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Re: MX2 2015 Integration Marathon This is incorrect, because you did not notice the jump at x=0

Re: HSC 2015 4U Marathon - Advanced Level no the angle between a fixed line and a fixed plane is a constant, which is defined as the angle between this line and its projection onto the plane.

Re: HSC 2015 4U Marathon - Advanced Level $ My answer is that $ \max\frac{PR+PQ}{QR}={\rm cosec}\frac{\alpha}{2} $ (where $ \alpha $ is the angle between the line $ PQ $ and the plane $ p $ ) and that the maximum is attained when $ R $ lies in the line $ PH $ such that $ PR=PQ. $ Here $ H $...

Re: HSC 2015 4U Marathon $ Let $ z=x+iy, $ then $ z-2=(x-2)+iy, z+2=(x+2)+iy. $ Taking tangent on both sides of $ {\rm arg}(z-2)+{\rm arg}(z+2)=\pi, \frac{\frac{y}{x-2}+\frac{y}{x+2}}{1-\frac{y}{x-2}\times\frac{y}{x+2}}=0, $ which is simplied to $ xy=0. $ By inspection, the locus is the...

Re: HSC 2015 4U Marathon Here is my try 1+z=1+cos\theta+i\sin\theta=1+\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos\frac{\theta}{2}=2\cos^2\frac{\theta}{2}+2i \sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos \frac{\theta}{2}{\rm...

Re: HSC 2015 4U Marathon No the perpendicular distance is totally different from the distance between two points. The question actually asked us to find one point P from the line y=2x-5 so that the sum of AP and BP (where A(0,0) and B(0,2), AP and BP mean distances between points NOT perp...

Re: HSC 2015 4U Marathon The last part of the question is interesting. I like it.

Re: HSC 2015 4U Marathon because that is given

Re: HSC 2015 4U Marathon 16x^5-20x^3+5x-\frac{1}{\sqrt2}=0

Re: HSC 2015 4U Marathon good try but i'm afraid it is invalid. |z|+|z-2i|=|z-4+2i|+|z-2i|=|4-2i-z|+|z-2i|\geq|4-2i-z+z-2i|=|4-4i|=4\sqrt2, $ where equality holds true when the vectors $ 4-2i-z $ and $ z-2i $ are in the same direction. By solving simultaneous equations ( to find the point of...

Re: HSC 2015 4U Marathon NEXT QUESTION $ If the complex number $ z $ satisfies that $ |z|=|z-4+2i|, $ find the minimum value of $ |z|+|z-2i|.

Re: MX2 2015 Integration Marathon oh my bad, I really meant the solution given by InteGrand, where the substitution introduced a singularity. my solution does not contain any singularity

Re: MX2 2015 Integration Marathon the integrand has a discontinuity (either a finite jump or infinity) in the interval of integration or at either endpoint of the interval, like tangent of x in any interval containing \pi/2, 1/x in any interval containing 0

Re: MX2 2015 Integration Marathon $ if so, then what about $ \int_{-3}^3\frac{dx}{\sqrt[3]x}=0? $ true or false? $

well I don't have the Terry Lee book so I can't check out the solution. but if the information I got from the thread is precise and complete, then I am sure that the solution from the book was incorrect, the correct solution should be the intersection of the two circle inequalities, but not the...

Re: MX2 2015 Integration Marathon As for improper integral, consider the following statement, is it true or false? Justify your answer: $ Because $ y=\frac{1}{x} $ is an odd function, $ \int_{-3}^3\frac{1}{x}dx=0.

Re: MX2 2015 Integration Marathon Improper integral again? $ For $ \int_0^\pi\frac{dx}{2+\sin x}, $ to avoid the singularity, split the interval into two parts and take the substitution $ u=\pi-x, \int_0^\pi\frac{dx}{2+\sin x}=\int_0^{\frac\pi2}\frac{dx}{2+\sin...

how can arg(-3+9i/2) be acute?

Looking from the context, the 4p in the last line was a typo, should be 4q.

$ Your method is certainly valid--I would do exactly the same as you did. And your answer was pretty much the same as the solution, if you drew the circles $ x^2+y^2=1 $ and $ (x+2)^2+y^2=3, $ then the locus is the region inside the first circle and outside the second circle. $ The $...