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Haha, I'd have trouble thinking of something more annoying if I tried.

Q.26 Given n points in a plane, prove by Induction that the maximum number of lines joining any two points is (1/2)n(n-1), for n ≥ 2

Well, two things to look for are the following. If you see numbers out front in a question like: 1) Find the value of ⁿC₁ + 2ⁿC₂ + 3ⁿC₃ + ... + nⁿC_n then it's a safe bet that the question involves...

With the substitution method you want to change the variable that you're integrating 'with respect to'. i.e. you want du - not dx. If you let u = x³ + 1 and differentiate it ---> du/dx = 3x² giving du = 3x² dx ('dx' isn't strictly an algebraic term but...

A random interjection concerning your mention of Nietzsche. I though I might add that Nietzsche was very much aposed to anti-semitism and also nationalism in general (the former being more apparent in his personal relations and the latter in his works). Elizabeth Nietszsche did a good job of...

I had a sgt peppers album riff in my head during english - within you without you I think.

Nah, 2003 was a cruise.

Yeah, I got to the end and realised I'd barely touched it.

Circle geometry and conics pwned me. I just can't do conics. In any case I'll probably get 95-104 (about 14 of the marks definately lost are from conics and circle geo... my arch nemesis topics. Oh well, luck didn't swing my way - c'est la vie).

'nough said.

WLM is set to pwn us all.

Haha, my bad :p, yeh I meant y=1. well spotted.

Yeah, good luck guys. I hope y'all find the paper to your liking.

Give iv) ago by letting y = 1 and choosing an appropriate value of n (i.e. so that 4n+1 = 1001) EDIT: Also, for part iii) the reason you can integrate between 0 and y (where 0 ≤ y ≤ 1) is because in summing the geometric series in the form (x⁴ⁿ⁺²+1)/(1 + x²)...

for iii) integrate each part of the inequality w.r.t x between 0 and y 1/[1+x²] ≤ 1-x²+x⁴-...+x⁴ⁿ ≤ 1/[1+x²] + x⁴ⁿ⁺² ∫ 1/[1+x²] dx ≤ ∫ 1-x²+x⁴-...+x⁴ⁿ dx &le...

For part two: x² + 1 ≥ 1 ... and ... 0 ≤ x⁴ⁿ⁺² (even power) This gives you: 0 ≤ x⁴ⁿ⁺²/(1 + x²) ≤ x⁴ⁿ⁺² 1/(1+x²) ≤ (x⁴ⁿ⁺²+1)/(1 + x²) ≤ x⁴ⁿ⁺² + 1/(1+x²)...

Another question to attach to this one: Whenever they have these questions where you're summing a series they always make the assumption that |x| < 1 (where x is the common multiplied difference), why is that?

For part ii) &sum; vertical forces = 0 hence -------> Ncos&theta; + Fsin&theta; = mg (1) &sum; horizontal forces = mv²/r -----> Nsin&theta; - Fcos&theta; = mv²/r (2) sin&theta; x (1) - cos&theta; x (2) yeilds: F(sin²&theta; + cos²&theta; )...

When 0 < &theta; < 90&deg; then sin&theta; < tan&theta; Hence &radic;(rgsin&theta; ) <&radic;(rgtan&theta; ), that is, it doesn't have enough speed so it will start to slip down to slope, hence friction will be up the slope.

In case it helps this is why all that stuff works: log_e2 = log_e2 (which is obvious) log_e2 = (log_e2)(log_ee) (because log_ee = 1) log_e2 = log_ee^ln2 (using log laws) &there4; 2 =...