If b>a, then the equation (x-a)(x-b)=1 has
Both roots in [a, b]
Both roots in (-∞,a)
Both roots in (b, +∞)
One root in (-∞, a) and the other in (b, +∞)
The equation is x2-(a+b)x+ab-1=0
∴ discriminant =(a+b)2-4(ab-1)=(b-a)2+4>0
∴ both roots are real. Let them be α,β where
α=(a+b)-(b-)2+42, β=(a+b)+(b-a)2+42
Clearly, α<(a+b)-(b-a)22=(a+b)-(b-a)2=a
(∵b>a)
and β>(a+b)+(b-a)22=a+b+b-a2=b
Hence, one root α is less than a and the other root β is greater than b.