Let f: R - \ (- 1)/(2)\ → R and g: R - \ (- 5)/(2)\ → R be defined...

Given: $g\left(x\right) = \dfrac{\left|x\right| + 1}{2x + 5}$.

Also, $g\left(x\right) \ne \dfrac{- 1}{2}$

$⇒ \dfrac{\left|x\right| + 1}{2x + 5} \ne \dfrac{- 1}{2}$

Case-I: $x \ge 0$

$⇒ g\left(x\right) = \dfrac{x + 1}{2x + 5} = \dfrac{- 1}{2}$

$⇒ 2x + 2 = - 2x - 5$

$⇒ 4x = - 7$

$⇒ x = \dfrac{- 7}{4} \to \text{Rejected as}x \ge 0$

Case-I: $x < 0$

$⇒ \dfrac{- x + 1}{2x + 5} = \dfrac{- 1}{2}$

$⇒ - 2x + 2 = - 2x - 5$

$⇒ 2 = - 5 \to \text{Not possible}$

So, domain of $f\left(g\left(x\right)\right) =$domain of $g\left(x\right)$.

$⇒ \text{Domain of}fog\left(x\right) = R - \left\{- \dfrac{5}{2}\right\}$