amath qn

• suju

  1 Jun 10, 10:17

  Given that y = (x2 + 2x -c)/ (x - 1) and x is real, find the range of values of c for which y can take all real values.

   

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• Straightstick

  1 Jun 10, 10:53

  My guess is that c=3 for y to take all real values. if u perform long division, you get y=x+3 +(3-c/x-1)

  however, at x=1, eqn is undefined since denominator cannot be zero. So only way to turn the curve into line y=x+3 is to set c=3. for y=x+3, y can take all real values.

  but strictly speaking, i feel that my concept is flawed. the graph of ((x2+2x-3))/(x-1)) is actually y= x+3 but at x=1, eqn should be undefined. ie,when u draw y=x+3, x=1 should not be included into the graph. therefore,i am also in doubt of my answer. furthermore, qn asks for range of values of c. so i could most likely be wrong. lol.

   

• SBS261P

  1 Jun 10, 13:47

  my answer is c > = 3. dunno if my method is correct

  y = (x^2 + 2x - c) / (x - 1)

  y (x - 1 ) = x^2 + 2x - c

  yx - y = x^2 + 2x - c

  x^2 + 2x - yx - c + y = 0

  x^2 + (2 - y) x + (y - c) = 0

  Since x is real, discriminant > = 0

  (2 - y)^2 - 4 (1) (y - c) > = 0

  4 - 4y + y^2 - 4y + 4c > = 0

  y^2 - 8y + (4c + 4) > = 0

  Consider that this curve is always > = 0. Hence the discriminant of this equation is < = 0 (either real when it is 0, or nt real when it is > 0)

  so (-8)^2 - 4 (1) (4c + 4) < = 0

  64 - 16c - 16 < = 0

  48 - 16c < = 0

  48 < = 16c

  c > = 3

• eagle

  1 Jun 10, 14:01

  Originally posted by SBS261P:

  my answer is c < = 3. dunno if my method is correct

  y = (x^2 + 2x - c) / (x - 1)

  y (x - 1 ) = x^2 + 2x - c

  yx - y = x^2 + 2x - c

  x^2 + 2x - yx - c + y = 0

  x^2 + (2 - y) x + (y - c) = 0

  Since x is real, discriminant > = 0

  (2 - y)^2 - 4 (1) (y - c) > = 0

  4 - 4y + y^2 - 4y + 4c > = 0

  y^2 - 8y + (4c + 4) > = 0

  Since the equation needs to be >=0, it means it has either equal roots or no real roots

  so (-8)^2 - 4 (1) (4c + 4) < = 0

  64 - 16c - 16 < = 0

  48 - 16c < = 0

  48  < = 16c

  c  > = 3

  I was waiting for someone to answer :)

  Almost correct

  Editing in red

   

  Give me a while to post a graph for further explanations

• SBS261P

  1 Jun 10, 14:03

  yeah i change liao. i realised my mistake!

• SBS261P

  1 Jun 10, 14:05

  i realise when i sketch the graph, den i realise the working had some mistakes. =P

• eagle

  1 Jun 10, 14:05

  The graph here

  For c =2, c=3 and c=4

  

• eagle

  1 Jun 10, 14:06

  Originally posted by SBS261P:

  yeah i change liao. i realised my mistake!

  Yup I saw.

  Good question :)

• wee_ws

  1 Jun 10, 15:45

  Hi,

    It's A-level standard, rather than O-level :P

    This type of question is often asked about in graphs of rational functions.

    Thanks.

  Cheers,
  Wen Shih