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2014 HSC Mathematics Solutions (1 Viewer)

Nukeboy

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Just draw (sinx-1)(tanx+2) between 0 and 2pie. There are 3 roots, hence 3 solutions.

Just because your calculator says math error doesn't necessarily mean it doesn't count. There are times where algebra is incorrect.
Really? Wouldn't it be undefined at that point?
 

photastic

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Really? Wouldn't it be undefined at that point?
No, the two functions are multiplied, hence the asymptotes of the standard tanx is no longer at pie/2. I've been using different graphical applications and they are showing 3 roots, same with wolfram alpha.
 

Nukeboy

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No, the two functions are multiplied, hence the asymptotes of the standard tanx is no longer at pie/2. I've been using different graphical applications and they are showing 3 roots, same with wolfram alpha.
Oh wow I hope you are right. But would they expect you to consider all that in 2u?
Change in asymptotes etc
 

ymcaec

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Really? Wouldn't it be undefined at that point?
Just draw (sinx-1)(tanx+2) between 0 and 2pie. There are 3 roots, hence 3 solutions.

Just because your calculator says math error doesn't necessarily mean it doesn't count. There are times where algebra is incorrect.
Hmm I see your point. Didn't pick it up while I was rushing through it.
It could be an open circle on the graph (i.e. approaching 0) so there might only be two solutions.
 

ymcaec

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Oh wow I hope you are right. But would they expect you to consider all that in 2u?
Change in asymptotes etc
Don't know if it's a trick question or not haha. (doesn't look like it though)
 

Nukeboy

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Hmm I see your point. Didn't pick it up while I was rushing through it.
It could be an open circle on the graph (i.e. approaching 0) so there might only be two solutions.
So how would you determine the answer? Like Aaron said, wolfram alpha does show a root at pi/2. But my calc and a graphing calc says otherwise.
Would they end up putting 2 correct answers? That would be great :)
 

photastic

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Hmm I see your point. Didn't pick it up while I was rushing through it.
It could be an open circle on the graph (i.e. approaching 0) so there might only be two solutions.
Confirming that these graphs can't pick up open circles, therefore I think the answer is B algebraically but hoping C cos I spent so much time graphing it lol :p
 
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chrisdelinet

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I've seen many science papers have 2 or even 0 answers but this is something the mathemagicians will decide.
agreed, i hope there will be a huge debate going over a bostes atm. it'll be war with protractors flying, calculators smashing, circles destroyed, etc.
 

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The answer for question 7 is (B) without a doubt.

I'm not too sure why there is so much debate going on here over that, as x clearly cannot be pi/2. I addressed this silly error in the multiple choice of my BOS Trials for 2U as well, this very same trap.
 

bongoli

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The answer for question 7 is (B) without a doubt.

I'm not too sure why there is so much debate going on here over that, as x clearly cannot be pi/2. I addressed this silly error in the multiple choice of my BOS Trials for 2U as well, this very same trap.
Why can't x= pi/2 ?
 

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