Prelim 2016 Maths Help Thread (3 Viewers)

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Green Yoda

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Oh yeah got it..I made silly mistakes while drawing in radian measure..I am really not used to that haha
 

Green Yoda

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help with this:
shade the region -|x|+2>=-√4-x^2


Should I be looking it it terms of the y values or the x values?
 

InteGrand

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help with this:
shade the region -|x|+2>=-√4-x^2


Should I be looking it it terms of the y values or the x values?
If we're talking about sharing that as a region in the x-y plane, since y doesn't appear in that inequation, y can be anything at all. So you just need to find the set of all x-values satisfying that equation, and then essentially draw vertical lines at each of these places (which will end up shading some part of the plane).

Also there should be brackets under the square root, I initially thought it was just sqrt(4), before looking at the actual number and realising it would be just said 2 if that were the case.
 

Green Yoda

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If we're talking about sharing that as a region in the x-y plane, since y doesn't appear in that inequation, y can be anything at all. So you just need to find the set of all x-values satisfying that equation, and then essentially draw vertical lines at each of these places (which will end up shading some part of the plane).

Also there should be brackets under the square root, I initially thought it was just sqrt(4), before looking at the actual number and realising it would be just said 2 if that were the case.
hmm I am not understanding this completely..What I do understand now is that we look at the x values..but how would I find the link to that inequality??
 

InteGrand

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hmm I am not understanding this completely..What I do understand now is that we look at the x values..but how would I find the link to that inequality??
Basically solving the inequation for x is the first step. Have you done that yet? Once you've done that, it's fairly easy.
 

Green Yoda

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Basically solving the inequation for x is the first step. Have you done that yet? Once you've done that, it's fairly easy.
nope haven't done that..?
In the question they told us to draw the two graphs and then shade the region..so they are asking us to do it graphically.
But I still dont seem to understand what the questions means.. Are they asking us to find for which values of x in -|x|+2>=-√(4-x^2)
 

InteGrand

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nope haven't done that..?
In the question they told us to draw the two graphs and then shade the region..so they are asking us to do it graphically.
But I still dont seem to understand what the questions means.. Are they asking us to find for which values of x in -|x|+2>=-√(4-x^2)
Well essentially you'd first find those x-values that satisfy the inequation. Whether you do this algebraically or graphically is up to you. Once you've found these values, the region would be found by drawing vertical lines at each of these x-values.
 

Green Yoda

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Well essentially you'd first find those x-values that satisfy the inequation. Whether you do this algebraically or graphically is up to you. Once you've found these values, the region would be found by drawing vertical lines at each of these x-values.
The value are 2 and -2 as thats where they intersect..Now what do I do now? and What do you mean by vertical lines?
 

em_paget

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would someone be able to explain the 2t value things to me?
its part of theoretical triginometry
thanks :)
 

InteGrand

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So -|x|+2>=-√4-x^2
basically means y>=-|x|+2 and y<=-√(4-x^2)
It doesn't mean that. I was just saying that would have made a better Q. If that was what was being meant, it would have been written as -√(4-x^2) ≤ y ≤ -|x|+2 .
 

Green Yoda

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It doesn't mean that. I was just saying that would have made a better Q. If that was what was being meant, it would have been written as -√(4-x^2) ≤ y ≤ -|x|+2 .
what does it mean then??
would then shading -|x|+2>=-√(4-x^2) be the same as y>=-|x|+2 and y<=-√(4-x^2)??
 

InteGrand

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what does it mean then??
would then shading -|x|+2>=-√(4-x^2) be the same as y>=-|x|+2 and y<=-√(4-x^2)??
No, because the former has no mention of y. It is simply an inequation in x, which has a solution: -2 ≤ x ≤ 2. If we wanted to sketch this on the number line, we'd sketch -2 ≤ x ≤ 2 on the number line, i.e. we'd shade in the part of the number line between -2 and 2 and include the endpoints. But if we wanted to sketch this on the x-y plane, we'd need to shade in all possible places in the x-y plane where -2 ≤ x ≤ 2. This becomes the vertical band between the vertical lines x = -2 and x = 2 (including these boundary lines).

It's like how an equation like x = 3 in the x-y plane is represented by sketching all places where x = 3. This becomes a vertical line because we just set x = 3 and can take any y-value (which means we can go up and down by any amount, but not across, since x needs to be 3). This is why something like x = 3 represents a vertical line in the x-y plane.
 

Green Yoda

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No, because the former has no mention of y. It is simply an inequation in x, which has a solution: -2 ≤ x ≤ 2. If we wanted to sketch this on the number line, we'd sketch -2 ≤ x ≤ 2 on the number line, i.e. we'd shade in the part of the number line between -2 and 2 and include the endpoints. But if we wanted to sketch this on the x-y plane, we'd need to shade in all possible places in the x-y plane where -2 ≤ x ≤ 2. This becomes the vertical band between the vertical lines x = -2 and x = 2 (including these boundary lines).

It's like how an equation like x = 3 in the x-y plane is represented by sketching all places where x = 3. This becomes a vertical line because we just set x = 3 and can take any y-value (which means we can go up and down by any amount, but not across, since x needs to be 3). This is why something like x = 3 represents a vertical line in the x-y plane.
Another silly question; But I now get that it's between +-2 but why do we only share between the two curves
http://www.wolframalpha.com/input/?i=-|x|+2>=-sqrt(4-x^2)

why not outside them as we are not bound by the y values??
 

bobmyself

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Given that a<0, find the solution to the inequality 1<abs(ax+1)<2, leaving your answer in terms of a.
 
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