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# Inverse trig function OCR C3 Help!!? Watch

1. Ive already done i and ii, it's just iii which I don't quite understand. The mark scheme doesn't help either.

Functions f and g are defined by f(x) = 2sin(x) for ?1/2Pi ? x ? 1/2Pi, g(x) = 4-x^2 for xER.
(i) State the range of f and the range of g.
(ii) Show that gf(0.5) = 2.16, correct to 3 significant figures, and explain why fg(0.5) is not defined.
(iii) Find the set of values of x for which f-1g(x) is not defined.

For iii) I'm aware that the inverse of f can only take x value between 2 and -2 and that the range is 1/2Pi and -1/2Pi

This means that g(x)>2 and g(x)<-2 for which f-1 is not defined.

trying to work it out I get.

4-x^2>2 => x= ±(sqrt)3
4-x^2<-2 => ±1

I get as far as this... the mark scheme however wants us to combine the value to get a range of x for which f-1 is undefined;

It says: Attempt solution involving four x values M1 to produce at least two sets of values Obtain x <-(sqrt)3, -1<x<1, x>(sqrt)3

this is the part I don't understand.

Could anyone just explain this part please.
Thanks.
2. (Original post by Jin3011)
Ive already done i and ii, it's just iii which I don't quite understand. The mark scheme doesn't help either.

Functions f and g are defined by f(x) = 2sin(x) for ?1/2Pi ? x ? 1/2Pi, g(x) = 4-x^2 for xER.
(i) State the range of f and the range of g.
(ii) Show that gf(0.5) = 2.16, correct to 3 significant figures, and explain why fg(0.5) is not defined.
(iii) Find the set of values of x for which f-1g(x) is not defined.

For iii) I'm aware that the inverse of f can only take x value between 2 and -2 and that the range is 1/2Pi and -1/2Pi

This means that g(x)>2 and g(x)<-2 for which f-1 is not defined.

trying to work it out I get.

4-x^2>2 => x= ±(sqrt)3
4-x^2<-2 => ±1

I get as far as this... the mark scheme however wants us to combine the value to get a range of x for which f-1 is undefined;

It says: Attempt solution involving four x values M1 to produce at least two sets of values Obtain x <-(sqrt)3, -1<x<1, x>(sqrt)3

this is the part I don't understand.

Could anyone just explain this part please.
Thanks.
what year is the paper from
3. (Original post by Jin3011)
Ive already done i and ii, it's just iii which I don't quite understand. The mark scheme doesn't help either.

Functions f and g are defined by f(x) = 2sin(x) for ?1/2Pi ? x ? 1/2Pi, g(x) = 4-x^2 for xER.
(i) State the range of f and the range of g.
(ii) Show that gf(0.5) = 2.16, correct to 3 significant figures, and explain why fg(0.5) is not defined.
(iii) Find the set of values of x for which f-1g(x) is not defined.

For iii) I'm aware that the inverse of f can only take x value between 2 and -2 and that the range is 1/2Pi and -1/2Pi

This means that g(x)>2 and g(x)<-2 for which f-1 is not defined.

trying to work it out I get.

4-x^2>2 => x= ±(sqrt)3
4-x^2<-2 => ±1
Solving these inequalities you can not get equalities.
4-x^2>2 => 2>x^2 =>taking root |x|<(sqrt)2 which means -(sqrt)2<x<(sqrt)2

4-x^2<-2 => x^2>6 => |x|>(sqrt)6 meaning x>(sqrt)6 or x<-(sqrt)6
4. Guys I asked this ages ago. It's weird that I was able to solve this when returning to college the next day, though I can't even understand what's going on since finishing A-levels =].
5. lol what is it with people and bumping year plus old threads.

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