H2 Math Qn

• Okazaki

  2 Sep 11, 18:26

  1. A curve has equation y=(ax^2 + 2)/(x-1). Show that if C has no stationary points,then -2
• FireIce

  2 Sep 11, 19:05

  ur qn not complete?

• wee_ws

  2 Sep 11, 19:15

  Hi,

    The graph of y = f(x) is symmetrical about the x-axis means that the points (x, f(x)) and (x, -f(x)) lie on the graph. Thanks.

  Cheers,
  Wen Shih

• Okazaki

  2 Sep 11, 19:25

  Hi thanks.what about the first part?How do I prove if a function have no stationary point, then the unkown constant in the function would lie in a certain range Of values?
• wee_ws

  2 Sep 11, 19:38

  Hi,

    Your question is missing some information.

    If the graph has no stationary points, we need to consider the equation dy/dx = 0. Typically, we will encounter a quadratic equation and we will then conclude that its discriminant < 0.

    Thanks.

  Cheers,
  Wen Shih

• Okazaki

  2 Sep 11, 19:49

  Oh ok thanks!I was confuse all the while why discriminant is involved.tyty
• frekiwang

  3 Sep 11, 09:32

  Q1

  dy/dx=[(2ax)(x-1)-(ax^2+2)1]/(x-1)^2=(ax^2-2ax+2)/(x-1)^2

  When dy/dx=0,

  ax^2-2ax+2=0

  The curve has no stationary point if this equation has no real roots.

  b^2-4ac<0

  4a^2-8a<0

  4a(a-2)<0

  0<a<2

   

• frekiwang

  3 Sep 11, 09:36

  At x=a,

  a=t^2+3

  t=+/- sqrt(a-3)

  y = t(x)=+/-sqrt(a-3)a

  Which implies for any x=a, y has two possible values symmetrical about x-axis. Therefore the graph is symmetrical about x-axis.

   

• aunty ling

  14 Oct 11, 22:27

  ~ min 30 posts for website links ~