(i) Here, we have f(x): R → R and g(x): R → R
(a) f + g
As we know, (f + g)(x) = f(x) + g(x)
(f + g)(x) = x3 + 1 + x + 1
= x3 + x + 2
Therefore, (f + g) (x): R → R
∴ f + g: R → R is given by (f + g) (x) = x3 + x + 2
(b) f – g
As we know, (f – g) (x) = f(x) – g(x)
(f – g) (x) = x3 + 1 – (x + 1)
= x3 + 1 – x – 1
= x3 – x
Therefore, (f – g) (x): R → R
∴ f – g: R → R is given by (f – g) (x) = x3 – x
(c) cf (c ∈ R, c ≠ 0)
As we know, (cf) (x) = c × f(x)
(cf)(x) = c(x3 + 1)
= cx3 + c
Therefore, (cf) (x) : R → R
∴ cf: R → R is given by (cf) (x) = cx3 + c
(d) fg
As 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)
Therefore, (fg) (x): R → R
∴ fg: R → R is given by (fg) (x) = (x + 1)2(x2 – x + 1)
(e) 1/f
As 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.
Therefore, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x3 + 1)
(f) f/g
As 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.
By 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) Now, we have f(x): [1, ∞) → R+ and g(x): [–1, ∞) → R+ as real square root is defined only for non-negative numbers.
(a) f + g
As 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
As 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)
As 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
As 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
As 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
As 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, ∞)
Thus, f/g: [1, ∞) → R is given by (f/g) (x) = √[(x - 1)/(x + 1)]