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October 17, 2026, 05:44:08 am

Author Topic: VCE Methods Question Thread!  (Read 6292036 times)  Share 

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JackSonSmith

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Re: VCE Methods Question Thread!
« Reply #9075 on: March 05, 2015, 09:41:31 pm »
0
If the graph of y = 2x^2 – kx + 3 touches the x-axis then the possible values of k are:

A   k = 2 or k = –3
B   k = 1
C   k = – 3 or k = –1/2
D   k = 1 or k = 3
E   k = 2 root6 or k = –2 root6
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StupidProdigy

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Re: VCE Methods Question Thread!
« Reply #9076 on: March 05, 2015, 10:00:04 pm »
+2
If the graph of y = 2x^2 – kx + 3 touches the x-axis then the possible values of k are:

A   k = 2 or k = –3
B   k = 1
C   k = – 3 or k = –1/2
D   k = 1 or k = 3
E   k = 2 root6 or k = –2 root6
E
Working in attachment (using discriminant) :)
« Last Edit: March 05, 2015, 10:18:19 pm by StupidProdigy »
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JackSonSmith

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Re: VCE Methods Question Thread!
« Reply #9077 on: March 05, 2015, 10:43:15 pm »
+1
E
Working in attachment (using discriminant) :)

Thanks fellow 2014 Psychology student.
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JackSonSmith

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Re: VCE Methods Question Thread!
« Reply #9078 on: March 05, 2015, 10:56:33 pm »
0
Find the coordinates of the points of intersection of the line with equation y = x + k and the parabola with equation y = x^2 – 4x, where k > 0.

I let x+k = x^2 - 4x. and got (k,k^2)  and (5, 5+k) as the intercepts. They were not correct,
the answer is (5 +/- root(25+4k) / 2 , 5 + 2k +/- root(25+4k) /2 )

I can see that this has something to do with the quadratic formula, could someone please explain the theory behind this please?
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dankfrank420

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Re: VCE Methods Question Thread!
« Reply #9079 on: March 05, 2015, 10:58:36 pm »
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In composite functions, is the domain of the inside function always the domain of the composite function?

kinslayer

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Re: VCE Methods Question Thread!
« Reply #9080 on: March 06, 2015, 09:59:01 am »
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In composite functions, is the domain of the inside function always the domain of the composite function?

Not always. You may have to restrict the domain in order to restrict the range of the "inside" function so that it falls within the domain of the "outside" function.

For example, take f(x) = x and g(x) = sqrt(x). Then the domain of f(x) is R but the domain of g(f(x)) = sqrt(x) is .


kinslayer

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Re: VCE Methods Question Thread!
« Reply #9081 on: March 06, 2015, 10:09:31 am »
+1
Find the coordinates of the points of intersection of the line with equation y = x + k and the parabola with equation y = x^2 – 4x, where k > 0.

I let x+k = x^2 - 4x. and got (k,k^2)  and (5, 5+k) as the intercepts. They were not correct,
the answer is (5 +/- root(25+4k) / 2 , 5 + 2k +/- root(25+4k) /2 )

I can see that this has something to do with the quadratic formula, could someone please explain the theory behind this please?

Move everything to the left of the equals sign so that you have something that looks like ax^2 + bx + c = 0 and you can apply the quadratic formula. Or you can complete the square:




JackSonSmith

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Re: VCE Methods Question Thread!
« Reply #9082 on: March 06, 2015, 10:16:27 pm »
0
If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, the possible values of b are:
A   5 and 7
B   –5
C   –3 and –5
D   ±5 and ±3
E   ±8
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #9083 on: March 06, 2015, 10:20:55 pm »
+1
If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, the possible values of b are:
A   5 and 7
B   –5
C   –3 and –5
D   ±5 and ±3
E   ±8

Plug in those values and see which work.

JackSonSmith

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Re: VCE Methods Question Thread!
« Reply #9084 on: March 06, 2015, 10:28:19 pm »
0
Plug in those values and see which work.

Is there a more mathematical way to solve this?
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #9085 on: March 06, 2015, 10:40:12 pm »
+1
If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, the possible values of b are:
A   5 and 7
B   –5
C   –3 and –5
D   ±5 and ±3
E   ±8

Using quadratic formula, x = 1/2 (-b +- sqrt(b^2 - 60))
So for b to be an integer, b^2 - 60 must be a perfect square. It also means |b| > 7 for any solutions
Out of the given choices, only E could possibly work.
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Apink!

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Re: VCE Methods Question Thread!
« Reply #9086 on: March 07, 2015, 10:34:47 am »
0
Hi there,
I have a question and I was wondering if someone could help me?

the simultaneous linear equations:
2x - my = 0
2 (1-m)x+2y =m
where m is a real constant, have no unique solution:

So I solved for m which was -1 and 2 and substituted it in. They turned out to be parallel lines (hence no unique solution for both) but then if the one of the answers turned out to be the same line (infinite number of solutions), would that be still no unique solutions?

I think so, because even though there are solutions, they weren't be unique (i.e the solutions are all same)
Could someone please clarify for me?

Also, what are critical points when doing addition of ordinates?
I was sketching a graph of truncus + cubic and the answer sheet said a mark is for critical points but I don't know what they are.

Thank you!
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Gentoo

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Re: VCE Methods Question Thread!
« Reply #9087 on: March 07, 2015, 11:17:14 am »
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I see what you're saying, but an infinite number of solutions (the same line) couldn't really be described as 'no unique solutions'. The word 'unique' in this instance refers to the intrinsic difference in one solution to the next (e.g. x=3 vs x=4 vs x=5) of which there are an unending amount, so in reality there's an infinite amount of unique solutions.

As for critical points... Er, axis intercepts, asymptotes (which aren't points though lol), turning points? They all come to mind. They could also be points where one of the two graphs has an x-intercept, so the addition of ordinates graph would have a shared point that's the same as the OTHER graph that didn't have the intercept (this is one of he main strategies to sketching the graphs - although it doesn't really make said points more important, so idk really). It's a term I haven't really seen before.

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Re: VCE Methods Question Thread!
« Reply #9088 on: March 07, 2015, 12:36:26 pm »
0
when doing inequalities with squares , if you take the positive square root the inequality doenst change, but negative square root it does?

e.g. x^2 <= 25

x<= 5 and x >= -5

lzxnl

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Re: VCE Methods Question Thread!
« Reply #9089 on: March 07, 2015, 12:50:46 pm »
+1
Critical points technically are points in the function's domain where the derivative is zero or undefined. (e.g. cusp, vertical tangent). If they're being mathematically precise, this is what they should mean. However it's VCE...

when doing inequalities with squares , if you take the positive square root the inequality doenst change, but negative square root it does?

e.g. x^2 <= 25

x<= 5 and x >= -5

Yeah pretty much. Think of it as |x|^2 <= 25
|x| <= 5


I see what you're saying, but an infinite number of solutions (the same line) couldn't really be described as 'no unique solutions'. The word 'unique' in this instance refers to the intrinsic difference in one solution to the next (e.g. x=3 vs x=4 vs x=5) of which there are an unending amount, so in reality there's an infinite amount of unique solutions.

Unique means 'only one'. There is no unique solution if there is more than one solution. There is no unique solution to the equation x^2 = 1 because there are two. Similarly, there is no unique solution to the equation sin x = sin x because this is trivially true for all x.
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