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    2 Unit Revising Marathon

    Re: 2 Unit Revising Marathon HSC '10 You'd be surprised ;)
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    2 Unit Revising Marathon

    Re: 2 Unit Revising Marathon HSC '10 In July 2001 a couple take out a loan of $800 000 from a bank to buy a house. The bank offers a reducible interest rate at 6% per annum. The couple is to repay the loan in equal fortnightly instalments of $25 000, but does not have to pay any interest for...
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    3 Unit Revising Marathon HSC '10

    \text{The binomial expansion is given by} \\ \sum_{k=0}^n ^nC_k \ x^{n-k}(-1)^k, \ n>k \\ \text{(i) By integrating the expansion above, show that} \\ \frac{(x-1)^{n+1}}{n+1}=\frac{(-1)^{n+1}}{n+1}+\sum_{k=0}^n \frac{^nC_k \ x^{n-k+1}(-1)^k}{n-k+1} \\ \text{where n and k are positive integers and...
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    2 Unit Revising Marathon

    Re: 2 Unit Revising Marathon HSC '10 Wrong.
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    3 Unit Revising Marathon HSC '10

    \text{(a) Prove by mathematical induction that } ln(n!)>n \text{ for all integers} \ n \ge 6 \\ (b) \text{Hence show that } \frac{1}{n!}<\frac{1}{e^n} \ for \ n\ge6 \\ (c) \text{Hence show that } \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}++\frac{1}{4!}... = \frac{103}{60} + \frac{1}{e^5(e-1)}
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    2 Unit Revising Marathon

    Re: 2 Unit Revising Marathon HSC '10 If a(x) and b(x) are even functions, a(-x)=a(x) and b(-x)=b(x) If c(x) and d(x) are odd functions, c(-x)=-c(x) and d(-x)=-d(x) (i) Let f(x)=a(x).b(x) Thus f(-x)=a(-x).b(-x) = a(x).b(x) = f(x) (ii) Let f(x) = a(x).c(x) Thus f(-x)=a(-x).c(-x) = a(x).-c(x) =...
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    Harder Circle Geometry

    lol, most of the harder circle geom problems in Ext 2 involve a cyclic quadrilateral. So once you see that everything becomes easy.
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    4 Unit Revising Marathon HSC '10

    That is from Cambridge! Oh wait no... I recall seeing it in my friend's First year lecture notes for USyd for some calculus topic...
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    2 Unit Revising Marathon

    Re: 2 Unit Revising Marathon HSC '10 2U (ie. long) method: All lines parallel to y=x+3 are of the form y=x+k, where k is a constant. Thus if y=x+k is a tangent to x^2+y^2=4 the quadratic formed by subsituting y=x+k into x^2+y^2=4 will have a discriminant of zero. Now sub y=x+kinto x^2+y^2=4...
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    4 unit maths terms...joking :)

    Their book is shocking for Mechanics :|
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    3 Unit Revising Marathon HSC '10

    Whoops... hahah... How could I forget? XD What you have found is the volume generated when area enclosed between the curve, y=7 and the Y AXIS is rotated about the y axis... RTQ... :P
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    4 unit maths terms...joking :)

    hey! = hey(hey-1)(hey-2)...(3)(2)1 :D
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    3 Unit Revising Marathon HSC '10

    d/dx a^x = x.lna Question: Find the volume generated when the area between the curve y=3+\sqrt{x}, the line x=16 and the x axis is rotated one revolution about the y axis. Edit: \text{Required Volume = Vol. Cylinder} - \int_3^7x^2dy \\ =\pi16^2.7-\int_3^7y-3dy \\ etc
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    Integration - Cambridge

    Post the question.
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    Taking Absolute Cases

    Yep also |x|=\sqrt(x^2)
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    Taking Absolute Cases

    That's how I was taught it.
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    Maths question!

    Could've also used Mathematica :P
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    "Turning Points" - Quality Sources?

    Stationary points, ie. local mins and maximums
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    Taking Absolute Cases

    Well, it isn't really negative either... if you know what I mean.
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    Taking Absolute Cases

    Well the definition of an absolute value is such that it is always positive. Thus |x| = x, if x>=0 (ie. if x is positive) and |x| = -x, if x<0 (ie. if x is negative) You can't really say that zero is negative, hence why equality can exist in the positive case.
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