Q. 11 C4.5( 2 Votes )

Let f be a real f

Answer :

We have, f(x) =


Clearly, domain of f = [2, ∞] and range of f = [0, ∞)


We observe that range of f is not a subset of domain of f


Domain of (fof) = {x: x ϵ Domain of f and f(x) ϵ Domain of f}


= {x: x ϵ [2, ∞) and ϵ [2, ∞)}


= {x: x ϵ [2, ∞) and ≥ 2}


= {x: x ϵ [2, ∞) and x – 2 ≥ 4}


= {x: x ϵ [2, ∞) and x ≥ 6}


= [6, ∞)


Clearly, range of f = [0, ∞) Domain of (fof)


Domain of ((fof)of) = {x: x ϵ Domain of f and f(x) ϵ Domain of (fof)}


= {x: x ϵ [2, ∞) and ϵ [6, ∞)}


= {x: x ϵ [2, ∞) and ≥ 6}


= {x: x ϵ [2, ∞) and x – 2 ≥ 36}


= {x: x ϵ [2, ∞) and x ≥ 38}


= [38, ∞)


Now,


(fof)(x) = f(f(x)) = f =


(fofof)(x) = (fof)(f(x)) = (fof) =


fofof : [38, ∞) R defined as


(fof)(x) = f(f(x)) = f =


(fofof)(x) = (fof)(f(x)) = (fof) =


fofof : [38, ∞) R defined as


(fofof)(x) =


(fofof)(38) =


=


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