VCE Methods Question Thread!

[Sine]

Ah ok thanks. I guess we're looking at two different documents though because my one has neither of the conditions you said above o.O.
-> https://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007mm2.pdf

I think that is the non-CAS version of the exam. In the past there was maths methods and maths methods (CAS) now it is all CAS based.

Your answers are correct for the questions you were answering. However, whenever they ask you to answer in terms of a, b, c etc you need to include those in your answer and put it in the correct form they are asking for.

[aspiringantelope]

I think that is the non-CAS version of the exam. In the past there was maths methods and maths methods (CAS) now it is all CAS based.

Your answers are correct for the questions you were answering. However, whenever they ask you to answer in terms of a, b, c etc you need to include those in your answer and put it in the correct form they are asking for.

Ah all right. Thanks for clearing it up

[pans]

x / (x-1) becomes 1+1 / (x+1). Then how do you find the simplified version for (2x-3) / (4x+6) and (7-4x) / (2x-3)

[Danzorr]

x / (x-1) becomes 1+1 / (x+1). Then how do you find the simplified version for (2x-3) / (4x+6) and (7-4x) / (2x-3)

This can be achieved by long dividing

[pans]

This can be achieved by long dividing

is there a faster way or no

[Danzorr]

is there a faster way or no

I don’t think so

[fun_jirachi]

x / (x-1) becomes 1+1 / (x+1). Then how do you find the simplified version for (2x-3) / (4x+6) and (7-4x) / (2x-3)

I think you mean \(\frac{x}{x-1} = 1 + \frac{1}{x-1}\). To do this, I assume you added one and subtracted one in the numerator, then simplified. You can apply similar tricks for the other two (multiplying and dividing by the same number). Not sure what the context of this question is, but long dividing is definitely not necessary (and is redundant if the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator).

For example:
\[\frac{2x-3}{4x+6} = \frac{1}{2} \times \frac{4x-6}{4x+6} = \frac{1}{2} \times \left(1 - \frac{12}{4x+6}\right).\]

Hope this makes sense - again, don't long divide if the degree of the polynomial in the numerator is less than or equal to the degree of the polynomial in the denominator. In fact, there are plenty of methods, and in no case unless told otherwise should you be jumping straight to long division (should be used for interest and as a last resort, it's often more trouble than it's worth)

[pans]

Thank you for the replies to my previous post!!!

Got another Q hehe
1. y=kx-5 intersects y=x^2 -6x at two points for {-2root5-6>k} U {k>2root5-6}
2. y=x-3 and y=x^2 +kx-1 do not intersect for {1-2root2<k<1+2root2}

I know i need to equate them, find the discriminant (=0, <0, >0), but the ending part (actually finding k) confuses me.

For Q1, the answer is k^2+12k+16>0, but how does this become {-2root5-6>k} U {k>2root5-6}??? shouldnt it be -2root5-6>k and 2root5-6>k if k^2+12k+16>0

For Q2, its (k-1)^2-8<0 for k, shouldnt that be 1-2root2<k and 1+2root2<k .

BRO THIS WHOLE THING IS CONFUSING ME, how does finding the discriminant help find k, and how does <, >, =0 even relate to the parametric k??

[fun_jirachi]

For \(k^2 + 12k + 16 > 0\), we have solutions at \(x = 2\sqrt{5} - 6, -2\sqrt{5} - 6\). Clearly we have a cup parabola (quadratic with positive leading polynomial), and hence we can deduce values greater than zero will lie outside the roots, and not in between them. Note that the range you've defined \(\{-2\sqrt{5}-6 > k\} \cap \{2\sqrt{5}-6 > k\} = \{2\sqrt{5} - 6 > k\}\); this is why the solution given is correct and you are not. Look at it again, and think about why this is the case. Try graphing \(k^2 + 12k + 16 > 0\) as well.

Similarly, we again have a cup parabola, but this time we want values less than zero on said parabola. These values will lie in between the roots.  Apply the same methodology.

Not sure what you're referring to with your last question; if you can expand on that a bit more I'll come back and help you with that too
 

[pans]

Not sure what you're referring to with your last question; if you can expand on that a bit more I'll come back and help you with that too

OMG I FINALLY GET IT AFTER READING YOU EXPLANATION! last q also clarified! THANK YOU SO MUCH

[pans]

Solve the inequality 0< (2x-1)/(x+1) <2
ans: 0<y<2, x>1/2
dont understand....

[fun_jirachi]

What don't you understand? What have you tried? Questions like these are often inconclusive for everyone reading them (and also for me personally trying to answer them).

If it's a solution you want, just ask for it - depending on how hard the question is, how you phrase your query and the amount of work you've already put in I will consider posting up a solution.

There are two clear methods for a question like this. The easiest way to solve this question is graphically - for what values of \(x\) does the curve \(y = \frac{2x-1}{x+1}\) lie between 0 and 2? You can of course solve it algebraically using your preferred method for solving inequalities. It might also help to split up the inequality into two separate inequalities to be solved separately.

To make sense of the solution, the range of \(y\) values is literally given in the question. The range of \(x\) values is what you are trying to find using one of the two methods I've described.

[1729]

Solve the inequality 0< (2x-1)/(x+1) <2
ans: 0<y<2, x>1/2
dont understand....

The question is essentially asking you to find the domain and range of \(f\left(x\right)=\frac{\left(2x-1\right)}{\left(x+1\right)}\) where \(0<f\left(x\right)<2\)

Remembering that the range is just all possible output values, it will simply be \(0<y<2\)

Now to find the domain solve the inequality \(0<\frac{\left(2x-1\right)}{\left(x+1\right)}<\:2\)

To solve the inequality above just think of it as two seperate inequalities and then find the intersection fo the two so it satisfies.

You have two inequalities, the first one being \(0<\frac{\left(2x-1\right)}{\left(x+1\right)}\) and the second one \(\frac{\left(2x-1\right)}{\left(x+1\right)}<\:2\)

The first inequality should result in \(x>-1\) and the second one should be \(x>1/2\), the intersection of this is simply \(x>1/2\)

Edit: Fun_jirachi literally by 30 seconds

[JIN1N]

Hey can someone explain how to do this question? its from tssm 2016 exam 2. Answer is A but idk how.

[fun_jirachi]

I think it might be a typo given it looks like A is suggesting you take the area of the rectangle and subtract the integral within the same bounds. Note that integrating with respect to x or y doesn't actually matter, since they are just using the change of variables x = y. Pretty sure it should be \(27 - \int_{-8}^1 \sqrt{-y+1} \ dy\).

Hope this helps