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  1. O

    Trivial Q 2

    Just a note : .25 is actually irrelevant, it could be any number eg. I{0--->pi/2}(1/(1+tan^.1(x))dx =I{0--->pi/2}(1/(1+cot^.1(x))dx thus 2I=I{0--->pi/2}(1)dx giving pi/4.
  2. O

    Trivial Q 2

    -------------------------------------------------------------------------------- Originally posted by underthesun I would have done I{a-->b}f(x)dx = I{a-->b}f(a-x)dx -------------------------------------------------------------------------------- ezzy85 quote: ye, thats what i was...
  3. O

    Trivial Q 2

    __________________________________________________ Affinity quote: well.. alternatively, you could do a x=arctan(y^4) substitution. __________________________________________________ Did you get the right answer pi/4 using that substitution ? Didn't think of that. Tell us, Affinity, where...
  4. O

    Trivial Q 2

    Give Spice Girl's integral notation a run, btw where's Spice? I{0--->pi/2}f(x)dx integral upper limit pi/2, lower limit 0 wrt x, and f(x)=1/(1+tan^.25 (x))
  5. O

    Trivial Q

    The sum is a constant but not the points. Anyway, nice work.
  7. O

    Trivial Q

    If 4 distinct points of the curve y=4x^4+14x^3+6x-10 are collinear, then their mean x-coordinates is a constant k. Find k. Trivial if you find the key!
  8. O

    Kermit's second problem.

    Following proves the general case which should be sufficient. A general parabola (x-h)^2=4a(y-k) could be written in parametric form as (h+2at , k+at^2). If a< 0 and (h,k) is in the first quadrant then the parabola could represent a projectile path. Let the end points of a horizontal...
  9. O

    Kermit's second problem.

    Nice one Affinity. Kermit reckons easier to prove general case -took off his Physics hat and put on the mathematics one instead.
  10. O

    Kermit's Problem

    ___________________________________________________ Quote Exeter: if the frog was frictionless it wouldnt be able to jump forward, only up. ___________________________________________________ Forgot to mention "frictionless except soles of feet". While in the topic of Frog Physics...
  11. O

    elegant problem 3

    Usually asked as part i) prove a+b>sqrt(ab) etc.
  13. O

    Kermit's Problem

    Very good, ND. Just a note on the Physics of Polynomials!? (as promised). 0<=d <= (v/g)*sqrt(v^2 - 2gh) give the boundaries of d. Could it then be deduced that maximum d=(v/g)*sqrt(v^2 - 2gh) is actually possible? The answer is yes. Since 1/2 v^2>gh (conservation of energy eqn., otherwise...
  14. O

    elegant problem 3

    btw, nice soln, Archman.
  15. O

    Kermit's Problem

    A frictionless* frog jumps from the ground with speed V at an unknown angle to the horizontal. It swallows a fly at a height h. Show that the frog should position itself within a radius of V/g SQRT(V^2-2gh) of the point below gulp point. This problem should come with a warning and a...
  16. O

    elegant problem 3

    Thanks McLake. &lt;
  17. O

    elegant problem 3

    Thanks Blackjack.
  19. O

    elegant problem 3

    what can I say? How did Affinity decipher this Q, and proceed to solve it. What went wrong with my typing? Typed it twice, edit it once. Just to test whether it does it again : a,b both positive and a+b &lt; ab. Prove a+b &gt; 4.
  20. O

    elegant problem 3

    a,b both positive and a+b &lt; ab. Prove a+b &gt; 4.
  21. O

    Elegant Problem 2

    underthesun: nice solid approach again. Archman: don't self efface, a good solution displaying some fancy combination algebraic footwork.
  22. O

    Elegant Problem 2

    Find the sum :nC1 (1^2) + nC2 (2^2)+...+nCn (n^2)
  23. O

    Elegant Problem

    underthesun: first to cross the line, solid approach. QUOTE "however, as the tangent meets the curve y = x3 with a "stationary point" manner, p(x) should have a double root at x = t." Or simply : point of tangency corresponds with repeated roots. Affinity :What a creative alternative!
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