IGCSE maths paper 4 variant 2. huge discusssion

[The_Boss]

the answer for the functions was g(x) only because the f(x) cannot be equall to the inverse when the values of "X"s are same. so the answer is SOLELY g(X)

 

[The_Boss]

21 AND 42 HOW ?????????
THEY WERE 38 AND 40

No he is talking about the qusetions previous to the ones u r talking about.

 

[qwertypoiu]

Actually, the way I remember it:
f(x) = 1 - x
g(x) = 1/x

If you find the inverse of both of these you'll see:
f^-1(x) = 1 - x
g^-1(x) = 1/x

You can even try inputting random numbers in the function, taking the output, then inputting that output into the function to see whether it comes back to the input:

f(12) = 1 - (12) = -11
f(-11) = 1 - (-11) = 12

g(5) = 1/(5) = 0.2
g(0.2) = 1/(0.2) = 5

Therefore, it's both f(x) and g(x)

 

[Arshad1]

Isnt 1D suppose to be $90 as it said round to the nearest cent.
3rd question last part is something like 2177.5(OR 2117.5) as for cost the answer should be to 2 d.p.
9th question(probability) last part the answer is 781/1024
10b the answer is x-1=g^-1(X) 10th question last one was g(x)
Last page the answers were : 1/3 , 1/5, 1/4, 1/6 , 0.363

Hello bro.. I think the answer you have mentioned for the last page answers for the 4th diagram is incorrect. The shape given was a hexagon and a triangle was drawn inside it. So you can only draw a maximum of 3 more triangle inside the shape of the hexagon. So the total no of triangles that can be drawn inside the hexagon will only be 4. As only one triangle was shaded, the shaded area would be 1/4 and not 1/6. Hope i am clear in my explanation.

 

[qwertypoiu]

Hello bro.. I think the answer you have mentioned for the last page answers for the 4th diagram is incorrect. The shape given was a hexagon and a triangle was drawn inside it. So you can only draw a maximum of 3 more triangle inside the shape of the hexagon. So the total no of triangles that can be drawn inside the hexagon will only be 4. As only one triangle was shaded, the shaded area would be 1/4 and not 1/6. Hope i am clear in my explanation.

It's clearly 1/6

 

[qwertypoiu]

Isnt 1D suppose to be $90 as it said round to the nearest cent.

Actually, nearest cent means nearest $0.01.
So the correct answer would be $89.99, as $90 is to the nearest DOLLAR.

 

[Arshad1]

View attachment 42658

It's clearly 1/6

Anyway, why can't it be divided like this?

 

[qwertypoiu]

Anyway, why can't it be divided like this?

View attachment 42671

Because those triangles are NOT the same size

 

[The_Boss]

Actually, nearest cent means nearest $0.01.
So the correct answer would be $89.99, as $90 is to the nearest DOLLAR.

Oh my bad. Yeah its suppose to be 89.99

 

[The_Boss]

Actually, the way I remember it:
f(x) = 1 - x
g(x) = 1/x

If you find the inverse of both of these you'll see:
f^-1(x) = 1 - x
g^-1(x) = 1/x

You can even try inputting random numbers in the function, taking the output, then inputting that output into the function to see whether it comes back to the input:

f(12) = 1 - (12) = -11
f(-11) = 1 - (-11) = 12

g(5) = 1/(5) = 0.2
g(0.2) = 1/(0.2) = 5

Therefore, it's both f(x) and g(x)

Mate you mixed up the functions, f(x)=1/x and g(x)=1-x
And the condition for the question was that the VALUE OF X SHOULD BE SAME IN THE INVERSE AND ORIGINAL funtion, so its clearly g(x) not both

Prove: g(0)=g^-1(0)
1-0=1-0
1=1

 

[qwertypoiu]

Mate you mixed up the functions, f(x)=1/x and g(x)=1-x
And the condition for the question was that the VALUE OF X SHOULD BE SAME IN THE INVERSE AND ORIGINAL funtion, so its clearly g(x) not both

Prove: g(0)=g^-1(0)
1-0=1-0
1=1

Okay so;
f(x) = 1/x

and just prove that:
f(2) = f^-1(2)

To prove the above, we must also know what f^-1(x) is. Just try working out what f^-1(x) comes to. You'll see f^-1(x) = 1/x. So:

f(2) = f^-1(2)
1/2 = 1/2
0.5 = 0.5

You already agree with g(x), it's f(x) as well.

 

[The_Boss]

Okay so;
f(x) = 1/x

and just prove that:
f(2) = f^-1(2)

To prove the above, we must also know what f^-1(x) is. Just try working out what f^-1(x) comes to. You'll see f^-1(x) = 1/x. So:

f(2) = f^-1(2)
1/2 = 1/2
0.5 = 0.5

You already agree with g(x), it's f(x) as well.

Well ****! Gonna loose marks

 

[Arshad1]

Well ****! Gonna loose marks

But I don't think they asked for two functions bro! And the question was also only for one mark!
So are two answers really required?

 

[The_Boss]

But I don't think they asked for two functions bro! And the question was also only for one mark!
So are two answers really required?

Well i also had a question that in f(x) if the value is zero it doesnt work so isnt the answer suppose to be g(X)?

 

[qwertypoiu]

But I don't think they asked for two functions bro! And the question was also only for one mark!
So are two answers really required?

Actually, it said WHICH of the functions.... so it's not necessarily only one answer.... as for it being only 1 mark, I think it was to make it less obvious that there were two answers...

Well i also had a question that in f(x) if the value is zero it doesnt work so isnt the answer suppose to be g(X)?

The question actually defined f(x) to be:
f(x) = 1/x , x ≠ 0

That actually means f(x) is not defined at x=0, so one should not consider the point when x=0.

But even if you were to take x =0, we would prove:

f(0) = f^-1(0)
undefined = undefined.

So that's sort of proven as well

Though really, x=0 should not be considered.

 

[Arshad1]

Well i also had a question that in f(x) if the value is zero it doesnt work so isnt the answer suppose to be g(X)?

Just relax bro! Hope the examiner accepts one answer..
Moreover, it will be better if you remember the question fully and type it here.
Did the question say "for any value of x"? If it said so, then f(x) would not be the answer.

 

[qwertypoiu]

Just relax bro! Hope the examiner accepts one answer..
Moreover, it will be better if you remember the question fully and type it here.
Did the question say "for any value of x"? If it said so, then f(x) would not be the answer.

The question did not say "for any value of x"....

And as I've already said previously, even if x=0 were to be considered, both f(x) and f^-1(x) give the value undefined... which is equal.

 

[Arshad1]

The question did not say "for any value of x"....

And as I've already said previously, even if x=0 were to be considered, both f(x) and f^-1(x) give the value undefined... which is equal.

Hmm.. Okay!

 

[The_Boss]

Actually, it said WHICH of the functions.... so it's not necessarily only one answer.... as for it being only 1 mark, I think it was to make it less obvious that there were two answers...



The question actually defined f(x) to be:
f(x) = 1/x , x ≠ 0

That actually means f(x) is not defined at x=0, so one should not consider the point when x=0.

But even if you were to take x =0, we would prove:

f(0) = f^-1(0)
undefined = undefined.

So that's sort of proven as well

Though really, x=0 should not be considered.

Hmm. thanx, u sound like a good mathematician.

 

[Arshad1]

Hmm. thanx, u sound like a good mathematician.

Ya..I think qwertypoiu is gonna get a world rank in maths.