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Think of a general situation: let y = tan^-1[f(x)]. When you see this you do as substitution so: y = tan^-1u ............................ u = f(x) dy/du = 1/(1 + u²) .................du/dx = f'(x) hence dy/dx = 1/(1 + f(x)²) . f'(x) What you...

Glad I could help.

This is something where I think people would use a variety of methods. What I did was consider a cross section parallel to the x-axis and perpendicular to the y-axis. I then drew a graph of part of this cross section: the quarter-circle x = √(R² - z²) (z ≥ 0)...

I'll explain ii) first. It's tough to describe without an adequate graph but I'll try. Let the centre of the sphere be O (as in the diagram). Where they have that point at the top indicating the distance 'r', let that point be R. Let the point where that arrow indicating distance 'r' touches...

Due to my tendency to make stupid mistakes in a mass of working, for me longer ussually = harder.

Yeah, that was the first thing I thought of but I realised it could get messy so I came up with an easier way to do it (I say easier without having actually tried both).

= 81.32k.9 - (81(32k) - 5N).4 = 9.(81)(32k) - 4.(81)(32k) + 5(4N) = 5.(81)(32k) + 5(4N) = 5T ... where T is an integer. The step that you're missing each time is just rearranging terms so that you can take out a factor of five. It was the same deal above for divisibility by 20.

Maybe... give it a go.

To transform the S_k case into the S_k+1 case. 3^2k x 9 = 3^2(k+1) 2^2k x 4 = 2^2(k+1) In the question you have 81(3^2k) - 2^2k and you have to make up the powers to 2(k+1) = 2k + 2 somehow so I multiplied...

That's just an expansion of (a + b + c)² ≤ 3(a² + b² + c²) On the left side you have: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca On the right side you have: 3(a² + b² +...

Skipping past that n= 1 stuff... Let S_n be the proposition that 81(32n) - 22n is divisible by 5 i.e. that 81(32n) - 22n = 5M (where M is an integer) Assume true for n = k 81(32k) - 22k = 5Q (multiply through by [4][9]) 4.81(3^2k+2) - 9.(2^2k+2) =...

I like this quote: Another student, 17-year-old Kim, said: "It's just too time-consuming, life-deteriorating and fun-depleting."

= 41(20N+1) - 1 = 20(41N) + 41 - 1 = 20(41N) + 40 = 20(41N + 2) = 20R, where R is an integer

Aucun problème.

Also make sure you account for the fact that 0 < y &le; 2(1 + &radic;2)

Hmm, my geuss will be 0 if n is odd and ⁿC_(n-2)/2.(-1)^(n+2)/2 if n is even.

Say you let t = tanx, then cos2x = (1 - t²)/(1+t²) and sin2x = 2t/(1 + t²) Then cos4x = cos²2x - sin²2x = {(1 - t²)² - (2t)²}/(1+t²)² = (1 - 6t² + t⁴)/(1 +...

Goes to show, different things for different people. I found 2002 harder than 2003 (2003 was pretty cruisy but 2002 tripped me up in places).

Good timing, I just did the 2003 exam. I've got the MANSW solutions and the inflexion points are one thing I noticed (because I missed them :p). Good to hear they weren't in the marking guidelines.

Which book/test is it from?