ATAR Notes: Forum
Archived Discussion => Mathematics and Science => 2013 => Exam Discussion => Victoria => Mathematical Methods CAS => Topic started by: BubbleWrapMan on November 06, 2013, 01:05:39 pm
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My solutions for both exams are below.
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for question 9ci, if we write g(x)=1/3|f(x+1)| does it still count as correct?
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for question 9ci, if we write g(x)=1/3|f(x+1)| does it still count as correct?
No, you had to use the actual function of f(x) and make the adjustments to it. Your final answer shouldn't be left with those transformations and "f(x)". If the function is given (which it was), it must be used to get full marks for the question. :)
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Sweet solutions thanks man
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For 9 ci
I left my answer as a hybrid function, would that be ok?
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For 9 ci
I left my answer as a hybrid function, would that be ok?
I believe you would still get the marks as long as the two rules are correct and a domain was included for each. :)
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Sick cheers Damoz!
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q8 should be 2+ 4/pi right?
check the negative sign on your solutions.
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q8 should be 2+ 4/pi right?
check the negative sign on your solutions.
Pretty sure it's negative
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I'll do it without simplifying any negatives from the start and showing more working than needed.
-0-\left[-\frac{4}{\pi}\cos\left(\frac{\pi x}{4}\right)\right]_{0}^{2} & =2\left(1\right)-\left(-\frac{4}{\pi}\cos\left(\frac{\pi}{2}\right)-\left(-\frac{4}{\pi}\cos\left(0\right)\right)\right)<br />\\ & =2-\left(-\frac{4}{\pi}\left(0\right)+\frac{4}{\pi}\left(1\right)\right)<br />\\ & =2-\frac{4}{\pi}<br />\end{alignedat})
Which is what Tim has.
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Also the expected value of
wouldn't be more than 2 since
from the density function.
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It can't be 2+4/pi because it's asking for the mean which surely has to be within the domain of [0,2], also 2-4/pi approximates to just under 1 because 4>pi which makes sense as the mean of a cos distribution if you look at the graph, the first quarter of a cos graph, the mean should be a little to the left of the middle value of 1 because that's where most of the area is
Beaten twice sigh
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My mistake!
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Exam 2 solutions in first post
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Exam 2 solutions in first post
Thank you very much!
OMG! I guessed the Equation of the second tangent equation, and I got it correct!!!! :D :D :D :D :D :D :D :D :D :D :D :D
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Not sure but for 10d)iii. Could the ans be 1/2 at k=8
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Not sure but for 10d)iii. Could the ans be 1/2 at k=8
For 4diii. you mean? 16/3 is correct.
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Sorry yes that
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Why cant it be 8? Since k is between [5,8]
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k=8 gives the maximum area.
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Oh shoot i mixed them around. Sigh...another few marks gone
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dat feeling when you find the correct tangent for last question but then enter it wrong in calculator and lose the next 6 marks :( :( :( :( :( :( :( :( :( :(
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dat feeling when you find the correct tangent for last question but then enter it wrong in calculator and lose the next 6 marks :( :( :( :( :( :( :( :( :( :(
ORRR that feeling when you GUESS the tangent equation (and get it correct) but don't continue on with the "guessed" equation. :'(
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haha yeah, got the gradient and stuffed up the area function, forfeiting the last 6 marks. I realised my mistake but only had time to save 4 of the 6
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haha yeah, got the gradient and stuffed up the area function, forfeiting the last 6 marks. I realised my mistake but only had time to save 4 of the 6
You'll get half marks for the rest for giving method though. I did the same thing :( >:(
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You'll get half marks for the rest for giving method though. I did the same thing :( >:(
Half marks? The only time you get half marks is if one assessor gives a mark and the other doesn't. But if its one assessor, the full thing has to be correct to get the one mark, otherwise they won't give you the mark.
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Half marks? The only time you get half marks is if one assessor gives a mark and the other doesn't. But if its one assessor, the full thing has to be correct to get the one mark, otherwise they won't give you the mark.
Sorry, let me rephrase. I meant method marks. Even if your solution to 4d)i is incorrect and you find the min/max by making deriv=0 then you will get the method marks but not the answer mark.