Additional Mathematics (0606) Doubt

[XxChampXx]

The function g is defined by g: x -> x^2 – 8x + 7 for the domain 3<=x<=k.
(iv) Determine the largest value of k for which g^–1 exists. [1]

Can someone please explain this and the functions part(especially one-one functions) of additional mathematics?

 

[qwertypoiu]

The function g is defined by g: x -> x^2 – 8x + 7 for the domain 3<=x<=k.
(iv) Determine the largest value of k for which g^–1 exists. [1]

Can someone please explain this and the functions part(especially one-one functions) of additional mathematics?

Complete the square:

x^2 – 8x + 7 = (x-4)^2 - 9

As you can see the intercept is (4,-9)

For a function to have an inverse it must be one-one.
If one side of the domain has been chosen to be x>3, then the maximum k for which the domain is such that the function is one-one is k=4.
So your domain will be 3<x<4
Any more than 4, like 3<x<5 will be not one-one as the curve goes upwards and thus for one output two different inputs may be traced.
Any less than 4 like 3<x<3.5 would be completely correct, to satisfy the condition of having an inverse, but this would not be maximum.

 

[XxChampXx]

View attachment 54015

Complete the square:

x^2 – 8x + 7 = (x-4)^2 - 9

As you can see the intercept is (4,-9)

For a function to have an inverse it must be one-one.
If one side of the domain has been chosen to be x>3, then the maximum k for which the domain is such that the function is one-one is k=4.
So your domain will be 3<x<4
Any more than 4, like 3<x<5 will be not one-one as the curve goes upwards and thus for one output two different inputs may be traced.
Any less than 4 like 3<x<3.5 would be completely correct, to satisfy the condition of having an inverse, but this would not be maximum.

Thank you so much

 

[XxChampXx]

View attachment 54015

Complete the square:

x^2 – 8x + 7 = (x-4)^2 - 9

As you can see the intercept is (4,-9)

For a function to have an inverse it must be one-one.
If one side of the domain has been chosen to be x>3, then the maximum k for which the domain is such that the function is one-one is k=4.
So your domain will be 3<x<4
Any more than 4, like 3<x<5 will be not one-one as the curve goes upwards and thus for one output two different inputs may be traced.
Any less than 4 like 3<x<3.5 would be completely correct, to satisfy the condition of having an inverse, but this would not be maximum.

Can you explain this also. Thank you in advance.


6 Given that each of the following functions is defined for the domain –2<=x<=3, find the range of

(i) f : x -> 2 – 3x, [1]

(ii) g : x -> |2 – 3x| , [2]

(iii) h : x -> 2 – |3x| . [2]

State which of the functions f, g and h has an inverse.

 

[qwertypoiu]

Can you explain this also. Thank you in advance.


6 Given that each of the following functions is defined for the domain –2<=x<=3, find the range of

(i) f : x -> 2 – 3x, [1]

(ii) g : x -> |2 – 3x| , [2]

(iii) h : x -> 2 – |3x| . [2]

State which of the functions f, g and h has an inverse.

To find the range evaluate the function at the two extremes of your domain, and also find its minimum/maximum point.

i) f(x)=2-3x

f(-2) = 2 - 3(-2) = 8
f(3) = 2 - 3(3) = -7

Therefore range is -7<=f(x)<=8

ii) g(x)=|2 - 3x|

g(-2) =|2 - 3(-2)|= 8
g(3) = |2 - 3(3)| = 7

Now consider where the minimim point must be. The minimum point of a modulus function is where the inside is zero, ie 2-3x=0, so x = 2/3

g(1.5) =|2 - 3(2/3)| = 0

So range is 0<=g(x)<=8


iii) h(x) = 2 - |3x|
h(-2) = 2 - |3(-2)| = -4
h(3) = 2 - |3(3)| = -7

Now consider the max point of this function. It would be when the inside of modulus is zero. ie. 3x=0 so x=0
h(0) = 2 - |3(0)| = 2

So range of function is -7<=h(x)<=2



For a function to have an inverse, it must be one-one. Only f(x) is one-one in this case, the rest have a turning point.

 

[XxChampXx]

To find the range evaluate the function at the two extremes of your domain, and also find its minimum/maximum point.

i) f(x)=2-3x

f(-2) = 2 - 3(-2) = 8
f(3) = 2 - 3(3) = -7

Therefore range is -7<=f(x)<=8

ii) g(x)=|2 - 3x|

g(-2) =|2 - 3(-2)|= 8
g(3) = |2 - 3(3)| = 7

Now consider where the minimim point must be. The minimum point of a modulus function is where the inside is zero, ie 2-3x=0, so x = 2/3

g(1.5) =|2 - 3(2/3)| = 0

So range is 0<=g(x)<=8


iii) h(x) = 2 - |3x|
h(-2) = 2 - |3(-2)| = -4
h(3) = 2 - |3(3)| = -7

Now consider the max point of this function. It would be when the inside of modulus is zero. ie. 3x=0 so x=0
h(0) = 2 - |3(0)| = 2

So range of function is -7<=h(x)<=2



For a function to have an inverse, it must be one-one. Only f(x) is one-one in this case, the rest have a turning point.

That was really helpful. Thank you so much.

 

[The_Boss]

That was really helpful. Thank you so much.

I might be late but here is a quick explanation of functions, domain and range.
Function is a relation between two variables, like u input one variable to get the other variable, then the function are the step done to PROCESS the first variable to the second variable. Think of them as a computer INPUT----->Process----> Output
A ONE-ONE function is when specific output refers to a SINGLE SOLITARY INPUT. Like in straight line graphs where every X value has a single Y value. So straight line graphs have inverse.But think about cubic and 4th power graphs they tend to cross up and down so its probable that a single x value might have different y values. so to make such graph a one-one function u limit the number of x values u can put and the number you can put(x values-Input) are called domain.
The corresponding Y values are the range. So to find range just put in the maximum and minimum values of input and find corresponding y values to find range.

 

[The_Boss]

And yeah for additional mathematics (IGCSE one) dont forget to use notations like F(x) instead of y for range and x for domain.
And try to include sine,cosine,tan,sec,cos,cot,ln,e graphs with domain and range. Like what is domain for ln and e and what is range for cos and sec

 

[XxChampXx]

I might be late but here is a quick explanation of functions, domain and range.
Function is a relation between two variables, like u input one variable to get the other variable, then the function are the step done to PROCESS the first variable to the second variable. Think of them as a computer INPUT----->Process----> Output
A ONE-ONE function is when specific output refers to a SINGLE SOLITARY INPUT. Like in straight line graphs where every X value has a single Y value. So straight line graphs have inverse.But think about cubic and 4th power graphs they tend to cross up and down so its probable that a single x value might have different y values. so to make such graph a one-one function u limit the number of x values u can put and the number you can put(x values-Input) are called domain.
The corresponding Y values are the range. So to find range just put in the maximum and minimum values of input and find corresponding y values to find range.

And yeah for additional mathematics (IGCSE one) dont forget to use notations like F(x) instead of y for range and x for domain.
And try to include sine,cosine,tan,sec,cos,cot,ln,e graphs with domain and range. Like what is domain for ln and e and what is range for cos and sec

Thank you so much.

 

[XxChampXx]

Can someone help me solve Oct Nov 2011 QP12 q8 part (ii)? I really don't understand the dividing of the line part where you have to use the section formula. Thank you in advance for your help.

Link:- https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230cS1pd0xaVWFGN0k

 

[The_Boss]

Can someone help me solve Oct Nov 2011 QP12 q8 part (ii)? I really don't understand the dividing of the line part where you have to use the section formula. Thank you in advance for your help.

Link:- https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230cS1pd0xaVWFGN0k

Alright these type of question are pretty hard unless u know a little bit of circle equation and their concept in graphs.
SO first part says find the equation its pretty easy, you can do that. Now (ii) part says to find coordinates of D. For this write the equation of the line in terms of y=MX+C.
Then second part is to find co-ordinates of C and M once found use length formula to equalize the known values, that is 2(CM) =square root of(Dx-Mx)2
And the just solve the quadratic,

 

[The_Boss]

Can someone help me solve Oct Nov 2011 QP12 q8 part (ii)? I really don't understand the dividing of the line part where you have to use the section formula. Thank you in advance for your help.

Link:- https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230cS1pd0xaVWFGN0k

 

[XxChampXx]

View attachment 54306 View attachment 54305

These questions are really complicated. Thank you so much.

 

[The_Boss]

So done with add maths mate?

 

[XxChampXx]

So done with add maths mate?

Still doing past papers. You?

 

[XxChampXx]

View attachment 54306 View attachment 54305

Can you help in in May June 2012 QP12 question 11 part (i). Thank you in advance.

Link:- https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230cS1pd0xaVWFGN0k

 

[The_Boss]

We'll I have my mechanics as lvl exam soon,so I am preparing for it right now.
Anyway did u take add maths privately or ur school/institute provides courses for it?

 

[The_Boss]

Can you help in in May June 2012 QP12 question 11 part (i). Thank you in advance.

Link:- https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230cS1pd0xaVWFGN0k

Why are u ONLY doing graphs question? Just give me an hour or so I will help ya out once I reach home.

 

[The_Boss]

Do u understand relative velocity?? It is kind of confusing! and the diagram questions with them.

Well I only found relative velocity difficult when it concerns two things moving relative to each other like in collision and interception problems but in water,air motion its pretty easy. If you want I can help you out but after 3rd as my Mechanics exam will be over on 3rd.

 

[XxChampXx]

We'll I have my mechanics as lvl exam soon,so I am preparing for it right now.
Anyway did u take add maths privately or ur school/institute provides courses for it?

Yeah my school provides it.