The HARD Maths Pop Quiz

JC 2K3

Right, so here's the idea, works the same as any other pop quiz topic - someone asks a question and whoever answers it correctly first gets to ask another. Now, as you can see from the title, the subject is Maths(HL) and I've specified that it's hard, so please post only challenging questions, ie. hard part (c)s from exam papers, mock papers, other exams you may have been given, Maths books or even questions you make up yourself(so long as they're on the HL course).

Perhaps if this topic develops fast then after a while we can make a compilation of all the hard questions and use them for self testing.....

Anyway, to set the ball rolling, from 1998 Q3 (c):
"Let z = cos(A) + isin(A)

Express 2/(1+z) in the form 1 - itan(kA) ; k E Q, z != -1"


(not so much hard as easy to go about the question in the wrong way, I guess. It just sticks in my mind as I, as well as a friend, struggled with it last week)

cson

Derive TanX from first principles. In our maths class today our teacher asked "What are the first principles I've been asking you to watch out for...Sinx, Cosx and....?" To which my mate replied "Tanx".

I shall leave now. Mathsy folk never seem to understand humour. Or at least my humour.

/Gets coat and leaves

Mushy

cson wrote:

Derive TanX from first principles. In our maths class today our teacher asked "What are the first principles I've been asking you to watch out for...Sinx, Cosx and....?" To which my mate replied "Tanx".

I shall leave now. Mathsy folk never seem to understand humour. Or at least my humour.

/Gets coat and leaves

Damn, I laughed..I'll get my coat too. Also will think about leaving honours Maths*


*obviously not serious

Jrembin

f(x) = tanx
f(x+h) = tan(x+h)
[f(x+h)-f(x)] = [tan(x+h) - tanx]

=[sin(x+h)]/[cos(x+h)] - sinx/cosx

=[(sinxcosx+cosxsinh)cosx - sinx(cosxcosh-sinxsinh)]

---

[(cosxcosh-sinxsinh)cosx]

I don't want to type this out but I multipled out the brackets and then changed cos^2(x) + sin^2(x) to 1 and ended up with

sinh/(cos^2(x)cosh-sinxsinhcosx)

[f(x+h)-f(x)]/h

= [sinh/(cos^2(x)cosh-sinxsinhcosx)]/h

=(sinh/h)*1/(cos^2(x)cosh-sinxsinhcosx)

=(sinh/h)*1/[cosx(cosxcosh-sinxsinh)]

=(sinh/h)*1/[cosxcos(x+h)]

lim h > 0 =

1 * 1/(cosxcos[x+0])

= 1/cos^2(x)

= sec^2(x)

Nobody answered JC's question so I won't ask one.

cocoa

sigh, spose i may as well give JC's Q a go...

2/[1+cosA + isinA] = 2[1+cosA - isinA]/[(1+cosa)^2 +sin^2A]
=2[1+cos-A + isin-A]/[2(1+cosA)]
=[1 + cos-A+isin-A]/[1+cos-A]
=1 - isinA/[1+cosA]
=1 -i{[(2TanA/2)/(1}+tan^2A/2]/[1+(1-tan^A/2)/(1+tan^2A/2)]}
=1-i{[(2tanA/2)(1+tan^2A/2)/(1+tan^2A/2)]/[(1+tan^2A/2 + 1-tan^2A/2)/(1+tan^2A/2)]}
=1-itanA/2
k=1/2

if you can read that mess without cringing then i salute you...
ok, new question...
proove that 1/log_a^b + 1/log_b^a equal to or greater than 2

ninebeanrows

I hate De Moivres, i just can't do the god damn thing!!

Q3 is lovely but with De Moivres likely to appear i've had to rule it out! Hope to god it doesn't come up this year

JC 2K3

What do you find hard about de moivre's? Seriously, once you get your head around it it's grand. It comes up most years.

Anyway:
proove that 1/log_a^b + 1/log_b^a equal to or greater than 2

log_a^b = log_b^b/log_b^a = 1/log_b^a

log_b^a = x

If x + 1/x >/= 2
x + 1/x - 2 >/= 0
x^2 - 2x + 1 >/= 0
(x - 1)^2 >/= 0
Which is true.

Therefore log_b^a +1/log_b^a >/= 2
and 1/log_a^b + 1/log_b^a >/= 2

Someone else post a Q, I can't think of one right now.

dan719

Prove that n(2^n-1)= n + 1(n all over one) + 2(n all over 2)+ r(n all over r)+...n

where I am using (n all over 1) etc as binomial/combinations notation.

Also find a fraction which can be expressed in partial fraction form as (x+(a+b) divided by x=a) + (a+b all divided by x+b)

I am so glad I saw this thread!lol:D