Welcome, Guest. Please login or register.

July 23, 2019, 04:49:01 pm

Author Topic: help with a question please  (Read 3150 times)  Share 

0 Members and 1 Guest are viewing this topic.

Collin Li

  • VCE Tutor
  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 4966
  • Respect: +17
help with a question please
« Reply #15 on: November 04, 2007, 07:28:16 pm »
0
Quote from: "Freitag"
Coblin, i graphed the function and it appears to me the solutions for k are never for k > 0 ( only one solution exists). Wtf am i doing wrong ><


I fixed my solution, it neglected one thing. It is a considerably complicated question.

enwiabe

  • Putin
  • ATAR Notes Legend
  • *******
  • Posts: 4424
  • Respect: +527
help with a question please
« Reply #16 on: November 04, 2007, 07:29:52 pm »
0
Yup I realised too after and went to go fix it and made an arithmetic mistake. Good work Coblin. :P

melanie.dee

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 477
  • Respect: +1
help with a question please
« Reply #17 on: November 04, 2007, 07:31:07 pm »
0
righto sweet thanks a lot guys i get it now, i never wouldve worked that out on my own haha, in fact i completely forgot the discriminant ever existed, ah i shouldve also listened in yr 11 clearly

eta; ok fucki just got the first bit and now theres more haha? :( i will go reread the explanations

Freitag

  • Victorian
  • Forum Regular
  • **
  • Posts: 77
  • Respect: 0
help with a question please
« Reply #18 on: November 04, 2007, 07:37:12 pm »
0
Quote from: "coblin"
Quote from: "Freitag"
Coblin, i graphed the function and it appears to me the solutions for k are never for k > 0 ( only one solution exists). Wtf am i doing wrong ><


I fixed my solution, it neglected one thing. It is a considerably complicated question.


Too complex for VCAA, that's fo sho.
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

melanie.dee

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 477
  • Respect: +1
help with a question please
« Reply #19 on: November 04, 2007, 07:38:41 pm »
0
i just read the last part.. there is no effing way i wouldve known what to do myself there.

i hope thats not a vcaa standard question because if so, im screwed! haha

Collin Li

  • VCE Tutor
  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 4966
  • Respect: +17
help with a question please
« Reply #20 on: November 04, 2007, 07:43:24 pm »
0
Definitely not a standard question, but I wouldn't dismiss it as being too hard. It is just complicated and technical, and possibly too uninteresting and time consuming for VCAA to consider.

Freitag

  • Victorian
  • Forum Regular
  • **
  • Posts: 77
  • Respect: 0
help with a question please
« Reply #21 on: November 04, 2007, 07:47:15 pm »
0
Time consumption would be the reason I'd put it past VCAA.
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

Ahmad

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1296
  • *dreamy sigh*
  • Respect: +15
help with a question please
« Reply #22 on: November 04, 2007, 07:48:10 pm »
0
Here is my solution.

A = 2^x
y = A^2 - 5A = A(A - 5) = k (A > 0)

Now plot y = A^2 - 5A (quadratic)

Clearly, for two solutions for A > 0, A must be between 0 and 5. By sketching, it is clear maximum is 0 (not inclusive) for two solutions, and the minimum of the function (minimum of quadratic) is -25/4 (not inclusive).

Hence, k is in (-25/4, 0)
Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!

The collage of ideas. The music of reason. The poetry of thought. The canvas of logic.


Collin Li

  • VCE Tutor
  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 4966
  • Respect: +17
help with a question please
« Reply #23 on: November 04, 2007, 07:49:35 pm »
0
Quote from: "Ahmad"
Here is my solution.

A = 2^x
y = A^2 - 5A = A(A - 5) = k (A > 0)

Clearly, for two solutions for A > 0, A must be between 0 and 5. By sketching, it is clear maximum is 0 (not inclusive) for two solutions, and the minimum of the function (minimum of quadratic) is -25/4 (not inclusive).

Hence, k is in (-25/4, 0)


Ah yeah. I don't know what I was thinking when I did the final part without remembering that a is just simply greater than zero.

Ahmad

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1296
  • *dreamy sigh*
  • Respect: +15
help with a question please
« Reply #24 on: November 04, 2007, 07:51:35 pm »
0
Well, your solution is correct, just thought I'd offer a more graphical way of doing it. :)
Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!

The collage of ideas. The music of reason. The poetry of thought. The canvas of logic.


Freitag

  • Victorian
  • Forum Regular
  • **
  • Posts: 77
  • Respect: 0
help with a question please
« Reply #25 on: November 04, 2007, 07:51:51 pm »
0
Jesus Coblin, how could you have?
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

Freitag

  • Victorian
  • Forum Regular
  • **
  • Posts: 77
  • Respect: 0
help with a question please
« Reply #26 on: November 04, 2007, 07:52:23 pm »
0
Quote from: "Ahmad"
Well, your solution is correct, just thought I'd offer a more graphical way of doing it. :)


You stole my approach T_T

Except you got the same answer without a calc :/
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()