VCE Methods Question Thread!

[lzxnl]

could anyone explain the difference between combinations and permutations? i still don't know when to use which. i know that for permutations order matters, but how do you know when it matters or not?

Suppose I want to buy 3 books from 10, and the question is how many groups of 3 books can I choose. The assumption is that I don't care in what order the books are chosen; I only care about the names of the books. Here, you would treat these as combinations.

Another example is in choosing players for a football team. If I have to choose 22 people from 35 to make a football team, I really couldn't care less the order in which I picked them, so I treat these as combinations too.

Imagine now I have 10 letters and I am asked how many 5 letter words I can make from them. Obviously, letter order is important in a word, so this would be a permutation question.

Similarly, imagine I know that a 4 digit code has all different digits, and the question is how many different codes were possible. Digit is also important in a word and each digit is only used once, so this would also be a permutation question.

[mxcanator]

Hey guys, can anyone provide with worked solutions to this question:

If the function f has the rule f(x) = √x^2−16 and the function g has rule g(x) = x+5, find the largest domain for g such that f ◦ g is defined.

The answer to the question is: x ≥ -1 or x ≤ -9

It would be much appreciated if anyone could help answer! Thanks!

[xdmemeguy]

Hey guys, can anyone provide with worked solutions to this question:

If the function f has the rule f(x) = √x^2−16 and the function g has rule g(x) = x+5, find the largest domain for g such that f ◦ g is defined.

The answer to the question is: x ≥ -1 or x ≤ -9

It would be much appreciated if anyone could help answer! Thanks!

Can you rewrite the equations?

[MB_]

Hey guys, can anyone provide with worked solutions to this question:

If the function f has the rule f(x) = √x^2−16 and the function g has rule g(x) = x+5, find the largest domain for g such that f ◦ g is defined.

The answer to the question is: x ≥ -1 or x ≤ -9

It would be much appreciated if anyone could help answer! Thanks!

Can you post any working you have so far or show us which specific part of the question you're having trouble with? It's much better for your understanding if we assist you with specific issues rather than giving you the full solution.

Hint: you should know that the sqrt function has a domain [0,∞) so you have to consider when the expression under the sqrt sign is greater than or equal to 0

[Pudding]

1. In how many ways can the letters of the word 'TROUBLE' be arranged in groups of three.

I keep getting 140, but the answer is 210

[MB_]

1. In how many ways can the letters of the word 'TROUBLE' be arranged in groups of three.

I keep getting 140, but the answer is 210

Using the formula for permutations you should get 7!/4! = 210

[matthewzz]

For f(g(x)) to be defined, the range of g must be in [2,inf). Now the range of g, quite obviously, is [a,inf] so for this to fit inside the domain of f, we have a >= 2 (if a = 3, for instance, the range of g is then a subset of the domain of f, which is fine)

For g(f(x)) to be defined, the range of f must be in (-inf, 1], and the range of f is (-inf, a-2]. For (-inf, a-2] to be in (-inf, 1], we require a-2 <= 1, or a <= 3. Combining the two gives you 2<=a<=3.

You need to remember that f(g(x)) is defined even when the domain of f isn't equal to the range of g; the domain of f just has to include the range of g.

I know I'm a bit late but thank you for this! It was a big help!! 

[Jimmmy]

Q) For which values of m does the equation mx^2 - 2mx + 3 = 0 have one solution for x?

My Working: Using the discriminant (b^2 - 4*a*c), 4m^2 - 12m = 0 to get one solution for x. This can be factorised to 4m(m-3)=0, which is how I came to the answer that m = 0, or 3 for there to be one solution for x. However, the solutions say;

"(  m=0 , is not a solution as it gives  3=0 )".

As far as I can work out, subbing m=0 into the equation gives 4*0(0-3)=0, which can be then written as 0*-3=0, which shows 0=0 and thus m=0 is a satisfactory answer.

Am I missing something here, or is this just a book error?

[MB_]

Q) For which values of m does the equation mx^2 - 2mx + 3 = 0 have one solution for x?

My Working: Using the discriminant (b^2 - 4*a*c), 4m^2 - 12m = 0 to get one solution for x. This can be factorised to 4m(m-3)=0, which is how I came to the answer that m = 0, or 3 for there to be one solution for x. However, the solutions say;

"(  m=0 , is not a solution as it gives  3=0 )".

As far as I can work out, subbing m=0 into the equation gives 4*0(0-3)=0, which can be then written as 0*-3=0, which shows 0=0 and thus m=0 is a satisfactory answer.

Am I missing something here, or is this just a book error?

Subbing 0 into the original equation gives 3=0, you just subbed it back into the discriminant

[Jimmmy]

Subbing 0 into the original equation gives 3=0, you just subbed it back into the discriminant

Thanks MB, it's been a long day...

EDIT:

Subbing 0 into the original equation gives 3=0, you just subbed it back into the discriminant

It looks like the Cambridge Methods Textbook changed whoever was providing solutions to the chapter midway through 4A.  Not only did the solutions go from using Set Notation to Interval, guess what the answer to this question was; 

Q) For which values of k does the equation kx^2 - 2kx = 5 have for one solution of x?

Textbook Solution: Δ=0⇔k=0  or  k=−5  Doesn't k=0 provides the same issue as before, that 0=5?

That was Question 16)b) of 4A in the Cambridge Methods textbook, for those wondering, so be careful!

[Jimmmy]

I'm creating a consecutive post because another question is annoying me. 

Q) Show that the equation ax^2−(a+b)x+b=0 has a solution for all values of  a  and b.

I *tried* to use the discriminant, and ended up with -(a+b)^2 - 4ab, but the solutions work with b = (a+b), and not with the negative sign in front that exists in the question.

They can then easily progress to the perfect square (a-b)^2 ≥ 0 which as the discriminant shows there is always at least one solution. Meanwhile, I'm left with -(a^2 + 2ab + b^2) - 4ab.

Can anyone figure out where I've gone wrong?

[MB_]

I'm creating a consecutive post because another question is annoying me. 

Q) Show that the equation ax^2−(a+b)x+b=0 has a solution for all values of  a  and b.

I *tried* to use the discriminant, and ended up with -(a+b)^2 - 4ab, but the solutions work with b = (a+b), and not with the negative sign in front that exists in the question.

They can then easily progress to the perfect square (a-b)^2 ≥ 0 which as the discriminant shows there is always at least one solution. Meanwhile, I'm left with -(a^2 + 2ab + b^2) - 4ab.

Can anyone figure out where I've gone wrong?

You should be getting (a-b)^2 for the discriminant as you have (-(a+b))^2 - 4ab which once simplified becomes (a-b)^2

Also, instead of making multiple posts you can use the 'Modify' button to edit your posts

[Jimmmy]

You should be getting (a-b)^2 for the discriminant as you have (-(a+b))^2 - 4ab which once simplified becomes (a-b)^2

Also, instead of making multiple posts you can use the 'Modify' button to edit your posts

Cheers MB! Yeah I made a new post for this new question, but removed the unnecessary one and modified the previous post.

For some reason, I expanded before applying the negative into the brackets...like I said earlier, it's been a long day  but Methods homework is finished for the holidays two weeks earlier than I expected!   

[hello876]

 Write the rule for the transformed function.
f(x)=2^​x  is translated  1 unit in the negative x direction, dilated by a factor of 3 from the y axis, and translated by 5 units in the negative y direction

Sorry with this I'm sort of confused about when a dilation from the y axis would affect the x axis translation done before, and how is that different to when the x axis translation was done after the y axis dilation

[Yertle the Turtle]

Write the rule for the transformed function.
f(x)=2^​x  is translated  1 unit in the negative x direction, dilated by a factor of 3 from the y axis, and translated by 5 units in the negative y direction

Sorry with this I'm sort of confused about when a dilation from the y axis would affect the x axis translation done before, and how is that different to when the x axis translation was done after the y axis dilation

Try thinking this question out point by point. Say you translate, and then dilate, then the translation itself will be dilated from the axis, whereas if you dilate first, and then translate, the translation will stay constant. Let's say you have f(x)=x², and you translate 2 up and dilate by 2 from the x-axis. If you translate first, then dilate it comes to f(x)=2(x²+2)=2x²+4. However, if you dilate first it comes to f(x)=2x²+2, so I hope you can now see the importance of the order of your transformations. This is just a general rule, it may not always be important to do it right as it may not always affect the final function, but it is important to do it right as a principle.