IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: cs on September 06, 2010, 08:20:36 am
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Cie 9709 paper 01, November 2007
Q11 part (v), why in the last part, its - and not +-..? here's the question..
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to avoid this confusion-you can just invert the completed square in part 11i) to get -->
1. 2(x-2)^2 +3 (completed square form)
2. x, -2 , ^2 , multiply by 2 , +3 (what the x value ungergoes in completed square form)
3.Inverse= x-3 divided by 2, then square root and finally add 2.
4. you will see that the domain has to be x>(or equal) 3
5.and from this you can deduce that the Range(y values) have to be >(or equal to) 2
to be honest im not really answering your question but this is just an alternative.. ;)
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2(x-2)^2+3
Let g(x)=y y=2(x-2)^2+3
We make x the subject
2(x-2)^2=y-3
(x-2)^2=(y-3)/2
(x-2)=+-SQR((y-3)/2)
For the given domain, g(x) is a decreasing function therefore we keep the negative
sign.+- Would make it a many to one function
so the inverse would be after substituting x with y
y=2-SQR((x-3)/2)
Now range of g inverse is the domain of g(x)
Domain of g(x) is X<=2
range of g inverse is g^-1(x)<=2
Note:For a One-one function g(x): Domain of g(x)=range of g inverse
and range of g(x)=domain of g inverse
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This is because only only one to one function have inverse
many to one function do not have inverse.
The function g(x) is one to one function and g^-1(x) is the reflection of g(x) on line y=x,if you give +- together then g^-1(x) become many to one function which is not possible.Since g^-1(x) is the reflection of g(x) on line y=x it must also be one to one function
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Thanks studyorplay, Requiem(+rep) and asiftasfiq93.. !!