# CAT 2022 Set-3 | Quantitative Aptitude | Question: 11

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The minimum possible value of $\frac{x^{2}-6 x+10}{3-x}$, for $x<3$, is

1. $-2$
2. $2$
3. $\frac{1}{2}$
4. $-\frac{1}{2}$

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.

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