Question -
Answer -
(i) Given f (x) =ex, g(x) = loge x
Letf: R → (0, ∞); and g: (0, ∞) → R
Now we have tocalculate fog,
Clearly, the range of g is a subset of the domain of f.
fog: ( 0, ∞) → R
(fog) (x) = f (g (x))
= f (loge x)
= loge ex
= x
Now we have tocalculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (ex)
= loge ex
= x
(ii)f (x) = x2, g(x) = cos x
f: R→ [0, ∞) ; g: R→[−1, 1]
Now we have tocalculate fog,
Clearly, the range of g is not a subset of the domain of f.
⇒ Domain (fog) = {x: x∈domain of g and g (x) ∈ domain of f}
⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}
⇒ Domain of (fog) = R
(fog): R→ R
(fog) (x) = f (g (x))
= f (cos x)
= cos2 x
Now we have tocalculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x2)
= cos x2
(iii) Givenf (x) = |x|, g(x) = sin x
f: R → (0, ∞) ; g : R→[−1, 1]
Now we have to calculatefog,
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (sin x)
= |sin x|
Now we have tocalculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog : R→ R
(gof) (x) = g (f (x))
= g (|x|)
= sin |x|
(iv) Givenf (x) = x + 1, g(x) = ex
f: R→R ; g: R → [ 1, ∞)
Now we have calculatefog:
Clearly, range of g is a subset of domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (ex)
= ex +1
Now we have to computegof,
Clearly, range of f is a subset of domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x+1)
= ex+1
(v) Givenf (x) = sin −1 x, g(x) = x2
f: [−1,1]→[(-π)/2 ,π/2]; g : R → [0, ∞)
Now we have tocompute fog:
Clearly, the range of g is not a subset of the domain of f.
Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}
Domain (fog) ={x: x ∈ R and x2 ∈ [−1, 1]}
Domain (fog) ={x: x ∈ R and x ∈ [−1, 1]}
Domain of (fog)= [−1, 1]
fog: [−1,1] → R
(fog) (x) = f (g (x))
= f (x2)
= sin−1 (x2)
Now we have tocompute gof:
Clearly, the range of f is a subset of the domain of g.
fog: [−1,1] → R
(gof) (x) = g (f (x))
= g (sin−1 x)
= (sin−1 x)2
(vi) Givenf(x) = x+1, g(x) = sin x
f: R→R ; g: R→[−1, 1]
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
Set of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (sin x)
= sin x + 1
Now we have to computegof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= sin (x+1)
(vii) Givenf (x) = x+1, g (x) = 2x + 3
f: R→R ; g: R → R
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (2x+3)
= 2x + 3 + 1
= 2x + 4
Now we have tocompute gof
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= 2 (x + 1) + 3
= 2x + 5
(viii) Givenf (x) = c, g (x) = sin x2
f: R → {c} ; g: R→ [ 0, 1 ]
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
fog: R→R
(fog) (x) = f (g (x))
= f (sin x2)
= c
Now we have to computegof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (c)
= sin c2
(ix) Givenf (x) = x2+ 2 and g (x) = 1 – 1 /(1 – x)
f: R → [ 2, ∞ )
For domain of g: 1− x ≠ 0
⇒ x ≠ 1
⇒ Domain of g= R − {1}
g (x )= 1 –[1/(1 – x)] = (1 – x – 1)/ (1 – x) = -x/(1 – x)
For range of g
y = (- x)/ (1 – x)
⇒ y – xy = − x
⇒ y = xy − x
⇒ y = x (y−1)
⇒ x = y/(y – 1)
Range of g =R − {1}
So, g: R −{1} → R − {1}
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R − {1}→ R
(fog) (x) = f (g (x))
= f (-x/ (1 – x))
= ((-x)/ (1 – x))2 +2
= (x2 +2x2 + 2 – 4x) / (1 – x)2
= (3x2 –4x + 2)/ (1 – x)2
Now we have tocompute gof
Clearly, the range of f is a subset of the domain of g.
⇒ gof: R→R
(gof) (x) = g (f (x))
= g (x2 + 2)
= 1 – 1 / (1 – (x2 +2))
= – 1/ (1 – (x2 +2))
= (x2 +2)/ (x2 + 1)