Well, just spend 40 mins getting q8 out, christ. Awesome paper, thanks.

 

[Trebla]

Yeah I know- I didn't think of that in the moment though Though were you there when I outlined my solution to finding A(x) and B(x) for Q11? I think that is the most unco solution anyone did in the whole paper. Though I made up for it by getting the whole of Q14 pretty quickly which was good

Think you mentioned something about multiplying additional functions or whatever.

Carrot is still writing up the full solutions. I'm not sure when he will finish and upload them though

 

[AwksPear]

how do you do Q7....still dont get any of the answers LOL...thanks guys

 

[deswa1]

how do you do Q7....still dont get any of the answers LOL...thanks guys

There's a few ways. This is how I did it. Note that 4a+4b+4c=4 ->a+b+c=1

Now the largest surface area will come from where all the side lengths are equal (you can prove this but not neccessary for MC). Therefore a=b=c=1/3

Now A=6a^2 =6/9= 2/3

 

[AwksPear]

There's a few ways. This is how I did it. Note that 4a+4b+4c=4 ->a+b+c=1

Now the largest surface area will come from where all the side lengths are equal (you can prove this but not neccessary for MC). Therefore a=b=c=1/3

Now A=6a^2 =6/9= 2/3

LOL right....misinterpreted thought a+b+c=4...yea...thanks (Y)

 

[JaySimmo]

There's a few ways. This is how I did it. Note that 4a+4b+4c=4 ->a+b+c=1

Now the largest surface area will come from where all the side lengths are equal (you can prove this but not neccessary for MC). Therefore a=b=c=1/3

Now A=6a^2 =6/9= 2/3

but if all the sides are equal it is no longer a rectangular prism (as the question specifies)?

 

[barbernator]

Loving 16 c) i) now that I have time to fully comprehend it. Inductive reasoning

 

[deswa1]

but if all the sides are equal it is no longer a rectangular prism (as the question specifies)?

A cube is a rectangular prism- specifically a rectangular prism where all the side lengths are equal. Its like a square is still a rectangle

 

[deswa1]

Loving 16 c) i) now that I have time to fully comprehend it. Inductive reasoning

Is it actually doable? I'm still intimidated...

 

[barbernator]

Is it actually doable? I'm still intimidated...

Ive proved all the way up to there so far, except a) lol, and its not too bad. There is a bit of reasoning im not sure about, but otherwise good. but for that one, look at the far RHS of the inequality, and then consider if a0/a1 was the minimum. sub it in and u get a1(a0/a1) = < a0 which is true. So no matter what there must be a minimum value in which that holds, and by inductive reasoning it holds the whole way down the chain.

 

[asianese]

fuck. No wonder my rectangular prism didn't work. READTHEFKNQUESTION.

 

[Trebla]

The solution Q16 is actually not too bad when you think about it. I personally think it is not as difficult as some of the earlier questions.

 

[deswa1]

The solution Q16 is actually not too bad when you think about it. I personally think it is not as difficult as some of the earlier questions.

Really? Maybe it is and we all just ran out of time- I'll take a proper look tomorrow. Personally, I could only get a and b) i) 2,3 by assuming the result we had to prove in 1. Everything else though...

 

[barbernator]

oh and deswa, look at 16 c) ii) possibly the easiest 2 marks in the whole paper (srs) and iii) is exactly the same!!

 

[deswa1]

oh and deswa, look at 16 c) ii) possibly the easiest 2 marks in the whole paper (srs) and iii) is exactly the same!!

Haha effffff. I flicked to the very last part of Q16 just in case there was one of those limits questions or something that you could just read off but obviously there wasn't. If we had noticed those questions though...

 

[asianese]

I think the scare factor contributed to lots of the freak outs in this paper...haha

 

[Trebla]

I think the scare factor contributed to lots of the freak outs in this paper...haha

No guts. No glory.