Search results

Re: HSC 2015 4U Marathon - Advanced Level can you double check you really meant this one? $ if the series does converge, the limit $ \lim_{n\rightarrow\infty}\cos\frac{n}{2}\cos\frac{3n}{2} $ should be zero, but which i don't think has a limit. $

Re: HSC 2015 4U Marathon - Advanced Level whatsoever, z=z^5 is totally different from z^5=1, coz z=z^5 is kind of z^4=1

Well, if a line is a tangent to the parabola x^2=2ay ( again NOT x^2=4ay), what condition must be satisfied? If you can not solve it by tomorrow, i'll post my solution then. The reason I thought u misunderstood that line PQ is a tangent to x^2=4ay is that if (p+q)^2=4pq then indeed PQ is a...

Make sure you guys understood the question? it says the line PQ is a tangent to x^2=2ay, but NOT x^2=4ay

You guys are all wrong! If (p+q)^2=4pq were true, this means (p+q)^2-4pq=0, which implies (p-q)^2=0 and thus p=q, and this is a contradiction. And indeed, I can show (p+q)^2=8pq, as desired.

Since you have managed to work out all other parts, I believe you have had a quite good understanding of the question. So I only give you a few hints and don't need to waste time posting the full solution. 1. Find the equation of the locus as required by the question ( I think you have obtained...

Re: HSC 2015 3U Marathon NEXT QUESTION -- a mc on perm and comb Four shoes are selected from five distinct pairs of shoes. In how many ways can these four shoes include exactly one pair? A. 240 B. 140 C. 120 D. 60

Re: HSC 2015 2U Marathon My solution applies to more general situations, disregarding the position of the focus, we can find the length of any chord without finding the coordinates of points of intersection. $ Let $ P(x_1,y_1) $ and $ Q(x_2,y_2). $ Then $ y_1=2x_1+1, y_2=2x_2+1, $ and $ x_1...

Re: HSC 2015 4U Marathon - Advanced Level $ The inequality is equivalent to $ LHS=cda^2+dab^2+abc^2+bcd^2\leq4 $ by using common denominator for the desired result. $ Now $ LHS=(cda^2+abc^2)+(dab^2+bcd^2)=ac(ad+bc)+bd(ab+cd), $ in case (i) if $ ad+bc\leq ab+cd, $ then $ LHS\leq...

Re: HSC 2015 4U Marathon - Advanced Level This solution is invalid.

$ The expression $=\frac{\cos\frac{\alpha-\beta}{2}}{\cos\beta}-\frac{\sin\frac{\alpha+\beta}{2}\sin\beta}{\cos \beta}=\frac{\cos\left(\frac{\alpha+\beta}{2}-\beta\right)-\sin\frac{\alpha+\beta}{2}\sin\beta}{\cos\beta}= \frac{\cos\frac{\alpha+\beta}{2}\cos\beta+...

Re: HSC 2015 2U Marathon Next Question $ The line $ y=2x+1 $ cuts the parabola $ x^2=4y $ at two points P and Q. Without finding the coordinates of P and Q, find the length of chord PQ. $

in that identity, put theta=15 degree

top line, last term, should be plus sign instead of minus

Re: HSC 2015 3U Marathon 2^{11 \times 186} -1=2048^{186}-1, $ and try to factorise $ a^n-b^n. $ But this is not the hence method.$ $ The hence deduce method: from the proof of part (a), the conclusion can be actually strengthened to be $ 2^{11n}-1 $ is divisible by $ 2047(=2^{11}-1).

Re: HSC 2015 4U Marathon - Advanced Level This way I knew. but I thought Sy meant something else?

Re: HSC 2015 4U Marathon - Advanced Level How? Note that you have constant term other than zero.

Re: HSC 2015 3U Marathon $ You've actually found all three roots: from $ \sin3x=\frac23, 3x=41.81^\circ, 138.19^\circ, 761.81^\circ, $ hence the roots are $ u=\sin x=\sin13.94^\circ, \sin46.06^\circ, \sin253.94^\circ, $ of which the smallest positive value is $ \sin13.94^\circ.

Re: MX2 2015 Integration Marathon For the first integral I, reverse the quotient rule: \frac{4x^5-1}{(x^5+x+1)^2}=\frac{(-x^5-x-1)+x(5x^4+1)}{(x^5+x+1)^2}=\frac{(x^5+x+1)(-x)^\prime-(-x)(x^5+x+1)^\prime}{(x^5+x+1)^2}=\left(\frac{-x}{x^5+x+1}\right)^\prime. $ Therefore, $ I=-\frac{x}{x^5+x+1}+c.

Re: HSC 2015 4U Marathon - Advanced Level This solution is included in my solution, which is the case where \alpha=90degree, and the maximum is square root of 2, when R lies on the circle.