Given,

f(x) = 2x + 5 and g(x) = x² + x 

Clearly,

Both f(x) and g(x) are defined for all x ∈ R. 

Hence, 

Domain of f = domain of g = R 

i. f + g 

We know,

(f + g)(x) = f(x) + g(x) 

⇒ (f + g)(x) = 2x + 5 + x² + x 

∴ (f + g)(x) = x² + 3x + 5 

Clearly, 

(f + g)(x) is defined for all real numbers x. 

∴ The domain of (f + g) is R 

ii. f – g 

We know,

(f – g)(x) = f(x) – g(x) 

⇒ (f – g)(x) = 2x + 5 – (x² + x) 

⇒ (f – g)(x) = 2x + 5 – x² – x 

∴ (f – g)(x) = 5 + x – x² 

Clearly, 

(f – g)(x) is defined for all real numbers x. 

∴ The domain of (f – g) is R 

iii. fg 

We know,

(fg)(x) = f(x)g(x) 

⇒ (fg)(x) = (2x + 5)(x² + x) 

⇒ (fg)(x) = 2x(x² + x) + 5(x² + x) 

⇒ (fg)(x) = 2x³ + 2x² + 5x² + 5x 

∴ (fg)(x) = 2x³ + 7x² + 5x 

Clearly, 

(fg)(x) is defined for all real numbers x. 

∴ The domain of fg is R

 iv. \(\frac{f}{g}\) 

We know,

(\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\)

∴ (\(\frac{f}{g}\))(x) = \(\frac{2x+5}{x^2+x}\)

Clearly,

(\(\frac{f}{g}\))(x) is defined for all real values of x, except for the case when x² + x = 0.

x² + x = 0 

⇒ x(x + 1) = 0 

⇒ x = 0 or x + 1 = 0 

⇒ x = 0 or –1

When x = 0 or –1,(\(\frac{f}{g}\))(x) will be undefined as the division result will be indeterminate.

Thus, 

Domain of \(\frac{f}{g}\) = R – {–1, 0}