Let’s transform (x$^{2}$ – 6x + 10) / (3 – x) ; (x < 3) to simpler form :
(x$^{2}$ – 6x + 10) / (3 – x) ; (x < 3)
= ((3 – x)$^{2}$ + 1) / (3 – x) ; (3 – x > 0)
Let’s take y = (3 – x). So, (y$^{2}$ + 1) / y ; (y > 0) = (y + 1/y) ; (y > 0)
Now we will find min value of (y + 1/y), for y > 0. Let z = (y + 1/y) => dz/dy = (1 – 1/y$^{2}$). dz/dy is 0 when y=1.
Let’s check whether at y=1, z is minimum or maximum. d$^{2}$z/dy$^{2}$ = (2/y$^{3}$). So, d$^{2}$z/dy$^{2}$ is >0 for y>0, which means that z is minimum at y=1 because concave up graph.
Therefore at y=1, z=2. Ans is B.