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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level Year 10 just started 4u
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level f(x)=x-(1-x)^2+x^2-(1-x)^3+x^3-...-(1-x)^n+x^n Here is one, should be some more, how many are you after?
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon Sketch the graph y=\dfrac{x^3-a}{x^2-a} for: i) a>1 ii) 0<a<1
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level Hmm what do you mean by "distinct mod 3"? and do you mind showing me your method, sounds interesting :)
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level Here is my attempt edit: damn just realized there is an error on like 3rd last line, how do i prove l is the positive case?
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level I got (1+lcos2xl)/2?
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon 6ln2-5/2? Just use u=ln(...) And hence x=e^u+e^2u then IBP
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon Yeh the formula should work even for n>=3 but to end up with and integral we can easily evaluate, we would want to sub a number in the form n=3k+2
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon On the 8th line u made a tiny mistake, 1+2(n-2)/9=(2n+5)/9 instead of (2n+3)/9 As for the n greater than equal to 5, it is not as easy to integrate I_1 as compared to integrating I_2, thats probably why.
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon Very nice question :)
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level Here is my attempt, its a bit long, im sure you have a quicker method.
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level Managed to get rid of trial and error to: Since m and n are integers m+n=4k (1) and (m+n)^2-mn=49k (2) (k>0) Subbing (1) into (2): 16k^2-m(4k-m)=49k M^2-4mk+16k^2-49k=0 M=2k+(49k-12k^2)^0.5 N=2k-(49k-12k^2)^0.5 Discriminant must be positive (and...
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon - Advanced Level Find lim n-> -infinity (x^2+4x+7)^0.5+x
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon a bit trial and error but all i could think of
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    HSC 2014 MX2 Marathon ADVANCED (archive)

    Re: HSC 2014 4U Marathon This is what i thought of: We know that m+n and mn is an integer and hence (m+n)^2-mn (the denominator) is an integer So the expression can be written in the form p/q where p and q are integers. Now p/q=4/49=8/98=12/147=16/196... (multiplying top and bottom by...
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