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help with a question please

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Collin Li:
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:
Time consumption would be the reason I'd put it past VCAA.

Ahmad:
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)

Collin Li:

--- 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)
--- End quote ---


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:
Well, your solution is correct, just thought I'd offer a more graphical way of doing it. :)

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