BOSTES 4U conspiracy?? (1 Viewer)

[faisalabdul16]

2007 paper was nice!

Question 8 - the n sided polygon thingy, looks like a scary question but pretty straightforward

 

[faisalabdul16]

Realise, what did you get for BOS trials?

 

[Luckytree]

Hopefully, integration last question much like 2003 and 2010!

 

[Kurosaki]

I want combinatorics and probability in the last question

 

[Carrotsticks]

I want combinatorics and probability in the last question

^^ The most hated Extension 2 student in NSW.

 

[Cleavage]

meconics pt 2 hyperbola

watch it happen

 

[RealiseNothing]

I'm loving MC

are you kidding me? Q16 c) is flawed and so dodgy :/ but a) and b) were straightforward

Why was it flawed/dodgy?

I want combinatorics and probability in the last question

Sounds familiar...

 

[Kurosaki]

^^ The most hated Extension 2 student in NSW.

aDimitri, Cleavage and mreditor16 like this.

Wow, so many haters =P.

 

[faisalabdul16]

Why was it flawed/dodgy?
Sounds familiar...

I think mreditor was referring to the solutions to this question

<a href="http://www.codecogs.com/eqnedit.php?latex=$&space;Show&space;that&space;if&space;\sqrt{n&plus;\frac{1}{4}}>&space;k&space;-&space;\frac{1}{2}$&space;\&space;$then&space;the&space;integers&space;n&space;and&space;k&space;satisfy&space;\sqrt{n}>&space;k&space;-\frac{1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?$&space;Show&space;that&space;if&space;\sqrt{n&plus;\frac{1}{4}}>&space;k&space;-&space;\frac{1}{2}$&space;\&space;$then&space;the&space;integers&space;n&space;and&space;k&space;satisfy&space;\sqrt{n}>&space;k&space;-\frac{1}{2}" title="$ Show that if \sqrt{n+\frac{1}{4}}> k - \frac{1}{2}$ \ $then the integers n and k satisfy \sqrt{n}> k -\frac{1}{2}" /></a>

When you square both sides you get

<a href="http://www.codecogs.com/eqnedit.php?latex=n>k(k-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?n>k(k-1)" title="n>k(k-1)" /></a>

Solutions add 1 to the RHS without explaining why (the dodgy part I assume). I was told by friends and a teacher that difference in the two integers in an equality must be greater than 1 or something, hence adding one to the RHS doesn't change the inequality. LHS still > RHS. So that means 1/4 can be added and won't change the inequality.

<a href="http://www.codecogs.com/eqnedit.php?latex=n>&space;k(k-1)&plus;1&space;\\&space;\therefore&space;n>&space;k(k-1)&plus;\frac{1}{4}\\&space;n>&space;(k-\frac{1}{2})^{2}&space;\\\\&space;\therefore&space;\sqrt{n}>(k-\frac{1}{2})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?n>&space;k(k-1)&plus;1&space;\\&space;\therefore&space;n>&space;k(k-1)&plus;\frac{1}{4}\\&space;n>&space;(k-\frac{1}{2})^{2}&space;\\\\&space;\therefore&space;\sqrt{n}>(k-\frac{1}{2})" title="n> k(k-1)+1 \\ \therefore n> k(k-1)+\frac{1}{4}\\ n> (k-\frac{1}{2})^{2} \\\\ \therefore \sqrt{n}>(k-\frac{1}{2})" /></a>

 

[Kurosaki]

 

[RealiseNothing]

I think mreditor was referring to the solutions to this question

<a href="http://www.codecogs.com/eqnedit.php?latex=$&space;Show&space;that&space;if&space;\sqrt{n+\frac{1}{4}}>&space;k&space;-&space;\frac{1}{2}$&space;\&space;$then&space;the&space;integers&space;n&space;and&space;k&space;satisfy&space;\sqrt{n}>&space;k&space;-\frac{1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?$&space;Show&space;that&space;if&space;\sqrt{n+\frac{1}{4}}>&space;k&space;-&space;\frac{1}{2}$&space;\&space;$then&space;the&space;integers&space;n&space;and&space;k&space;satisfy&space;\sqrt{n}>&space;k&space;-\frac{1}{2}" title="$ Show that if \sqrt{n+\frac{1}{4}}> k - \frac{1}{2}$ \ $then the integers n and k satisfy \sqrt{n}> k -\frac{1}{2}" /></a>

When you square both sides you get

<a href="http://www.codecogs.com/eqnedit.php?latex=n>k(k-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?n>k(k-1)" title="n>k(k-1)" /></a>

Solutions add 1 to the RHS without explaining why (the dodgy part I assume). I was told by friends and a teacher that difference in the two integers in an equality must be greater than 1 or something, hence adding one to the RHS doesn't change the inequality. LHS still > RHS. So that means 1/4 can be added and won't change the inequality.

<a href="http://www.codecogs.com/eqnedit.php?latex=n>&space;k(k-1)+1&space;\\&space;\therefore&space;n>&space;k(k-1)+\frac{1}{4}\\&space;n>&space;(k-\frac{1}{2})^{2}&space;\\\\&space;\therefore&space;\sqrt{n}>(k-\frac{1}{2})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?n>&space;k(k-1)+1&space;\\&space;\therefore&space;n>&space;k(k-1)+\frac{1}{4}\\&space;n>&space;(k-\frac{1}{2})^{2}&space;\\\\&space;\therefore&space;\sqrt{n}>(k-\frac{1}{2})" title="n> k(k-1)+1 \\ \therefore n> k(k-1)+\frac{1}{4}\\ n> (k-\frac{1}{2})^{2} \\\\ \therefore \sqrt{n}>(k-\frac{1}{2})" /></a>

Well isn't that obvious?



then obviously



I don't really see how the question is dodgy, it's completely valid (all four parts).

 

[faisalabdul16]

Well isn't that obvious?



then obviously



I don't really see how the question is dodgy, it's completely valid (all four parts).

Yeah i know, however you don't know what n and k is in this case.
n can be 5 and k can be 4.99 (just for example) and then it wouldn't work, however with inequalities the difference between the two integers is greater than 1.

--------------------------
This is what I think editor found dodgy because solutions didn't explain why the 1 could be added -

imo, q16 was straightforward, but I can see that part of the question confusing a lot of people

 

[RealiseNothing]

Yeah i know, however you don't know what n and k is in this case.
n can be 5 and k can be 4.99 (just for example) and then it wouldn't work, however with inequalities the difference between the two integers is greater than 1.

--------------------------
This is what I think editor found dodgy because solutions didn't explain why the 1 could be added -

imo, q16 was straightforward, but I can see that part of the question confusing a lot of people

It tells you that n and k are integers though so you don't really have to worry about cases like the example you gave.

 

[Luckytree]

I think the reasoning was that 1 could be added on the right side so that it can be the equality sign/and inequality sign, because n and k differ at least 1, where n is greater or equal to k and also n,k are integers
and then the equality sign disappears because you took out 3/4

But i think you needed to have the equality/and inequality sign when you added by 1

 

[harrypotterfan]

Yeah i know, however you don't know what n and k is in this case.
n can be 5 and k can be 4.99 (just for example) and then it wouldn't work, however with inequalities the difference between the two integers is greater than 1.

--------------------------
This is what I think editor found dodgy because solutions didn't explain why the 1 could be added -

imo, q16 was straightforward, but I can see that part of the question confusing a lot of people

Are you serious? It was such an easy question... I literally did the whole thing in like 2 minutes LOL

 

[mreditor16]

Are you serious? It was such an easy question... I literally did the whole thing in like 2 minutes LOL

 

[mreditor16]

RN and FA16, that wasn't the dodgy part. dw about it guys.

 

[RealiseNothing]

RN and FA16, that wasn't the dodgy part. dw about it guys.

What was dodgy then?

 

[faisalabdul16]

Are you serious? It was such an easy question... I literally did the whole thing in like 2 minutes LOL

teach me

hahahah

 

[dan964]

Just did about half of 2005, man that paper took long. Then again I only spent 2 hr 15 on it, and got up to Q6.
Why is there always a question proving cos 36 and cos 72, they have popped up at least 3 times. At least twice in the past 8 years. I like those question.


#I love a good induction question. The one in 2005 was good, but I did it shiftly by using a reduction formula as a shortcut, and saying I have proven this statement to be true using this means.

#I love a good resisted motion question, especially in 3U. Those a=kv or a=kv^2 questions are so methodical.