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May 25, 2019, 02:52:39 am

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#### Collin Li

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##### help with a question please
« Reply #15 on: November 04, 2007, 07:28:16 pm »
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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

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##### help with a question please
« Reply #16 on: November 04, 2007, 07:29:52 pm »
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Yup I realised too after and went to go fix it and made an arithmetic mistake. Good work Coblin.

#### melanie.dee

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##### help with a question please
« Reply #17 on: November 04, 2007, 07:31:07 pm »
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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

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##### help with a question please
« Reply #18 on: November 04, 2007, 07:37:12 pm »
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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

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##### help with a question please
« Reply #19 on: November 04, 2007, 07:38:41 pm »
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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

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##### help with a question please
« Reply #20 on: November 04, 2007, 07:43:24 pm »
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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

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##### help with a question please
« Reply #21 on: November 04, 2007, 07:47:15 pm »
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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

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##### help with a question please
« Reply #22 on: November 04, 2007, 07:48:10 pm »
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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.

$-=\mathbb{A}\mathfrak{hmad Issa}=-$

#### Collin Li

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« Reply #23 on: November 04, 2007, 07:49:35 pm »
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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

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##### help with a question please
« Reply #24 on: November 04, 2007, 07:51:35 pm »
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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.

$-=\mathbb{A}\mathfrak{hmad Issa}=-$

#### Freitag

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« Reply #25 on: November 04, 2007, 07:51:51 pm »
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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

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##### help with a question please
« Reply #26 on: November 04, 2007, 07:52:23 pm »
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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 )