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MathematicsFunctionsCompositeHard2 minPYQ_2020
MathematicsHardsingle choice

For a suitably chosen real constanta, let a function,f : R--aRbe defined byfx=a-xa+x. Further supposed that for any real numberx-a,andf(x)-a, fofx=x. Thenf-12is equal to :

Options:

Answer:
D
Solution:

fof(x)=a-fxa+f(x)=x

a-ax1+x=f(x)

a(1-x)1+x=a-xa+x        (a=1)

so f(x)=1-x1+x

f-12=3

 

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Composite
2mℹ️ Source: PYQ_2020

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