Login

Welcome, Guest. Please login or register.

March 28, 2024, 11:18:12 pm

Author Topic: 3U Maths Question Thread  (Read 1230229 times)  Share 

0 Members and 4 Guests are viewing this topic.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #780 on: October 18, 2016, 09:34:52 pm »
0
If our working is a little dodgy (I.e made some leaps in my head) but the answer is correct can you still get full marks?
Yep leaps are fine unless it's one of those show that questions. If it says show that (or prove that etc.) you should only skip the trivial steps.

The HSC can't penalise you if you're showing an advanced level of algebraic skills and manipulation, until it becomes overboard, which isn't easy.

jamgoesbam

  • Adventurer
  • *
  • Posts: 15
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #781 on: October 18, 2016, 10:05:17 pm »
0
Hi again,
Could someone please explain these questions? Thanks in advanced! :)

In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?

Answer: 2/11

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #782 on: October 18, 2016, 11:16:26 pm »
+1
Hi again,
Could someone please explain these questions? Thanks in advanced! :)

In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?

Answer: 2/11

Hey! Okay, so for the first question, honestly the least confusing way to think about it is to draw a table of possibilities. Like, with all the stuff we know i  Extension, sometimes we go back to basics and wonder why the hell we were so confused. Here is the table with the possibilities on the outside, and the score on the inside.



The bits in green represent the throws where a '3' appeared; that's our restricted sample space! There are 11 elements in that restricted sample space, and two of those have a score greater than 5 (that's six). So, \(P=\frac{2}{11}\) ;D

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #783 on: October 18, 2016, 11:20:35 pm »
0
Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! ) 
I hope that was helpful. (Y)

Please quote the post you are referring to. This is definitely not a recent post you are alluding to, and potentially very outdated.

Mahan

  • Trailblazer
  • *
  • Posts: 31
  • Maths is beautiful !!!!!
  • Respect: 0
Re: 98 in 3U Maths: Ask me Anything!
« Reply #784 on: October 18, 2016, 11:28:06 pm »
0
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D

Solve the following equation:
2cos2ϴ=1-3sin2ϴ   0≤ϴ≤360

Thank you!

Neutron

Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! ) 
I hope that was helpful. (Y)
Mahan Ghobadi

Maths Tutor- ESL (80)| 2 Unit maths (96)(2013) | 3 Unit maths (99)| 4 Unit maths(95)| Physics (88)| music1(93)

Get more answers for your questions, as well as weekly tips and blog posts, from my friends and I at: HSC http://bit.ly/HSC-Help
VCE  http://bit.ly/VCE-Help

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #785 on: October 18, 2016, 11:30:22 pm »
+1
Hi again,
Could someone please explain these questions? Thanks in advanced! :)

In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?

Answer: 2/11

And now for your Calculus questions!

So the first one starts with some calculus:



Now in the range given, \(\sin{\theta}\) is always positive (and so is \(\theta\), clearly), so our condition is fulfilled!

Unfortunately for the rest of your questions I don't have the expression! Mind uploading what you have so far? I'll take it from there (or maybe this well help enough) ;D

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 98 in 3U Maths: Ask me Anything!
« Reply #786 on: October 18, 2016, 11:32:00 pm »
0
Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! ) 
I hope that was helpful. (Y)

Welcome to the forums! But that question is nearly 9 months old, love that you are keen to help out but be sure to click to the more recent pages to see the unanswered questions! :)

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #787 on: October 18, 2016, 11:35:12 pm »
0



jamgoesbam

  • Adventurer
  • *
  • Posts: 15
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #788 on: October 19, 2016, 12:51:26 pm »
0
And now for your Calculus questions!

So the first one starts with some calculus:



Now in the range given, \(\sin{\theta}\) is always positive (and so is \(\theta\), clearly), so our condition is fulfilled!

Unfortunately for the rest of your questions I don't have the expression! Mind uploading what you have so far? I'll take it from there (or maybe this well help enough) ;D

Hi Jamon,
Thanks for the help so far! This was the first part of the question I forgot to upload. I couldn't get Part D, and I checked some online answers which I still didn't understand (also attached below). Could you please explain what it all means? :'D Thanks heaps!! :)

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #789 on: October 19, 2016, 10:17:06 pm »
+1
Hi Jamon,
Thanks for the help so far! This was the first part of the question I forgot to upload. I couldn't get Part D, and I checked some online answers which I still didn't understand (also attached below). Could you please explain what it all means? :'D Thanks heaps!! :)

Awesome!! Okay, so if we want \(\frac{dA}{d\theta}=0\), we really just want this (the denominator can't be zero and neither can w):



Let's try and use what we proved above. If we go back to the function \(g(\theta)\), \(g(0)=0\). But the function is always increasing, that's what we proved, so that means that for any other value of \(\theta>0\), \(g(\theta)>0\).

So that means above, we resort to:



That's what the answer is getting at. Does it help? ;D

nimasha.w

  • Adventurer
  • *
  • Posts: 22
  • Respect: 0
  • School: taree high school
  • School Grad Year: 2016
Re: 3U Maths Question Thread
« Reply #790 on: October 21, 2016, 09:37:00 pm »
0
hi! can i please have some help on finding the latest time in part ii?

teapancakes08

  • Forum Regular
  • **
  • Posts: 81
  • "Smile for another day"
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #791 on: October 21, 2016, 10:37:31 pm »
0
Have any tips into showing the working for this question?

2. b) "Given that PQ passes through (0, 6a) find the equation of locus of T"

I worked out the first part which was (a) Find out the point of intersection of T of the tangents (2ap, ap^2) and Q (2aq, aq^2) on the parabola x^2 = 4ay" in which the answer was T ( a(p+q), apq) – found by finding the normal and using simultaneous equations. Would I use T to help solve question 2b?
Certified awkward turtle; HSC '17

Subjects:
English Advanced
Japanese Continuers
Mathematics
Mathematics Extension 1
Physics
Visual Arts

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #792 on: October 22, 2016, 12:28:45 am »
+1
hi! can i please have some help on finding the latest time in part ii?

Howdy! Sure thing!

Just as a side note, if you need help with a latter part of the question, try and attach your working so far! That way we can just pick up from where you left off :)

So the max height and time you obtained in A:



Now for Part B, let's get an expression for the angle of travel:



Now we assume that we are working with negative acute angles here; doesn't make sense otherwise. So we are looking at values of \(\theta\) between -45 and -60 degrees, meaning \(\tan{\theta}\) ranges from -1 to \(-\sqrt{3}\).



So that should be your answer! :)

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #793 on: October 22, 2016, 12:59:00 am »
0
Have any tips into showing the working for this question?

2. b) "Given that PQ passes through (0, 6a) find the equation of locus of T"

I worked out the first part which was (a) Find out the point of intersection of T of the tangents (2ap, ap^2) and Q (2aq, aq^2) on the parabola x^2 = 4ay" in which the answer was T ( a(p+q), apq) – found by finding the normal and using simultaneous equations. Would I use T to help solve question 2b?

Let me lend a hand! So you are correct, T is essential, since it has the coordinates we need. But we want more information to help eliminate our parameters.

Let's try and use the new info we were given. Let's look at the gradient of PQ:



But it passes through (0,6a), so let's get another expression for that:



Equating these:



So this is a new bit of information for us to use to find our locus! Remember, the idea is to tie x and y back in to the T coordinates. Let me show you; start by considering the y-coordinate:




And actually, believe it or not, that's it. We have an equation that defines a set of points that obey the given conditions! Any point on the line \(y=-4a\) will satisfy our requirements, and so THAT is the locus of the Point T :)



fizzy.123

  • Forum Regular
  • **
  • Posts: 72
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #794 on: October 22, 2016, 01:15:24 am »
0
Hi, this isnt an actual maths question, but i was wondering what mark out of 70 you will need for Maths ext 1 for it to be scaled to a 85+?