Question -
Answer -
(i) f(x) = x3 + 1 and g(x) = x+ 1
We have f(x): R → R and g(x): R → R
(a) f + g
We know, (f + g) (x) = f(x) + g(x)
(f + g) (x) = x3 + 1+ x + 1
= x3 + x + 2
So, (f + g) (x): R → R
∴f + g: R → R is given by (f + g) (x) = x3 + x + 2
(b) f – g
We know, (f – g) (x) = f(x) – g(x)
(f – g) (x) = x3 + 1– (x + 1)
= x3 + 1 – x – 1
= x3 – x
So, (f – g) (x): R → R
∴f – g: R → R is given by (f – g) (x) = x3 – x
(c) cf (c ∈ R, c ≠ 0)
We know, (cf) (x) = c × f(x)
(cf)(x) = c(x3 + 1)
= cx3 + c
So, (cf) (x) : R → R
∴cf: R → R is given by (cf) (x) = cx3 +c
(d) fg
We know, (fg) (x) = f(x) g(x)
(fg) (x) = (x3 + 1)(x + 1)
= (x + 1) (x2 – x +1) (x + 1)
= (x + 1)2 (x2 – x + 1)
So, (fg) (x): R → R
∴fg: R → R is given by (fg) (x) = (x + 1)2(x2 –x + 1)
(e) 1/f
We know, (1/f) (x) = 1/f (x)
1/f (x) = 1 / (x3 +1)
Observe that 1/f(x) is undefined when f(x) = 0 or when x = – 1.
So, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x3 + 1)
(f) f/g
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = (x3 + 1)/ (x + 1)
Observe that (x3 +1) / (x + 1) is undefined when g(x) = 0 or when x = –1.
Using x3 + 1 = (x +1) (x2 – x + 1), we have
(f/g) (x) = [(x+1) (x2–x+1)/(x+1)]
= x2 – x + 1
∴f/g: R – {–1} → R is given by (f/g) (x) = x2 – x + 1
(ii) f(x) = √(x-1) and g (x) = √(x+1)
We have f(x): [1, ∞) → R+ andg(x): [–1, ∞) → R+ asreal square root is defined only for non-negative numbers.
(a) f + g
We know, (f + g) (x) = f(x) + g(x)
(f+g) (x) = √(x-1) + √(x+1)
Domain of (f + g) = Domain of f ∩ Domain of g
Domain of (f + g) = [1, ∞) ∩ [–1, ∞)
Domain of (f + g) = [1, ∞)
∴f + g: [1, ∞) → R is given by (f+g) (x) = √(x-1) + √(x+1)
(b) f – g
We know, (f – g) (x) = f(x) – g(x)
(f-g) (x) = √(x-1) – √(x+1)
Domain of (f – g) = Domain of f ∩ Domain of g
Domain of (f – g) = [1, ∞) ∩ [–1, ∞)
Domain of (f – g) = [1, ∞)
∴f – g: [1, ∞) → R is given by (f-g) (x) = √(x-1) – √(x+1)
(c) cf (c ∈ R, c ≠ 0)
We know, (cf) (x) = c × f(x)
(cf) (x) = c√(x-1)
Domain of (cf) = Domain of f
Domain of (cf) = [1, ∞)
∴cf: [1, ∞) → R is given by (cf) (x) = c√(x-1)
(d) fg
We know, (fg) (x) = f(x) g(x)
(fg) (x) = √(x-1) √(x+1)
= √(x2 -1)
Domain of (fg) = Domain of f ∩ Domain of g
Domain of (fg) = [1, ∞) ∩ [–1, ∞)
Domain of (fg) = [1, ∞)
∴fg: [1, ∞) → R is given by (fg) (x) = √(x2 -1)
(e) 1/f
We know, (1/f) (x) = 1/f(x)
(1/f) (x) = 1/√(x-1)
Domain of (1/f) = Domain of f
Domain of (1/f) = [1, ∞)
Observe that 1/√(x-1) is also undefined when x – 1 = 0 or x = 1.
∴1/f: (1, ∞) → R is given by (1/f) (x) = 1/√(x-1)
(f) f/g
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = √(x-1)/√(x+1)
(f/g) (x) = √[(x-1)/(x+1)]
Domain of (f/g) = Domain of f ∩ Domain of g
Domain of (f/g) = [1, ∞) ∩ [–1, ∞)
Domain of (f/g) = [1, ∞)
∴ f/g:[1, ∞) → R is given by (f/g) (x) = √[(x-1)/(x+1)]