For x ∈ ℝ, two real valued functions f(x) and g(x) are such that, g...

We have been given that

$f\left(g\left(x\right)\right) = x + 3 - \sqrt{x}$

$⇒ f ( \sqrt{x} + 1 ) = x + 3 - \sqrt{x}$

Put $\sqrt{x} + 1 = t$

$\sqrt{x} = t - 1$

$x = ( t - 1{)}^{2}$

$⇒ f ( t ) = ( t - 1{)}^{2} + 3 - ( t - 1 )$

$⇒ f ( 0 ) = 1 + 3 + 1 = 5$

Hence this is the correct option.