Damn, but x can't equal to pi/2It's 3 solutions for the MC question :/
Inversetan(-2) has two in the domain and sinx had one
Damn, but x can't equal to pi/2It's 3 solutions for the MC question :/
Inversetan(-2) has two in the domain and sinx had one
b5 is 64-66 probsb5 cutoff boys??? i got 70/100 pls say i can b5 tyy
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.
Maybe OP can clarify?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.Really? Wouldn't it be undefined at that point?
Oh wow I hope you are right. But would they expect you to consider all that in 2u?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.
Really? Wouldn't it be undefined at that point?
Hmm I see your point. Didn't pick it up while I was rushing through it.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.
Don't know if it's a trick question or not haha. (doesn't look like it though)Oh wow I hope you are right. But would they expect you to consider all that in 2u?
Change in asymptotes etc
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.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 lolHmm 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.
Don't some papers have answers like "B OR C"?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
I've seen many science papers have 2 or even 0 answers but this is something the mathemagicians will decide.Don't some papers have answers like "B OR C"?
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.I've seen many science papers have 2 or even 0 answers but this is something the mathemagicians will decide.
wow, now that will be something ...Nek minute, its not B or C
Why can't x= pi/2 ?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.
Cos tan(pi/2) is undefined. Well there goes my atar.Why can't x= pi/2 ?
what a retarded questionCos tan(pi/2) is undefined. Well there goes my atar.
Already lost 10 marks, without sillies, fmlwhat a retarded question
I just realised what you guys are on about with the tan pi/2 thing (((
#typicalbostes
Same ((((Already lost 10 marks, without sillies, fml