(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)]