If the quadratic equation $4x^{2}-2(a+c-1)x+ac-b=0(a>b>c).$

$\begin{array}{l} 4 x^{2}-2(a+c-1) x+a c-b=0 \\ 4 x^{2}-2 a x-2 c x+a c+(2 x-b)=0 \\ 2 x(2 x-a)-c(2 x+a)+(2 x-b)=0 \\ (2 x-a)(2 x-c)+2 x-b=0 \\ f\left(\frac{a}{b}\right)=a-b ; \quad f\left(\frac{c}{2}\right)=c-b . \end{array}$

Now, $f\left(\frac{a}{2}\right) f\left(\frac{c}{2}\right)=(a-b)(c-b)<0 \quad(a>b>c)$

Hence, exactly one of roots lies between c/2 and $a / 2$