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प्रश्न
Let f be a real function given by f (x)=`sqrt (x-2)`
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f2
Also, show that fof ≠ `f^2` .
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उत्तर
f (x) = `sqrt(x-2)`
For domain,
x − 2 ≥ 0
⇒ x ≥ 2
Domain of f = [ 2,∞ )
Since f is a square-root function, range of f =( 0,∞)
So, f : [2,∞) → ( 0,∞ )
(i) fof
Range of f is not a subset of the domain of f.
⇒Domain(fof)= { x : x ∈ domain of fand f (x) ∈ domain of f}
⇒ Domain (fof) = `{x :x in [2, ∞ ) and sqrt (x-2) in [ 2 ∞ )}`
⇒ Domain (fof) = `{x :x in [2, ∞ ) and sqrt (x-2)≥ 2 }`
⇒ Domain(fof) = { x : x ∈ [2,∞) and x−2 ≥4 }
⇒ Domain(fof) = { x : x ∈ [2,∞) and x ≥ 6}⇒ Domain(fof) = { x : x ≥ 6}
⇒ Domain(fof) = [ 6, ∞ )
fof : [6, ∞) → R
(fof) (x) = f (f (x))
= ` f (sqrt(x -2))`
= `sqrt (sqrt(x - 2) - 2)`
(ii) fofof= (fof) of
We have, f : [ 2,∞ ) → ( 0,∞ ) and fof : [ 6, ∞ ) → R
⇒ Range of f is not a subset of the domain of fof.
Then, domain((fof)of)={ x : x ∈domain of fand f (x) ∈ domain of fof }
⇒ Domain((fof)of) = `{ x : x in [ 2,∞) and sqrt (x-2) in [ 6 ,∞)}`
⇒ Domain ((fof)of) = ` x:x in [ 2 ∞ ) and sqrt(x-2) ≥ 6 }`
⇒ Domain ((fof)of) = { x : x ∈ [2,∞) and x − 2 ≥ 36}
⇒ Domain ((fof)of) = { x : x ∈ [2,∞) and x ≥ 38 }
⇒ Domain ((fof)of) = { x : x ≥ 38}
⇒ Domain ((fof)of) = [ 38, ∞ )
fof : [38,∞)→ R
So, ((fof)of) (x) = (fof) (f (x))
= (fof) `(sqrt(x-2))`
= `sqrt (sqrt (sqrt(x-2) -2 )-2)`
(iii) We have, (fofof) (x) = `sqrt (sqrt (sqrt(x-2) -2 )-2)`
So, (fofof) (38) = `sqrt (sqrt (sqrt(38-2) -2 )-2)`
=`sqrt (sqrt (sqrt(36) -2 )-2)`
=`sqrt (sqrt(6-2) -2 )`
= `sqrt (2 -2)`
= 0
(iv) We have, fof = `sqrt (sqrt(x-2) -2 )`
` f^2 (x) = f (x) xx f (x) = sqrt(x - 2) xx sqrt(x - 2) = x -2`
So, fof ≠ `f^2`
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