H2 math

• wee_ws

  25 Apr 09, 19:41

  Hi,

    Here is a question that I developed for my students as a form of practice:

    The hyperbola H has two oblique asymptotes which intersect at the origin. The points A (2, 0) and B (-4, -sqrt(3)) lie on H. Find the equations of the two oblique asymptotes of H. [3 marks]

    H undergoes a scaling of 2 units parallel to the y-axis. Sketch the resulting graph of H, indicating clearly its oblique asymptotes and the coordinates of the points corresponding to A and B. [3 marks]

    Do give it a try it. Thanks!

  Cheers,
  Wen Shih

• wee_ws

  25 Apr 09, 19:56

  Originally posted by Destined2REIGN:

  
  Thanks! is the answer: a=3 b = 2?
  
  Got another qn:
  
  A ractangular hyperbola has asymptopes whose intersection point is the same as the intersection point of the curve in the earlier question and passes through (1,1). Find the equation of the hyperbola.

  Earlier Qn: Hyperbola intersects at x axis at x=-1 and x= 1 and y = 0.5
  
  And asymptotes x=-2 and y=-x+2. So which intersection point is the qn referring to?

  Hi,

    With the values you found for a and b, do the given points satisfy your expression? In Polya's problem-solving approach, that is the last step i.e. look back at the completed solution. 

    Starting point: a rectangular hyperbola has what general form?

    Point of consideration: could the intersection point be the location where the asymptotes x = -2 and y = 2 - x meet?

    Thanks!

  Cheers,
  Wen Shih

• Destined2REIGN

  25 Apr 09, 21:00

  Originally posted by wee_ws:

   

    Point of consideration: could the intersection point be the location where the asymptotes x = -2 and y = 2 - x meet?

  If this is the case, is it safe to say the asymptotes of the hyperbola of this qn pass through (1,1) not the hyperbola itself?

• Destined2REIGN

  25 Apr 09, 21:00

  Double

• wee_ws

  25 Apr 09, 21:59

  Hi,

    I feel that it's better to clarify with your teacher, rather than speculating :) Thanks!

  Cheers,
  Wen Shih

• Destined2REIGN

  6 May 09, 01:59

  Originally posted by wee_ws:

  Hi,

    I feel that it's better to clarify with your teacher, rather than speculating :) Thanks!

  Cheers,
  Wen Shih

  Just got it clarified. There was a missing sentence in the question. Thanks for yr help. =)

• eagle

  6 May 09, 09:43

  Mr Wee, can I check with you if mathematical induction on divisibility is in syllabus...

  The SEAB syllabus list did not specifically mention what type of mathematical induction is tested...

• Rudin

  6 May 09, 10:29

  Hi Mr Wee,

  I have difficulties with this question.

  Let a and b be positive integers such that ab + 1 divides a² + b². Show that (a² + b²)/(ab + 1) is a perfect square.

  Thanks for your help!

• wee_ws

  6 May 09, 13:13

  Originally posted by eagle:

  Mr Wee, can I check with you if mathematical induction on divisibility is in syllabus...

  The SEAB syllabus list did not specifically mention what type of mathematical induction is tested...

  Hi,

    No divisibility test is assessed.

    Specifically, the syllabus assesses:

    1. MI to prove simple results and statements involving sequence or series,
    2. MI to prove the formula for the general term of a sequence defined by a recurrence relation,
    3. How to make conjecture and prove it by MI.

    Thanks!

  Cheers,
  Wen Shih

• wee_ws

  6 May 09, 13:18

  --- deleted, due to flaw in working ---

• wee_ws

  7 May 09, 08:24

  Hi Rudin,

    It is an Olympiad question from 29th IMO in 1988.

    You may be able to read up details of the solution by searching the web.

    I'll just outline the steps:

    1. Let k = (a^2 + b^2)/(ab + 1).

    2. We show that among all pairs of nonnegative integers p, q with the property that (p^2 + q^2)/(pq + 1) = k and p >= q, the one with p + q minimal has q = 0. In this case k = p^2, a perfect square.

    3. Let p, q (where p >= q >= 0) be the pair that minimizes p + q and suppose that q > 0. Consider (x^2 + q^2)/(xq + 1) = k which simplifies to the quadratic equation in x, i.e. x^2 - qkx + q^2 - k = 0. One root is p and the other root is qk - p, since their sum is qk in the quadratic equation.

    4. Show that 0 <= qk - p < p contradicts the minimality of p + q.

    Thanks!

  Cheers,
  Wen Shih

• Rudin

  7 May 09, 23:32

  Originally posted by wee_ws:

  Hi Rudin,

    It is an Olympiad question from 29th IMO in 1988.

    You may be able to read up details of the solution by searching the web.

    I'll just outline the steps:

    1. Let k = (a^2 + b^2)/(ab + 1).

    2. We show that among all pairs of nonnegative integers p, q with the property that (p^2 + q^2)/(pq + 1) = k and p >= q, the one with p + q minimal has q = 0. In this case k = p^2, a perfect square.

    3. Let p, q (where p >= q >= 0) be the pair that minimizes p + q and suppose that q > 0. Consider (x^2 + q^2)/(xq + 1) = k which simplifies to the quadratic equation in x, i.e. x^2 - qkx + q^2 - k = 0. One root is p and the other root is qk - p, since their sum is qk in the quadratic equation.

    4. Show that 0 <= qk - p < p contradicts the minimality of p + q.

    Thanks!

  Cheers,
  Wen Shih

  Thanks Mr Wee...

  Gosh this is hard...

• Destined2REIGN

  30 Jul 09, 21:19

  A)if Y=sin x / x, prove that d2y/dx2 + 2/x.dy/dx + y = 0
  
  I manipulated to get yx = sinx and differentiated to get dy/dx and d2y/dx2. So how do I go on from there?
  
  B) If y = ae^(Sin x), a = constant, prove d2y/dx2 = dy/dx (Cos x - Tan x)
  
  Qn A and B are unrelated.

• Forbiddensinner

  30 Jul 09, 22:16

  Originally posted by Destined2REIGN:

  A)if Y=sin x / x, prove that d2y/dx2 + 2/x.dy/dx + y = 0
  
  I manipulated to get yx = sinx and differentiated to get dy/dx and d2y/dx2. So how do I go on from there?
  
  B) If y = ae^(Sin x), a = constant, prove d2y/dx2 = dy/dx (Cos x - Tan x)
  
  Qn A and B are unrelated.

  A) You should get dy/dx = (1/x^2)(xcosx-sinx)

  and d^2y/dx^2 = (1/x^3)(-x^2(sinx)+2sinx-2xcosx)

  With y = sinx/x, you just need to put them in the [d2y/dx2 + 2/x.dy/dx + y] order, and you will see how it becomes to zero.

   

  B) dy/dx = acosxe^(sinx) and d^2y/dx^2 = a(cosx)^2.e^(sinx) - asinx.e^(sinx)

  d2y/dx2 = dy/dx (Cos x - Tan x) is the same as proving

  [d2y/dx2] / [dy/dx] = (Cos x - Tan x)

  You should be able to see what comes next.

   

• SBS1984E

  30 Jul 09, 22:30

  Originally posted by Forbiddensinner:

  A) You should get dy/dx = (1/x^2)(xcosx-sinx)

  and d^2y/dx^2 = (1/x^3)(-x^2(sinx)+2sinx-2xcosx)

  With y = sinx/x, you just need to put them in the [d2y/dx2 + 2/x.dy/dx + y] order, and you will see how it becomes to zero.

   

   

  erm. I must be doing smth wrong. I got dy/dx as 1/2x^2.tan y
  
  as for B) managed to solve it with your help. Thanks =)

• Forbiddensinner

  30 Jul 09, 22:33

  Originally posted by SBS1984E:

  erm. I must be doing smth wrong. I got dy/dx as 1/2x^2.tan y

  Did you use the correct formula?

  dy/dx = [v.du/dx - u.dv/dx] / v^2,

  where u is the numerator and v is the denominator.

• Destined2REIGN

  30 Jul 09, 23:35

  Originally posted by Forbiddensinner:

  Did you use the correct formula?

  dy/dx = [v.du/dx - u.dv/dx] / v^2,

  where u is the numerator and v is the denominator.

  I brought the power up then I used the muliplication rule. ie the keep one diff other + keep other diff one that one.

• Forbiddensinner

  31 Jul 09, 11:13

  Originally posted by Destined2REIGN:

  I brought the power up then I used the muliplication rule. ie the keep one diff other + keep other diff one that one.

  So what did you get for dy/dx and d^2y/dx^2, and where are you having problems with?

• SBS1984E

  1 Aug 09, 21:35

  I got dy/dx =1/2x^2.tan y

• Forbiddensinner

  1 Aug 09, 22:14

  Originally posted by SBS1984E:

  I got dy/dx =1/2x^2.tan y

  y = sinx / x

  In this case, u = sinx and v = x

  dy/dx = [ v.du/dx - u.dv/dx ] / v^2

  = [ x( cosx ) - sinx( 1 ) ] / x^2

  = 1/x^2 [ xcosx - sinx ]

• SBS1984E

  12 Aug 09, 23:28

  Checking of  ans: Intergration By Parts.
  
  1) x^2 sin x dx
  
  ANS: cos x (-x^2 +2) +2x sin x +C
  
  2)e^x cos x dx
  
  Ans: (Cos x e^x) /2 + (sinx e^x)/2 +C
  
  3)ln x dx
  
  Ans: ln x - x + C
  
  4) tan ^-1 x dx
  
  Ans: x tan^-1 x - 0.5ln (1 + x^2) + C
  
  Thanks =) Just want to check if my answers are correct. I dun have e answer key.

• wee_ws

  13 Aug 09, 08:08

  Hi,

    1) is correct.

    2) is correct.

    3) is incorrect. Answer is x ln x - x + c.

    4) is correct.

    Good work!

    In future, you may use Wolfram Mathematica Online Integrator (which I did so) to help you check:

    http://integrals.wolfram.com/index.jsp

    Thanks!

  Cheers,
  Wen Shih

• eagle

  13 Aug 09, 08:43

  Originally posted by SBS1984E:

  Checking of  ans: Intergration By Parts.
  
  1) x^2 sin x dx
  
  ANS: cos x (-x^2 +2) +2x sin x +C
  
  2)e^x cos x dx
  
  Ans: (Cos x e^x) /2 + (sinx e^x)/2 +C
  
  3)ln x dx
  
  Ans: ln x - x + C
  
  4) tan ^-1 x dx
  
  Ans: x tan^-1 x - 0.5ln (1 + x^2) + C
  
  Thanks =) Just want to check if my answers are correct. I dun have e answer key.

  Give you something to try
  Past prelim question

   

  integrate x^3 cos (x^2)

  

• SBS1984E

  15 Aug 09, 13:08

  Originally posted by eagle:

   

   

  integrate x^3 cos (x^2)

  

  
  I stumped at intergrating the cos x^2 part.

• wee_ws

  18 Aug 09, 19:48

  Hi SBS1984E,

    Cambridge examiners would usually begin the question with:

    Differentiate sin(x^2) with respect to x.

    Try to integrate x^3 cos (x^2) again by applying the above result.

    Thanks!

  Cheers,
  Wen Shih