Search results

Re: HSC 2013 4U Marathon Well done. Alternatively: Re-arranging: x^4 \left( \frac{x^2}{y^2} - 1 \right ) + y^4 \left( \frac{y^2}{x^2} - 1\right) (x^2-y^2) \left(\frac{x^6-y^6}{x^2y^2} \right ) If x > y, then the both brackets are positive, multiplied together to become positive If x <...

Re: HSC 2013 4U Marathon Yep that's it, well done. For (ii) that is pretty much what you do === $Prove for any real$ \ x,y \frac{x^6}{y^2} + \frac{y^6}{x^2} \geq x^4 + y^4

Re: HSC 2013 4U Marathon A lot of the inequalities I post here I take directly from the books: - Mathematical Circles (Russian Experience) - The Art and Craft of Problem Solving - Paul Zeitz Some other questions I 'make', by copying a Wikipedia article Some of the sums that I post come from...

Re: HSC 2013 4U Marathon $Prove using differentiation by first principles the following$ $i)$ \ \frac{d}{dx} \ e^x = e^x $ii)$ \ \frac{d}{dx} \ \sin x = \cos x $Using the fact that$ \ e^x =\lim_{n \to \infty}\left( 1 + \frac{x}{n} \right )^n $iii)$ \ \frac{d}{dx} \ \ln x = \frac{1}{x}

Use the facts that for some general quadratic equation: $In the quadratic equation$ \ \ ax^2+bx+c= 0 \ \ $with roots$ \ \ \alpha , \beta \alpha + \beta = -\frac{b}{a} \alpha \beta = \frac{c}{a} How did I figure out to use these formula? Looking at the question its talking specifically...

Oh if the co-efficient was ((k+1)/4) it woudl be: (k+1)^2 - 4 \times 1 \times \frac{k+1}{4} \geq 0 (k+1)^2 - (k+1) \geq 0 We can either expand and simplify, or we can take the common factor (k+1) (k+1) ((k+1) - 1) \geq 0 k(k+1) \geq 0 Then you solve for the inequality =)

So we need to find the values of k so that: (k+1)^2 - 4(k+1/4) \geq 0 (k^2+2k+1) - 4\times k - 4 \times \frac{1}{4} \geq 0 k^2+2k + 1 - 4k - 1 \geq 0 k^2-2k \geq 0 Its up to you to solve this inequality now

Re: HSC 2013 3U Marathon Thread $A plane curve is shown below$ $The curve has parametric equations$ x=\cos^3 \theta y= \sin^3 \theta $for$ \ 0 \leq \theta \leq 2\pi $i) Show that the tangent to the curve at$ \ P(\cos^3 \theta, \sin^3 \theta) \ \ $is given by$ y \cos \theta + x...

Re: HSC 2013 2U Marathon This proof would probably be accepted at 2U level, same concept is used in loan repayments/annuities etc However above that I don't think it would be accepted at 3U and above Better proof: P($even product$) = P($ at least 1 even result$) = 1- P($no even results$) =...

Re: HSC 2013 2U Marathon don't be fooled, you will not need to use any 3U knowledge here :P You don't need to actually differentiate sin(x) using first principles, I am asking that using the differentiation from first principle, i.e.: Using the fact that: \cos x = \lim_{h \to 0}...

Re: MX2 Integration Marathon By parts yields: x\sin(x^2) - \int 2x^2 \cos(x^2) \ dx your integral had only a (2x) instead of (2x^2) === The integral in fact cannot be found in terms of elementary function. ==== \int \frac{\cos^3 x - \sin x \cos^2 x + 1}{\sin x \cos^2 x + \cos^3 x + \sin...

Re: HSC 2013 4U Marathon Ah ok my bad

Re: HSC 2013 4U Marathon That is indeed a function with those roots, its not a polynomial however

http://community.boredofstudies.org/showthread.php?t=296191

Re: HSC 2013 2U Marathon \int \frac{1+\cot x}{1-\cot x} \ dx

Re: MX2 Integration Marathon Alternatively multiply top and bottom by (1-cos x) to yield 2 'standard' integrals \int x^3 e^{x^2} \ dx

Re: HSC 2013 2U Marathon $By using differentiation from first principles of the function$ \ \ f(x) = \sin x $Prove$ \lim_{t \to 0} \frac{\sin t}{t} = 1

Re: HSC 2013 4U Marathon $Prove carefully$ \ \lim_{x\to \infty} \frac{\ln x}{x} = 0

Re: MX2 Integration Marathon Yep I guess that works, you could also use the identity: a+1/a \geq 2 ===========

I'm not going to do any of the generic essay business, I'll make sure to memorise my quotes, read certain professional analysis of certain texts (like Hamlet) and get ideas from there, and write a couple of essays to give to the teacher for feedback to improve style and other stuff like that...