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^ It can't be 2 or above or else you could keep dividing.

A critical point is when the gradient of the tangent is undefined or "vertical". An example would be at the point (0,0) for y = sqrtx . It initially starts vertical at the origin

That's basically all you need. Except since you know it's reflected about the y-axis, see where it lies ON the y-axis. Sub x = 0, gives you y = 0. So it starts at origin. Unsure how to explain this part, but it has to be a stationary point at the origin, as symmetrical and differentiable curves...

Also glittergal and Braintic

Signing up for both ext 1 and 2. PM sent Thanks

Nitrates do contribute to eutrophification but to a lesser extent that phosphate ions.

Re: MX2 2015 Integration Marathon Nice solution Another way would've been to do IBP with the first ln(lnx) and then doing IBP again to the following integral which will cancel out the original 1/(lnx)^2

Re: MX2 2015 Integration Marathon Both are solvable

Re: MX2 2015 Integration Marathon $ Evaluate $ \int_0^{\frac{\pi}{2}} \ln(\sin x) dx \int \ln(\ln x) + \frac{1}{(\ln x)^{2}} dx

Re: MX2 2015 Integration Marathon it doesn't

Re: MX2 2015 Integration Marathon 1) $ If $ I_{n} = \int \cot^{n}x dx , $ then $ I_{0} + I_{1} + 2(I_{2} + ... + I_{8} ) + I_{9} + I_{10} = ? $ Express the answer in sigma notation form $ 2) $ If $ \int x^{\frac{-1}{2}} (2+3x^{\frac{1}{3}})^{-2} dx = A...

Who's Davidson Sole?

Re: 2015 permutation X2 marathon lol mine's way off. I did 1 - (n-1)^n / n^(n-1)

are you gonna do it this year??

Re: 2015 permutation X2 marathon new q: There are 'n' cars in a parking lot with 'n' exits. Any car can go into any exit. What is the probability that there is at least one exit which is not accessed by any car? Again, don't have the solution to this

Re: 2015 permutation X2 marathon You would only divide by like 3! if a question said: how many ways can 3 pairs of people up to play a game, right? as opposed to how may ways can you pick 3 pairs of people?

Re: 2015 permutation X2 marathon I was conflicted in the MC and couldn't decide whether we divide by 2 or not, so I didn't. I thought you'd divide by 2! only if the way that they are placed don't matter and something like 4,4,2 and 4,4,2 means 8 different people could be in the first two rooms...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread 2\cos2E - 7\sinE = 4 $ for $ 0 \leq E \leq 360 2(1-2\sin^{2}E) - 7sinE = 4 ( $ since $ \cos2E = 1-2sin^{2}E -4\sin^{2}E - 7sinE -2 = 0 4\sin^{2}E + 7sinE + 2 = 0 \delta = 49 - 4*2*4 = 17 ( $ what a weird number...

Re: 2015 permutation X2 marathon Sorry, but it was in a multiple choice that we got today in school. The choices were: 44100, 22050, and two other ones in the ten thousands but I can't remember them

Re: 2015 permutation X2 marathon new q: There are 10 people, and three rooms A, B, C. How many ways can all these ten people accommodate these rooms such that each room can only have a maximum of 4 people? not sure of the answer but I got: 44100