Explanation

(d)
Let $y = \ \frac{(x - a)(x - b)}{(x - c)}$
$\Rightarrow y(x - c) = x^{2} - (a + b)x + ab$
$\Rightarrow x^{2} - (a + b + y)x + ab + cy = 0$
Now, discriminant
$D = (a + b + y)^{2} - 4(ab + cy)$
$= y^{2} + 2y(a + b - 2c) + (a - b)^{2}$
Since, $x$ is real and $y$ assumes all real values, we must have $D \geq 0$ for all real values of $y$.
$\Rightarrow 4(a + b - 2c)^{2} - 4(a - b)^{2} \leq 0$
$\Rightarrow 4(a + b - 2c + a - b)(a + b - 2c - a + b) \leq 0$
$\Rightarrow 16(a - c)(b - c) \leq 0$
$\Rightarrow (c - a)(c - b) \leq 0$