VCE Methods Question Thread!

[zhen]

I finally understand now. Actually thank you so much. ❤
Also are u assuming 1/2 as the yvalue since if 1/2 is x on the second quadrant.
Or neither and half is just positive and first and second quadrant is positive as the top section of the graph is positive if your looking at it from the y-axis

I'm assuming that 1/2 is the y value, since we treat sine like the y value. But if it was cos(x)=1/2, we treat cos as the x value. On a graph the x values are positive in the first and fourth quadrant, so cos is positive in the first and fourth quadrant. Another way of viewing it is that (1,1) in the first quadrant and (1,-1) in the fourth quadrant have both positive x values. So cosine is positive in the first and fourth quadrant. Whereas for sine (1,1) in the first quadrant and (-1,1) in the second quadrant has both positive y values. So, sine is positive in the first and second quadrant.

[LPadlan]

Hey, can anyone give tips on how to solve application problems(worded problems)? Like what to look for and stuff like that? i can never solve it by myself

[MattBro]

Find the values of a for which the equation (a−3)x2+2ax+(a+2)=0 has no solutions for x.

get as far as woking out a=-6 using the discriminant but how know if it is a>-6 or a<-6

a<-6 is the answer by the way

I so bad at this stuff

Any hints?

Thanks

[zhen]

Hey, can anyone give tips on how to solve application problems(worded problems)? Like what to look for and stuff like that? i can never solve it by myself

For applications what I do is
1) Skim read the question.
2) Look for the relevant information, which are generally numbers or equations, since for application questions I feel like most of the time 50% of the question is useless.
3) Look for what the question is asking (is it asking for you to find the intercepts or turning points)
4) If an equation wasn't given make your own equation/equations relating all the information together
5) Solve the question by using the given equation or the equation you made.
6) Make sure I'm answering the question or reread for tricks or things I missed

These are the steps I'd take to answer application questions. But, if you want to improve all you need is more practice and exposure to these types of questions.

[zhen]

Find the values of a for which the equation (a−3)x2+2ax+(a+2)=0 has no solutions for x.

get as far as woking out a=-6 using the discriminant but how know if it is a>-6 or a<-6

a<-6 is the answer by the way

I so bad at this stuff

Any hints?

Thanks

This is how I did it. Hope it helps. 

[TheCommando]

What are u using to upload thoose documents?

[zhen]

What are u using to upload thoose documents?

I'm using my ipad and it uploads just fine.

[TheCommando]

I'm using my ipad and it uploads just fine.

How? Like how do u have that ( )^2

[Aaron]

Please keep the discussion relevant to Methods Questions/Answers. If you want to ask someone a question that is not related to methods, please PM them. I will split this off if the off topic discussion continues. Thanks.

You can PM here: https://atarnotes.com/forum/index.php?action=pm;sa=send

[TheCommando]

https://postimg.org/image/fb5465zi7/
How do i do 2b i though theere would only be 2 answers as trre is a solution for cos in the fourth quadrant where it is positive. Also how do they get pie/6 and -pie/6
If the range is [-pie,pie] i thought you did pie -pie/6 and -pie+pie/6. Can someone explain how this is wrong thanks

[zhen]

https://postimg.org/image/fb5465zi7/
How do i do 2b i though theere would only be 2 answers as trre is a solution for cos in the fourth quadrant where it is positive. Also how do they get pie/6 and -pie/6
If the range is [-pie,pie] i thought you did pie -pie/6 and -pie+pie/6. Can someone explain how this is wrong thanks

Cos is positive so it has to be in the first or fourth quadrant. Normally you would do x=π/6 (first quadrant) and x=11π/6 or 2π-π/6 (fourth quadrant). But the domain is [-π,π]. So, since x=2π-π/6=-π/6 (since 2π is going around the circle back to the original position it's the same as saying 0-π/6). Note that since -π+π/6 is in the third quadrant, where cos is negative. So, it can't be that.
So, the solution is x=-π/6 and x=π/6

[TheCommando]

Cos is positive so it has to be in the first or fourth quadrant. Normally you would do x=π/6 (first quadrant) and x=11π/6 or 2π-π/6 (fourth quadrant). But the domain is [-π,π]. So, since x=2π-π/6=-π/6 (since 2π is going around the circle back to the original position it's the same as saying 0-π/6). Note that since -π+π/6 is in the third quadrant, where cos is negative. So, it can't be that.
So, the solution is x=-π/6 and x=π/6

I understand how you got -pie/6 but not pie/6 i understand you got it from the first quadrant since cos is postive there
I thought the domain [-pie,pie] was a domain that existed from where pie usually is to 2pie

[TheCommando]

So i guess thats not true
That domain means there can only be 2 solutions and thats it
It doesnt say that the values have to be in this quadrant etc

[deStudent]

For this http://m.imgur.com/a/hlB9p
Part b) why aren't we allowed to take the plus and minus of the square root. Isn't x = sqrt(x^2 + 1), roughly if x gets large? Or does the +1 actually make a difference because it'll make the argument of log_e negative?

[zhen]

For this http://m.imgur.com/a/hlB9p
Part b) why aren't we allowed to take the plus and minus of the square root. Isn't x = sqrt(x^2 + 1), roughly if x gets large? Or does the +1 actually make a difference because it'll make the argument of log_e negative?

It's like you said the square root of x^2 is x. So the square root of x^2+1 will be greater than x. So x-sqrt(x^2+1) will be negative and the log of a negative number doesn't exist. So you discard the negative from the plus and minus sign.