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HSC 2015 MX1 Marathon (archive) (2 Viewers)

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InteGrand

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Re: HSC 2015 3U Marathon

davidgoes4wce said:
Curious to also know has a 'First Principles' question ever popped up in previous years' exam papers for MX1? 3 unit?

One of the students who I chat to says that their school has not covered it and the teachers have not put it into their school curriculum
Click to expand...
It has. See page 10 Q.7 (a) (i) of the 2009 HSC: http://www.boardofstudies.nsw.edu.a...doc/2009-hsc-exam-mathematics-extension-1.pdf.

And I'm pretty sure the teachers should have at least covered first principles differentiation in 2U.
 
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Re: HSC 2015 3U Marathon

Solve the equation:

 
leehuan

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Re: HSC 2015 3U Marathon

Assuming degrees.

90deg, 270deg

Shouldn't this be in the 2U thread?


Edit: Qn done above in rads
 
Speed6

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Re: HSC 2015 3U Marathon

leehuan said:
Assuming degrees.

90deg, 270deg

Shouldn't this be in the 2U thread?


Edit: Qn done above in rads
Click to expand...
Hmm I think this is leaning towards 3U???

Simple question but it can also be considered 2U..
 
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Re: HSC 2015 3U Marathon

2U are expected to know how to change a domain and take positive and negative roots so I wouldn't consider this harder 2U for the 3U student.

Something at the 3U exclusive level would be something like sin(x) + sqrt(3)cos(x) = 1, which uses the auxiliary angle method.

Harder 2U would be something rigorous involving tan^2(x) and sec^2(x) and cos^2(x) and etc all in the one go, which requires extensive use of Pythagorean identities and the ratio identities of tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x)
 
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Re: HSC 2015 3U Marathon

Speed6 said:
Hmm I think this is leaning towards 3U???

Simple question but it can also be considered 2U..
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It depends on how you approach the question

Ekman did it using a 3 unit method (i.e. Double angle)

He used the identity: cos(2x) = 1 - 2sin^2(x)
 
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Re: HSC 2015 3U Marathon

Fitzpatricks book definies the locus "A problem to be solved by eliminating the parameter from some parametric equations (ususally x(t)), y(t)) to form a new Cartesian equation (y=f(x))
 
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Re: HSC 2015 3U Marathon

Can everyone see the latex on tapatalk? I seem to be having issues rendering it.


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HSC 2015 3U Marathon

davidgoes4wce said:
I dont see Latex on yout tapatalk or the message you posted
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I mean for previous questions , I can't see any latex at all when I'm on my phone


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SammyT123

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Re: HSC 2015 3U Marathon

davidgoes4wce said:
Fitzpatricks book definies the locus "A problem to be solved by eliminating the parameter from some parametric equations (ususally x(t)), y(t)) to form a new Cartesian equation (y=f(x))
Click to expand...
Eliminate p by making it the subject .
X=...
Y=...
And link the equations


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