ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: melanie.dee on November 04, 2007, 06:52:09 pm
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im sure this is really simple, i just have no idea what to do, and i have no clue with methods lol
its on the non calc exam, so how to do it wthout a calc please
For what values of k, where k is a real constant, does the equation 4^x - 5(2^x) = k, have two distinct solutions?
a) what does real constant mean haha :oops:
b) how do i even do this haha
im sure its ridiculously simple and you're thinking what an idiot haha, but i have no idea.
ps. if i get about 24/40 on the first exam (just did one and got that. everything i attempt i get right, i just dont know how to do half of it ahha) and same sort of thing for the second exam, what study score am i looking at?
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The two distinct solutions generally comes up when using the quadratic formula, with det > 0, but I don't think that'll come up in this question =/
It's more like, for what value of k will two different values of x satisfy the equation.
I saw the question too, wasn't sure how to approach it.
I'm interested to see Ahmad nail it :P
( EDIT ) Oh, and a real constant means a number without any complex numbers.
Check http://en.wikipedia.org/wiki/Real_number
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ahah that makes me feel better than its not just me that didnt know how to approach it.. normally its just because im so shit at methods lol.
anyway.. coblin, ahmad?
eta; thanks for the real constant thing.. i thought it'd be something simple like that, but my complete lack of homework or listening all year means i have no idea about all the terminology and such lol
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Which exam is this from? I know I've done it, but I've done a few.
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its from the mav cas 2007 exam.. are doing cas? its pretty possible it'd be on the normal mav 07 exam too though or something
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Yes, I'm doing CAS.
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Yay for CAS ^_^
I can't work out how to do it either. Strangly, this doesnt actually bother me, or my accompanied realisation I can't do half the methods exponential/log stuff anymore xD
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There's no way VCAA would give you such a question.. In my opinion anyways.
By graphing the function (plug in random x values and form a graph) you can see that the graph will have two distinct solution when -6.25 < k < 0 . (I used my calculator).
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It's basically asking how do we ensure that:
4^x - 5(2^x) = k, has two distinct solutions?
Revision
If you recall from year 11, there was the concept of the "discriminant" (Δ).
Δ = b^2 - 4ac, where ax^2 + bx + c = 0
If Δ < 0, then there are no solutions
If Δ = 0, there is one solution
If Δ > 0, then there are two solutions.
This comes from the quadratic formula, where we have +/- root(b^2 - 4ac) = +/- root(Δ).
If Δ is negative, we can't take the square root
If Δ is zero, we get zero: one solution
If Δ is positive, we get two different solutions.
Okay, so lets actually do this:
4^x - 5(2^x) = k
=> (2^x)^2 - 5(2^x) = k [Here you must recognise that 4^x = (2^x)^2]
Let a = 2^x: a quadratic equation is evident:
=> a^2 - 5a = k
=> a^2 - 5a - k = 0
To fix 2 solutions: Δ > 0
Δ = (-5)^2 - 4(-k) = 25 + 4k
=> 25 + 4k > 0
=> 4k > -25
=> k > -25/4
This ensures there are 2 solutions to a^2 - 5a - k = 0.
(i.e.: a = u, and a = v are solutions to the equation, where u and v are some real numbers)
=> 2^x = u, and 2^x = v
=> x = log2(u), x = log2(v)
This means that we must ensure the two solutions, u and v are greater than zero (because you can't log negative numbers, or zero).
Use the quadratic formula: a^2 - 5a - k = 0
a = [ 5 +/- root(25 + 4k) ] / 2
We need to ensure:
=> [ 5 +/- root(25 + 4k) ] / 2 > 0
=> 5 - root(25 + 4k) > 0
[I removed the positive root solution, because that will always be larger than the negative root solution: remember, we just want to make sure both of the solutions are positive, so if the smallest solution is positive, both will be]
=> root(25 + 4k) < 5
=> root(25/4 + k) < 5/2
Sketch graph, with the root(x) function translated 25/4 units to the left. Convince yourself that for the function to be less than 5/2, k must be less than some value, which we will now find.
Solve: root(25/4 + k) = 5/2
=> 25/4 + k = 25/4
=> k = 0
So, k < 0.
This means the boundary is: k for (-25/4, 0)
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this makes me feel so much better, i thought it'd just be me who couldnt do it haha!
i hate the non calc or notes exam. hmph. not that i really know how to use the cas properly anyway haha, but i always manage to fiddle around on the calc and find a solutioin someway ha
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im sure this is really simple, i just have no idea what to do, and i have no clue with methods lol
its on the non calc exam, so how to do it wthout a calc please
For what values of k, where k is a real constant, does the equation 4^x - 5(2^x) = k, have two distinct solutions?
a) what does real constant mean haha :oops:
b) how do i even do this haha
im sure its ridiculously simple and you're thinking what an idiot haha, but i have no idea.
ps. if i get about 24/40 on the first exam (just did one and got that. everything i attempt i get right, i just dont know how to do half of it ahha) and same sort of thing for the second exam, what study score am i looking at?
For two disctinct solutions, you need a discriminant greater than 0, meaning b^2 - 4ac > 0
Now, let A = 2^x,
so A^2 - 5A = k
A^2 - 5A - k = 0
So you need b^2 - 4ac > 0
meaning 25 - 4(-k)(1) > 0
25 + 4k > 0
k > -25/4
So 4^x - 5(2^x) = k has two distinct real solutions for {k: k > -25/4}
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beaten to the punch by 3 minutes. :P
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Well, my det > 0 proved to be useful ^^. I just wasn't sure how to apply it.
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ahhh thanks both of you, i shall now proceed to read that and let it sink in aha
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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 ><
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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.
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Yup I realised too after and went to go fix it and made an arithmetic mistake. Good work Coblin. :P
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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
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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.
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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
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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.
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Time consumption would be the reason I'd put it past VCAA.
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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)
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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.
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Well, your solution is correct, just thought I'd offer a more graphical way of doing it. :)
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Jesus Coblin, how could you have?
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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 :/