VCE Methods Question Thread!

[vcestressed]

thanks, good luck with your results tomorrow !! 

[Jakeybaby]

thanks, good luck with your results tomorrow !! 

Cheers, appreciate it

[Izzy999]

Hey guys, could any of you help me with this question that I'm having trouble with?

Find the inverse function of each of the following:

f:R∖{3}→R, f(x)=1/(x−3)+1

I've found that the inverse is f^-1(x)=1/(x-1)+3, but the answers for the question say that f^-1(x)=3/(x-1)+2, but I'm not sure how that is.

Thanks!

[RuiAce]

Hey guys, could any of you help me with this question that I'm having trouble with?

Find the inverse function of each of the following:

f:R∖{3}→Rf:R∖{3}→R , f(x)=1x−3+1

I've found that the inverse is f^-1(x)=(1/x-1)+3, but the answers for the question say that f^-1(x)=(3/x-1)+2, but I'm not sure how that is.

Thanks!

Look at your question. You accidentally typed the domain and codomain twice however you're definitely missing a fraction sign there. And your brackets look misplaced for the answers.

If you want to communicate \(\frac{1}{x-1} +3\) you write it as 1/(x-1)+3

[Izzy999]

Look at your question. You accidentally typed the domain and codomain twice however you're definitely missing a fraction sign there. And your brackets look misplaced for the answers.

If you want to communicate \(\frac{1}{x-1} +3\) you write it as 1/(x-1)+3

Oh whoops, sorry about that. I didn't double check correctly when I typed up my question, obviously. Thanks

[RuiAce]

Image is too large on desktop

Wolfram agrees, so unless there is a typo they are wrong.

Edit: A typo is likely, because if I re-invert their answer I get 1/(x-2)+1

Image is too large on desktop

[Izzy999]

Ah okay, thank you!

[deStudent]

Hi guys, got 2 questions I need help on http://m.imgur.com/a/6JhU7

For the first image, is anyone else not getting what the answer has when using a Ti-Nspire? I'm getting x = -(k-mx-4) / 2(m-2) and y = k-mx / 2. I get what the answer has with my classpad but I plan on using the Ti in the exam.

For Q5, I don't quite understand the alternative method shown using matrices. More specifically lines 2 and 3. In line 2 how'd they get that matrix and why is it equal to 0. For line 3, how'd they get 15 - 2m - m^2?

[Syndicate]

Hi guys, got 2 questions I need help on http://m.imgur.com/a/6JhU7

For the first image, is anyone else not getting what the answer has when using a Ti-Nspire? I'm getting x = -(k-mx-4) / 2(m-2) and y = k-mx / 2. I get what the answer has with my classpad but I plan on using the Ti in the exam.

It works fine for me.Try inserting multiplication signs in front of every coefficient of x and y. 

Personally I hate the book's method and would rather equate the gradient of both equations in order to calculate when they are both equal.

My approach:

So now we know that there are either no solutions or infinitely many solutions when m = 4.

a) \(m \space \epsilon \space \mathbb{R} \text { \ {4}} \) and \( k \space \epsilon \space \mathbb{R} \)

b)  \( c_1 = c_2 \\ \frac{k}{2} = 2 \\ \therefore k  = 4 \) However the question says to work out the value(s) of m and k, when there would be no solutions. Which means \( m_1 = m_2 \space \text{and} \space c_1 \neq c_2 \)

Therefore m = 4 and \( k \neq 4\)

c) The two equations must be a scalar multiple of one another. Therefore \( m_1 = m_2 \space \text{and} \space c_1 = c_2\)

m = 4 and k = 4

For Q5, I don't quite understand the alternative method shown using matrices. More specifically lines 2 and 3. In line 2 how'd they get that matrix and why is it equal to 0. For line 3, how'd they get 15 - 2m - m^2?

I would suggest looking at cramer's rule.

Line 2 : [Seems like they made a mistake. Row 2, column 2 must be 5] They have let it equal to zero, in order to work out the determinant of the matrix. A determinant of 0 corresponds to parallel lines (with either infinitely many solutions or no solutions), and a non-invertible matrix.


In line 3: they have worked out the determinant of the matrix.

[deStudent]

It works fine for me.Try inserting multiplication signs in front of every coefficient of x and y. 

Personally I hate the book's method and would rather equate the gradient of both equations in order to calculate when they are both equal.

My approach:

So now we know that there are either no solutions or infinitely many solutions when m = 4.

a) \(m \space \epsilon \space \mathbb{R} \text { \ {4}} \) and \( k \space \epsilon \space \mathbb{R} \)

b)  \( c_1 = c_2 \\ \frac{k}{2} = 2 \\ \therefore k  = 4 \) However the question says to work out the value(s) of m and k, when there would be no solutions. Which means \( m_1 = m_2 \space \text{and} \space c_1 \neq c_2 \)

Therefore m = 4 and \( k \neq 4\)

c) The two equations must be a scalar multiple of one another. Therefore \( m_1 = m_2 \space \text{and} \space c_1 = c_2\)

m = 4 and k = 4

I would suggest looking at cramer's rule.

Line 2 : [Seems like they made a mistake. Row 2, column 2 must be 5] They have let it equal to zero, in order to work out the determinant of the matrix. A determinant of 0 corresponds to parallel lines (with either infinitely many solutions or no solutions), and a non-invertible matrix.


In line 3: they have worked out the determinant of the matrix.
(Image removed from quote.)

Thanks for the help again!

[deStudent]

Soz, one more question 

http://m.imgur.com/a/0nJ0i

For Q7) I seem to be missing a solution (m = 0). The method which I used was equating the gradients and solving for m, and then subbing them back in. I got m=4 (infinite sols) and m=-2 (no sols). I didn't get m = 0 which was the solution for part (b).

Why does my method not give all 3 solutions? I also tried subbing equation 1 in to eqn 2, but I still only got 2 solutions.

My answer for part b was: x = 4/m+2, y = 2(m+4)/m+2, where m = R except {-2,4}. Is this wrong? The book's answer didn't write the answer in terms of m, they just subbed in 0 and solved.

Q8) So I solved this and got y(-2k-3) = 0 and thus y = 0, but the answer included next to this 'k =/= -3/2'. I didn't realise this until part (b) where this was the answer. How do I recognise that I need to write that part about k, in part (a)?

Thanks

[Guideme]

Consider the Functions f:R--->R, f(x)=x^2-4 and g:R+U{0} --->R,g(x)=square root x      (Also: x ^(1/2))
 Explain why g(f(x)) does not exist

[Syndicate]

Consider the Functions f:R--->R, f(x)=x^2-4 and g:R+U{0} --->R,g(x)=square root x      (Also: x ^(1/2))
 Explain why g(f(x)) does not exist

ran f must be a subset of dom g

domain of g(x) is \( \mathbb{R}^+ U \text{ {0} } \) and the range of f(x) is [-4, infinity). This means range of f(x) is not a subset of the domain of g(x), since it is larger.

Since range of f(x) is not a subset of the domain of g(x), g(f(x)) cannot exist.

Soz, one more question 

http://m.imgur.com/a/0nJ0i

For Q7) I seem to be missing a solution (m = 0). The method which I used was equating the gradients and solving for m, and then subbing them back in. I got m=4 (infinite sols) and m=-2 (no sols). I didn't get m = 0 which was the solution for part (b).

Why does my method not give all 3 solutions? I also tried subbing equation 1 in to eqn 2, but I still only got 2 solutions.

My answer for part b was: x = 4/m+2, y = 2(m+4)/m+2, where m = R except {-2,4}. Is this wrong? The book's answer didn't write the answer in terms of m, they just subbed in 0 and solved.

a) The lines are only parallel when m = -2, 4. When m = 0, it returns a unique solution, which means 0 must be discarded. Regardless of what method the book has used, the only two correct solutions for m will be -2 and 4.

b) Once again, there are multiple methods to solve this question. Your method should be completely fine (unless stated to use a specific method), as long as you get the correct answers.

Q8) So I solved this and got y(-2k-3) = 0 and thus y = 0, but the answer included next to this 'k =/= -3/2'. I didn't realise this until part (b) where this was the answer. How do I recognise that I need to write that part about k, in part (a)?

I guess it doesn't really matter since the question hasn't stated for you to write the values for k. It has only asked you to work out the values for y and x. So, if you don't write it on the exam, you shouldn't exactly lose marks (please do confirm this one).

[samuelbeattie76]

I have attached a screenshot I need help with questions b and c. Any help is greatly appreciated. Thank you.

[nyggfany]