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
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: 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
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: 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