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Re: MX2 Integration Marathon Try this one \int(2x^2-1)e^{-x^2}dx

Re: HSC 2014 4U Marathon Use the bisector angle theorem, i.e., AB:AC=BP:CP, AB:AC=BQ:CQ So CP=240\times\frac13=80, CQ=240 Therefore, the radius is 160 because triangle APQ is right angled at A and PQ is the diameter.

Re: MX2 Integration Marathon Yup, nearly the same as my working. as a test of typseting LaTex, I got \frac{\pi}{\sqrt2} what was your answer

Re: HSC 2014 4U Marathon - Advanced Level Nice job!

Re: HSC 2014 4U Marathon - Advanced Level counterexamples are everywhere, a third times (a+b)=2, then a+b=6 which is greater than 2

Re: HSC 2014 4U Marathon - Advanced Level Are you sure? Your reasoning was quite wrong. (a+b) times a positive number=2, thus you deduce a+b less than 2?

Re: MX2 Integration Marathon My method--which I reckon kind of ugly--is: After twice applications of double angle formula, convert the integrand into a fraction of $\cos4x$. Then use the $t$-formula. Care must be taken for tangent function has infinity discontinuity at $\frac{\pi}{2}. So I...

Re: HSC 2014 4U Marathon - Advanced Level Try this one: Given that $a>0, b>0$ and $a^3+b^3=2$. Without using calculus prove that $a+b\leq2$.

Re: HSC 2014 4U Marathon Solve simultaneous equations to find points of intersection. Let the circle be $(x-h)^2+(y-k)^2=r^2$. Then sub $y=\frac{1}{x}$. Thine resulting quartic equation should have solutions $x_1, x_2, x_3, x_4$. Finally use product of roots.

gradient of tangent; so tan B=(dy)/(dx); by chain rule or implicit differentiation: (dy)/(dx)=(dy/dt)/(dx/dt)

The last integral is an application of integration by parts, though it is sophiscated because u and v must be carefully chosen. v = -(ln x + 1)^{-1}, u=x\ln{x}.