H2 Math : Algebra

• Theplayfulgirl brainy

  30 Jan 13, 18:34

     Hi

    There is a Algebra question. I don't know which topic it come from so..

    Anw, It is : By solving the equation: sin2x = cos 3x find sin18 o.

     I can solve it but how to find  sin 18o.

     My solution:

      sin2x =cos 3x 

  => sin 2x = sin (pi/2- 3x)

  =>     2x = pi/2- 3x +k2pi

     or  2x = pi/2 +3x+ k2pi

   

     Thank you

   

• Mad Hat

  30 Jan 13, 19:58

  sin 2x = cos 3x
  ⇒ sin 2x = cos(2x + x)
  ⇒ sin 2x = cos 2x cos x - sin 2x sin x
  ...
  ⇒ 4 sin²x + 2 sin x - 1 = 0
  ...
  ⇒ sin x = ¼(-1 ± √5)

  Then we notice that x = 18° satisfies sin 2x = cos 3x.
  And we know that sin 18° is positive. (First quadrant.)
  So sin 18° = ¼(-1 + √5).

   

   

• Theplayfulgirl brainy

  31 Jan 13, 00:28

   How stupid I am.>"<

    Thank you

• Theplayfulgirl brainy

  1 Feb 13, 13:57

    When I solve a math problem,there is something like this:

   Deduce that: 1/(N^2+N+1) -1/(4N^2+2N+1) < 1/N^2

   How can we induce like that

• Mad Hat

  1 Feb 13, 15:04

  Assuming that N is a positive number, notice that 1/(N² + N + 1) is already less than 1/N². (Because it has a greater denominator.)

  And then we subtract 1/(4N² + 2N + 1), which is a positive quantity.

  So the final answer will be even less than 1/N².

• Theplayfulgirl brainy

  1 Feb 13, 22:23

     Ah ha.

     The other question about function sketching:

    I have a function : ( -x^2 -x+1)/(x^2)

     Normally, we find that lim when x go to infinity is -1 => asymptote is y =-1.

     But when I draw it,both use hand and GC there is no that asymptote.

     Is it anything wrong with me or something special in this excercise?

• Theplayfulgirl brainy

  1 Feb 13, 23:14

   How to solve this: differentiation 2 prime of y respect to x:

    x = e^t 

    y = e^t (t-1)

    Thank you

• 8light

  2 Feb 13, 03:26

  y=-1 should indeed be an asymptote. Try going to a greater range of x to see if the asymptote appears.

  _________________________

  x = e^t  =>  t = ln x

  dx/dt = e^t

  dy/dt = t e^t

   

  Using the above information,

  dy/dx = (dy/dt) * (dt/dx) = (dy/dt) / (dx/dt) = t = ln x

  d2y/dx2 = 1/x 

• Mad Hat

  3 Feb 13, 17:19

  For y = −x²−x+1 / x², you should get this graph:

  

  From the shape, we can see a horizontal asymptote at the bottom, which is indeed the line y = −1. If you are not sure whether your GC is drawing a graph properly, you can check with this website, which is very easy to use:

  https://www.desmos.com/calculator

• Theplayfulgirl brainy

  5 Feb 13, 14:06

      1. About the graph, we can easily see that if x = 1, y =-1 so y=-1 cannot be asymptote

          It is so confusing

      2. I had new Math problem: 

    Find sum from 1 to infinity of 

            ln ( 1+ 1/x) .

    How to do it ?

• Mad Hat

  5 Feb 13, 16:39

  It's okay if the graph cuts an asymptote in the middle, as long as it doesn't cut at the extreme ends. Sometimes you will see a graph like this:

  

  The red dashed line still counts as a horizontal asymptote. The behaviour at the extreme ends (as x approaches positive or negative infinity) is the main thing, not what happens in the middle.

  For the ln question, re-express ln(1 + 1/x) as ln (x+1)/x
  = ln (x+1) − ln x.

  Then use method of differences.

   

   

• Theplayfulgirl brainy

  5 Feb 13, 23:54

    I 'm not really understand your solution in Ln excercise. Do you use Maclaurin series?? But I don't know how it relate now?  Can you elaborate more

• Mad Hat

  6 Feb 13, 00:41

  Originally posted by Theplayfulgirl brainy:

    Find sum from 1 to infinity of 

            ln ( 1+ 1/x) .

    How to do it ?

  Sorry maybe I misunderstood your question. I thought you were asking for this sum:
  ∞
  Σ ln (1 + 1/x)
  x=1

  Because for this sum, you can just continue:

  = ∞
  = Σ ln (x+1)/x
  = x=1
  = ∞
  = Σ  ln(x+1) - ln x
  = x=1

  And then you can use the method of differences. No need Maclaurin.

  If this is not what you meant, can you clarify what "sum from 1 to infinity" you meant?

• Theplayfulgirl brainy

  6 Feb 13, 17:14

    Yes. It is what I want. But I still not understand your method. 

• Mad Hat

  6 Feb 13, 22:42

  Oh I see, you are not familiar with the method of differences? Ok, I show the working. First, we calculate the sum of the first N terms. (Sum from 1 to N.) After that, we see what happens when N → ∞.

  N
  Σ ln (1+1/x)
  x=1

  = N
  = Σ ln (x+1)/x
  = x=1

  = N
  = Σ  ln(x+1) − ln x
  = x=1

  = (ln2−ln1) + (ln3−ln2) + (ln4−ln3) + (ln5−ln4) + ... + (ln(N+1)−lnN).

  = ln(N+1) − ln1. Because all the positive and negative terms in between cancel out. This is the method of differences. You can often use this method to calculate the sum of the first N terms. It's quite important at A-levels actually. Always come out.

  = ln(N+1).

  So we have shown that:

  N
  Σ ln(1+1/x)  =  ln(N+1).   Sum of the first N terms.
  x=1

  Now we see what happens as N → ∞. If you are familiar with the ln function, it is quite obvious that, as N → ∞, ln(N+1) → ∞. So:

  ∞
  Σ ln(1+1/x)  =  ∞.
  x=1

• Theplayfulgirl brainy

  7 Feb 13, 09:55

   Ak ha. Ok. Actually, I know this method but I don't know its name. :(

    Thank you

• Theplayfulgirl brainy

  26 Feb 13, 20:08

  http://www.flickr.com/photos/mabuthaodang/8506604619/

  Please help me question b and c.

  Thank you

• wee_ws

  26 Feb 13, 20:49

  Hi,

    For (b):

    1. Rewrite α/(2i) by multiplying top and bottom by i.

    2. Convert the complex number that is in exponential form to the trigonometric form.

    3. Obtain z = cos (α/2) + i {1 - sin (α/2)} as a result.
  

    4. Form an equation since |z| = sqrt(3), then simplify to obtain sin (α/2) = -1/2.

    5. Find α, taking note of the given range of values of α.

    Thanks.

  Cheers,
  Wen Shih
  wenshih.wordpress.com

• wee_ws

  27 Feb 13, 08:05

  Hi,

    For (c):

    1. Recognise that |z - (-1 + i)| = sqrt(2) represents a circle with centre at ? and radius of ?, passing through the origin.

    2. Recognise that |z| = |z -(-2)| represents a perpendicular bisector passing through the midpoint of (0, 0) and (0, 2).

    3. For each above, identify the region and take the intersection.

    4. To find the greatest and least values of |z - (4 + i)|, we are concerned with the maximum and minimum distance between the point (4, 1) and some point in the shaded region.

    Thanks.

  Cheers,
  Wen Shih
  wenshih.wordpress.com

• Theplayfulgirl brainy

  28 Feb 13, 18:39

   Thanks . Can you help me, one more?

  http://www.flickr.com/photos/mabuthaodang/8515758304/in/photostream

       Question 1 only

  http://www.flickr.com/photos/mabuthaodang/8514643815/in/photostream

      Question 5 only

  http://www.flickr.com/photos/mabuthaodang/8514643991/in/photostream

      Question 8

   Thank you

• Theplayfulgirl brainy

  28 Feb 13, 21:41

    Someone help me? I need it now

• Mad Hat

  1 Mar 13, 00:05

  Q1. Since f(2006) = 0, the min. value is at least 0. So the answer is either A or E. I suspect it's A but I don't know how to prove it in any simple way. Does anyone know? Is this an IB question? It's not A-level material.

  Q5. Since f^-1(³/₂) = 7, this means that f(7) = ³/₂. After that, it's easy to find w.

  Q8. This depends on the min value of (2004 cos2005x − 2006)². Can reason like this:

  −1  ≤  cos2005x  ≤  1
  ⇒  −2004  ≤  2004 cos2005x  ≤  2004
  ⇒  −4010  ≤  2004 cos2005x − 2006  ≤  −2

  So the min. value of (2004 cos2005x − 2006)² should be 4. Then just add the 2007.

• Theplayfulgirl brainy

  1 Mar 13, 08:00

   10 min before the exam...

    I have an idea for the question 8.

  (2004 cos2005x − 2006)^2 >=0

   => the min value when 2004 cos2005 =2006.

   => normal solution

   

• Theplayfulgirl brainy

  1 Mar 13, 18:39

     How to solve this:

      We have parabola : y =4ax^2. The region is limited by parabola, line y =b, x-axis is rotated 4 right angle about:

   a. x -axis

   b. y axis. 

           Find them.

   If the second one is 10 times the first one, find b followed a

• wee_ws

  2 Mar 13, 08:24

  Hi,

    Suppose a > 0 and b > 0 in the context of the problem.

    To find (a):

    1. Draw a diagram to identify the region.

    2. Find the x-coordinate of a suitable point of intersection between the parabola and the line y = b.

    3. Required volume = volume of cylinder (rectangular region)

                                       - integral of (pi)(y^2) dx (region under the parabola).

    To find (b):

    1. Required volume = volume of cylinder (rectangular region)

                                       - integral of (pi)(x^2) dy (region to the left of the parabola).

    Thanks.

  Cheers,
  Wen Shih
  wenshih.wordpress.com