Question 2.

Let $\frac{x^2 - 6x + 10}{3 - x} = p$

$x^2 - 6x + 10 = 3p - px$

$x^2 - (6 - p)x + 10 - 3p = 0$

Since the equation will have real roots,

$(6 - p)^2 - 4 \times (10 - 3p) \ge 0$

$p^2 - 12p + 12p + 36 - 40 \ge 0$

$p^2 \ge 4$

$p \ge 2 ,\ p \le -2$

Now, when $p = -2$, $x = 4$. Since it is given that x < 3, thus this value will be discarded.

Now, $\frac{1}{2}$ and $-\frac{1}{2}$ do not come in the mentioned range.

When $p = 2$, $x = 2$

Thus, the minimum possible value of p will be 2.

Thus, the correct option is C.

Alternate explanation:

Since x < 3,

3 - x > 0

Let $3 - x = y$. So, y > 0.

Now,

$\frac{x^2 - 6x + 10}{3 - x} = \frac{x^2 - 6x + 9 + 1}{3 - x}$

$\Rightarrow \frac{(3 - x)^2 + 1}{3 - x}$

Since $3 - x = y$, the equation will transform to $\frac{y^2 + 1}{y}$ or $y + \frac{1}{y}$

The minimum value of the expression $y + \frac{1}{y}$ for y > 0 will be at $y = 1$

i.e. Minimum value = $1 + 1 = 2$

Thus, the correct option is C.