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Consider the equation x^3 + Ax^2 + (B-m)x + (c-k) = 0 the roots of this equation are the x-coords of the intercepts of the cubic and the line y=mx+k Now this means that the roots are in arithmetic progression (to be equally spaced in a line). Let these roots be u-d, u, u+d. sum of...

For the first one, assuming that x is a constant (cos otherwise we don't have anything about it), then let u = tanx I(u^n)dn = u^n/lnu + C = (tanx)^n / ln(tanx) + C For the second one: I((log[x,e])^-1)dx = I(lnx/lne)dx ..........change of base = I(lnx)dx and integrate this by parts

yes the assumption is legitimate Re(0) = Im(0) = 0

if u get something in terms of n, you use integration by parts to reduce it to an expression of an integral in terms of n-1 e.g. I(x^n*e^x)dx = I_n = x^n*e^x - I((nx^n-1)*e*x = x^n*e^x - nI_(n-1) sometimes they may ask u to put it in terms of an integral in terms of n-2, in which...

it's not multi-choice, and it's from a HARD maths comp you're given an average of 30min to solve each question

Your teacher is either v slack, v senile, or v deluded. This question is from 2001 UNSW school maths comp, and it's harder than a usual 4u question. Anyway, the solutions are (1,1,1) and (2,3,4) - all permutations of. Soln: Notice that if x >= 3 and x <= y <= z then xyz >= 3yz >= xy +...

yeh nbut u can snap it...it's very thin on the side

I dunno, only heard about it... If you hold the bottom end really low, and push the pip. filler downwards on the upper end really hard, you can actually break the tip of the pipette and it can become really sharp as you're pushing it.

ouch i've heard of someone who actually ran the pipette thru his palm...euck

Now we know b is the conjugate of a. ~a is a-bar, or the conjugate of a, so ~a = b, ~b = a so since y is real (y+b)^n = (y+~a)^n = {~(y + a)}^n = ~{(y+a)^n} thus (y + a)^n and (y + b)^n are conjugates. Now solving the quadratic, we know a, b = 1 + i, 1 - i We know z - ~z = 2i*Im(z)...

Yup, the induction above shows it's true for all positive integers. All integers are real anyway.

If n can be any real number, the expression won't be an integer, let alone divisible by something...

You're supposed to use Im{nz/(1-z) - z^2(1-z^n)/(1-z)^2} = n E (sin@+sin2@...........+sink@) k=1 Plug z/(1-z) = i(cis(@/2))/2sin(@/2) in, and expand and simplify.

This is from noticing that sin has a period of 2pi, and sin(x+pi) = -sinx

Trick here is to do it for n being odd, and for n being even, separately. n even: n = 0 obviously true assume n=k then sin(kpi+x) = (-1)^ksinx = sinx (as (-1)^k = 1) sin({k+2}pi + x) = sin(kpi + x + 2pi) = sin(kpi + x) = sinx = (-1)^k+2sinx n being odd...

Btw, I just realised, They all say in the books that conc H2SO4 is a catalyst for the esterification reaction. In actual fact, H2SO4 absorbs the water produced, driving the reaction to the right. It affects the equilibrium position! How can it be simply a catalyst?

Go tell your teacher to get a life. Alternatively, go tell her to join BOS and write exam papers along with all those other bunch of froot loops.

where did you get upto?

Re: answer thanks hehe good luck

tan 2 = sin 2 /cos 2 = 2 sin cos / (cos^2 - sin^2) =2(sin/ cos)/(1 - (sin^2/cos^2 ) ..................divide by cos^2 = 2 tan / 1 - tan^2 as required