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Recent content by Kandinsky

  1. K

    Mathematics: Post your doubts here!

    (a) If M is a minimum point, then dy/dx = 0 at this point. dy/dx = d(4*e^(x/2) - 6x + 3)/dx = 4* d(e^(x/2))/dx - 6 = 2* e^(x/2) - 6; dy/dx = 0 when e^(x/2) = 6/2 or, x/2 = ln(3) x = ln(3) = ln(3*3) = ln(9). hence a = 9; (b) The area enclosed by the curve is the definite integral of y wrt x...
  2. K

    Mathematics: Post your doubts here!

    To solve this problem, you should be familiar with the following fact: a plane can be completely determined by 2 intersecting lines. In this case, imagine: A and B are points that do not belong to the line l (obvious from (a)) now suppose that you already have a plane P that contains A and...
  3. K

    Mathematics: Post your doubts here!

    I think the question is a differential equation. so if sin2theta in your notation is sin(2*theta), not sin^2(theta), then dx/ (x+1) = cot(2*theta) d(theta) integrating both parts ln|x+1| = 1/2* ln|sin(2*theta)| + C x + 1 = sqrt(C* sin(2*theta)) and hence x = sqrt(C*sin(2theta)) - 1 Let us...
  4. K

    Further Mathematics: Post your doubts here!

    You have listed the roots for (a) w^12 = 1. Now, you have (z+2)^12 = z^12 z = 0 is not a root of the given equation, hence we can divide by z^12, with justification you get {(z+2)/z } ^12 = 1 Making a substitution y = (z+2)/z; you obtain the equation from (a). However, the task is to show that...
  5. K

    How was your math paper 33?

    OK, that's a mistake of CIE. Take a look at the line before the answer. It is: ln y = 3 ln (x^2 +4) + ln 32 - 3 ln 4 Simplify this: ln y = ln {(x^2+4)^3 * 32 : (64)} ln y = ln {(x^2+4)^3 : 2} hence y = 1/2 (x^2+4)^3; Alternatively, substitute the wrong answer y = 1/2 * (x^2+4) without cube...
  6. K

    October/November 2012 CIE |Question Papers ||Marking Schemes||Grade Thresholds| |Examiner Reports|

    Thanks! =)
  7. K

    Mathematics: Post your doubts here!

    In fact, it will not have roots only if k = 0. Here is the explanation: k^2 * x^2 + 2k*x + 1 = 0 (kx)^2 + 2*(kx) + 1 = 0 it seems like it is full square (kx+1)^2 = 0 so, kx+1 = 0 this equation does not have roots if and only if k = 0;
  8. K

    Statement of Results

    MindStealth Thanks for info!
  9. K

    Statement of Results

    OK. Thank you =)
  10. K

    Statement of Results

    What about provisional results slip? It is written for example as A(a) Sorry, bro, I think, I annoyed you =)
  11. K

    Statement of Results

    So, for example, B in A2 maths will be written as B(b), and B in AS maths?
  12. K

    Statement of Results

    Not really. Suppose Y took 1 AS and 3 A levels. How are they distinguishable(AS from A) on the provisional results list?
  13. K

    Statement of Results

    Oh... OK then)
  14. K

    Statement of Results

    That's cool. Em, what subjects and at what levels did you take?
  15. K

    Statement of Results

    Btw, do you have your statement of results and certificate?
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