Q. 11 D5.0( 1 Vote )

# Let f be a real f

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, ∞)

Now,

(fof)(x) = f(f(x)) = f = fof: [6, ∞) R defined as

(fof)(x) = f2(x) = [f(x)]2 = = x – 2

f2: [2, ∞) R defined as

f2(x) = x – 2

fof ≠ f2

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