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idk what you are trying to say

if you graph y=\sqrt{x} and you see at x = 0 those kind of points are considered critical points, because the function 'ends' there any 'sharp' points ( like cusps) or single points are critical points more egs y=\left | x \right |\\\\ y=\sin^{-1}x the sharp point on y = absx and the points for...

give an example

I tend to finish the paper in an hour

ln(ex)>e^{-x}\\\\ ln(e)>e^{-1}\\\\ ln(2e)>e^{-2}\\ ....\\\\ ln(n-1)e>e^{-(n-1)}\\\\ ln(ne)>e^{-n} \\ \\ can\ you\ see\ how\ to\ finish\ now

q8 is a monster

\\Determine\ the\ min\ and\ max\ values\ of\ f(x)=3x^4-20x^3+36x^2\\\ and\ sketch\ f(x)\ for\ -1\leq x\leq 4\\\\ Obtain\ the\ condition\ that\ f(x)=k\ has\ four\ distinct\ roots.\\ Prove\ that\ when\ this\ condition\ is\ satisfied;\\ the\ difference\ between\ the\ greatest\ and\ least\ of\...

that is better and you can shorten it \\assume\\\\ 7^k(1+3k)-1=9M \\\\for\ n = k+1\\\\ 7^{k+1}+3(k+1)(7^{k+1})-1\\\\ =7^{k+1}(3k+4)-1\\ =7^k(21k+28)-1\\ =7^k[(1+3k)+(27+18k)]-1\\ =7^k(1+3k)-1+7^k(9)(3+2k)\\ =9M+9(7^k)(3+2k)\\\\ which\ is\ divisible\ by\ 9

yea and there is a very simple way to do it

\\ Prove\ by\ induction\ that\\ 7^n+3n(7^n)-1\ is\ divisible\ by\ 9\ for\ n\geq 1

actually its my bad I thought it was pi/3 not 2pi/3 yes solution is valid

they are saying that CA = wCB, which is wrong eg if you have p=iq, then p =cis(pi/2)q so you are rotating vector q 90 deg anti clockwise so it should be CB = wCA

if you look theres solutions at the back but it is wrong x by vector = rotate anticlockwise not clockwise

i thought you were talking about q8 2010

you need x in it somewhere so the obvious choice is to put dv = dx

havent you tried last years q8

yea i was stuck and just tried that the previous part is irrelevant though

how did you see this part so quickly? It took me a while to figure out

here's a good one\\The\ n+1^{th}\ term\ of\ a\ sequence\ is\ given\ by\\\\ u_{n+1}=u_{n}+u_{n}^2\\\\ and\ if\\\\ u_{1}=1/3\\\\ Evaluate\sum_{n=1}^{\infty }\frac{1}{1+u_{n}}