Let f (x) =[x], the greatest integer less than or equal to x and g(x) = {x}, fractional part of x. The solution of the equation 4g(x) = i (x)+ f (x); x > 0, i being identity function is

$\frac{5}{3}$

As $\{x\}= x −[x]$, the given equation can be written as

$4(x −[x]) = x +[x] ⇒ 3x = 5[x] ⇒ x=\frac{5[x]}{3}$

Next using $[x] = x-\{x\}$, we can write the given equation is

$4\{x\}=x+x-\{x\}⇒x=(\frac{5}{2})\{x\}$

As $0 ≤\{x\}<1$, we get $0 ≤x<\frac{5}{2}$

Thus $0 ≤\frac{5[x]}{3}<\frac{5}{2}⇒0 ≤[x]<\frac{3}{2}$

Since [x] ∈ I so [x] = 0 or 1 thus 3x = 0 or $3x = 5 ⇒x=\frac{5}{3}$

But x > 0 so $x=\frac{5}{3}$