Let f(x) = 2x + 5

Solution:

Given:

f(x) = 2x + 5 and g(x) = x2 + x

Clearly, f (x) and g (x) assume real values for all x.

Hence,

domain (f) = R and domain (g) = R.

$\therefore \quad D(f) \cap D(g)=R$

Now,

(i) (f + g) : R → R is given by (f + g) (x) = f (x) + g (x) = 2x + 5 + x2 + x = x2 + 3x + 5.

Hence, domain ( f + g) = R .

(ii) $(f-g): R \rightarrow R$ is given by $(f-g)(x)=f(x)-g(x)=(2 x+5)-\left(x^{2}+x\right)=5+x-x^{2}$

 Hence, domain ( f - g) = R.

(iii) $(f g): R \rightarrow R$ is given by $(f g)(x)=f(x) \cdot g(x)=(2 x+5)\left(x^{2}+x\right)$

= 2x3 + 2x2 + 5x2 + 5x

= 2x3 + 7x2 + 5x

Hence, domain ( f.g) = R .

(iv) Given:

g(x) = x2 + x

g(x) = 0 ⇒ x2 + x = 0 = x(x+ 1) = 0

⇒ x = 0 or (x + 1) = 0

⇒ x = 0 or x =  -1

Now,

$\frac{f}{g}: R-\{-1,0\} \rightarrow R$ is given by $\left(\frac{\mathrm{f}}{\mathrm{g}}\right)(\mathrm{x})=\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\frac{2 \mathrm{x}+5}{\mathrm{x}^{2}+\mathrm{x}}$

Hence, $\mathrm{d}$ omain $\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\mathrm{R}-\{-1,0\}$.