Search results

Sorry to post this in this forum- where else do I belong! Is it possible with the Adobe/Acrobat software (the purchased version) to do the following : say I want to print the 10 exponential growth and decay problems from 2-unit HSC 1990-1999 (assuming there was one a year), can I cut and...

ND : Yes Isaac is rarin to swap car stories with you. Congrats and good luck.

Xayma, the induction variable is m - introducing another variable k could be confusing.

Thanks Guardian. Just as an update, broadband as a teaching medium seems slow to catch on, considering the convenience. I had but one Sydney broadband student last year. It worked well, we communicated clearly, he was able to save all the whiteboard discussions we had. He had a reasonably...

Hi everyone, I promised myself not to get sucked into this( forum) this year. Maybe just this one question. The sigma sum notation could be confusing, particularly when there is an embedded summation(see Q8 HSC2000) or product like this one. First thing you have to do is have a feel for the...

Wishing you all the very best. Remember : write fast; don't look back; don't be too careful and lose time; better off investing your time on a new question with fresh points than checking an old one that in all probability has been answered sufficiently; it is not the olympics (meaning...

____________________________________________________ "Hmm... am i the only one that does maths ext 2 without doing physics?!!" ____________________________________________________ Now, now ... don't you lose confidence over this. I know students who got top band without Physics. All the...

Maybe not. But a question like this would have leading parts. The general proof of AM-GM for any n is hard, but do practice the "dance steps" up to n=4.

Gather the question is : a^4+b^4+c^4+d^4<=4 then prove a^-4+b^-4+c^-4+d^-4>=4. Straight application of AM-GM. Let A=a^4, B=b^4, C=c^4 and D=d^4 then, a^-4+b^-4+c^-4+d^-4= 1/A+1/B+1/C+1/D =(BCD+ACD+ABD+ABC)/ABCD>=(4(A^3.B^3.C^3.D^3)^.25)/ABCD AM>=GM =4/(ABCD)^.25>=4 since...

Interesting problem. Affinity did well. Following might be an improvement. triangle inequality could also be : |a-b|<=c where a,b,c are sides of triangle. Consider the triangle whose vertices in the Argand Plane are 1,cis(arg z), and z. Then using the inequality, ||z-cis(arg...

Just to say that I've always appreciated wogboy's thorough commentary on various topics, especially inverse trig.

excellent solution from maniacguy, stuck close to the plot of the nth-root unity story. Nice effort from skunk as well. Would be good exercise to prove the same for a regular n-gon inscribed in a "floating" unit circle with center zc, where zn-zc not necessarily =1.

This question is from Moriah College 2003 Trial Q8. P_1,P_2,P_3,...,P_n represent the complex numbers z1,z2,z3,...,zn (zn=1) and are the vertices of a regular polygon on a unit circle. Prove that (z1-z2)^2+(z2-z3)^2+(z3-z4)^2+...+(zn-z1)^2=0

Function with 1,2,3... turning points would have at most 2,3,4... roots respectively.

______________________________________ Quote ND: Shouldn't harder 2u be in 3u? :p ______________________________________ Perhaps. Thought might be interesting using the second derivative to count solutions.

Consider the eqn. 2^x=1+x^2 i) find two obvious solutions. ii) show that there is another solution between 4 and 5. iii) show that these are the only solutions.

_________________________________________________ Quote McLake: nice Q (I leave it for HSC students) _________________________________________________ Thanks McLake. Though it has a distinct recycled feel to it: Q8 b 1997. ND shows mastery of complex vectors - a topic many students...

____________________________________________________ quote freaking_out: actually, the fact that the question is posted by "OLDMAN", turns me off straight away. ____________________________________________________ Criticism noted. An examiner's dilemma : pose a challenge, but be fair. Only...

___________________________________________________ Quote ND : But isn't that just a single step? I mean, isn't it like saying show x^2+x=x(x+1)? It's just one factorisation. ___________________________________________________ True. But I've seen simpler. May not necessarily be above...

well done, underthesun. I wonder how much harder it would've been if I named the subject "harder polynomial" rather than "integral/polynomial" thus not giving the integral method away. harimau : often, integral limits are simpler to work with than adding a constant.