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Yes it is always true. Namely, if f(z)=g(z), then the conjugate of f(z)=conjugate of g(z).

and also this one about counting technique: How many whole numbers from 1000 to 9999 have 6 as the sum of their digits?

yes from the first graph to the second, it is a transformation indeed. by completing the squares, x^2-4x+4=(x-2)^2 and x^2-4x+3=(x-2)^2-1. so if the first function is f(x), then the second is f(x-2). and you just translate the first graph 2 units right.

for question 25, let the equation of the line be y-b=m(x-a), then use the condition that M is the midpoint of A and B to for an equation in m. noting that a and b are constants, solve the equation you can find m, thus the equation of the line. for the current one, following the idea of...

Re: HSC 2015 4U Marathon lol part (i) is too simple to be true and is irrelevant, maybe it means find P(a+b+c) i guess ? Edit: oh he actually meant P(a)+P(b)+P(c), and he wanted this result used for part (iii): P(x)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc, and sub. x=a, b, c and use P(a)+P(b)+P(c)=0...

Re: MX2 2015 Integration Marathon there is also algebra mistake: 1+x^6=(1+x^2)(1+x^4-x^2) not 1+x^6=(1+x^2)(1+x^2-x) so I would go I=3\int_0^\infty\frac{x^3}{1+x^6}dx=\frac{3}{2} \int_0^\infty\frac{x^2dx^2}{(1+x^2)(1+x^4-x^2)}=\frac{3}{2}\int_0^\infty\frac{udu}{(1+u)(1+u^2-u)} and...

Re: MX2 2015 Integration Marathon the algebra is wrong here I used the substitution x=\sin2\theta and got an answer of 4\sqrt{2}-\pi-2

Re: MX2 2015 Integration Marathon nice questions, I'll post my solution for question one. $ Denote $ I=\int_0^\infty\frac{\ln(1+x)}{(1+x)\sqrt{x}}dx. $ By the substitution $ x=\tan^2\theta, I=-4\int_0^{\frac{\pi}{2}}\ln\cos\theta d\theta $ Now let $ J=\int_0^{\frac{\pi}{2}}\ln\cos\theta...

draw a 2D diagram like this: ABCD is a rectangle where AB=8, BC=15 and BC is first on the ground level, now rotate the rectangle around point C so that point A is 12 above the ground. solve the 2D problem and we can manage to establish the equation 15\sin\theta+8\cos\theta=12 and the...

Re: MX2 2015 Integration Marathon split into partial fractions \frac{x^2}{1+x^4} = \frac{1}{2\sqrt2}\left(\frac{x}{1+x^2-\sqrt2x}-\frac{x}{1+x^2+\sqrt2x}\right)

Re: MX2 2015 Integration Marathon \tan^{-1}\left({\frac{1}{1+x(x-1)}\right)=\tan^{-1}x-\tan^{-1}(x-1)

when you have a fraction and don't use latex, you'd better add brackets to avoid confusing

Re: HSC 2015 3U Marathon there is an obvious solution which is t=0, by doing some calculus there is another one at around 10.35, but there is no formula for the solution thus you can not simply send it into calculator, but you can use Newton's method or other estimation approaches (even guess...

Re: MX2 2015 Integration Marathon most interesting question ever :haha: i am getting \frac{4}{3}+\frac{2\sqrt{3}\pi}{27} , kind of weird, need OP to check pls

and from here, i got the double roots are \alpha=1, \beta=5 , and the tangent is y=75x-30

hint, this is equivalent to that the quintic equation x^5-10x^4+22x^3+32x^2-20x+20=ax+b has one root x=-2 and two double roots, where y=ax+b is the mentioned tangent.

Re: MX2 2015 Integration Marathon 2-\frac{\pi}{2}

Re: HSC 2015 4U Marathon do u mean the answer IS 17pi/6 ?

Re: Find the value of k for which the line y=kx bisects the area enclosed by the curv did you draw a diagram? you should not integrate y=kx, but between the curve and the line