If x is real, the function $\dfrac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided
A. $a > b > c$
B. $a < b < c$
C. $a > c < b$
D. $a < c < b$

Hint: Equate the above function to y. Use the fact that the discriminant must be greater than or equal to 0 if the roots of the equation are real. Then find the relation between a, b and c such that each term of the equation we get after expanding the discriminant is greater than or equal to 0.

Formula used: Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$, ${p^2} + {q^2} - 2pq = {(p - q)^2}$. Arithmetic mean of m and n is$\dfrac{{m + n}}{2}$.

Complete step by step solution:
Let y = $\dfrac{{(x - a)(x - b)}}{{(x - c)}}$
$y(x - c) = (x - a)(x - b)$
$xy - cy = {x^2} - ax - bx + ab$
Writing the above equation in the standard form we get,
${x^2} - (a + b + y)x + ab + cy = 0$
(Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$).
Since x is real, the discriminant of the above equation must be greater than or equal to 0.
Therefore, ${(a + b + y)^2} - 4(ab + cy) \geqslant 0$
${a^2} + {b^2} + {y^2} + 2ab + 2ay + 2by - 4ab - 4cy \geqslant 0$
${y^2} + {a^2} + {b^2} - 2ab + 2(a + b - 2c)y \geqslant 0$
${y^2} + 2(a + b - 2c)y + {(a - b)^2} \geqslant 0$
We know that ${y^2} + {(a - b)^2}$ is always greater than or equal to 0 since both the terms are always greater than or equal to 0.
Since ${y^2} + {(a - b)^2} \geqslant 0$, for ${y^2} + 2(a + b - 2c)y + {(a - b)^2}$ to be greater than or equal to 0, $2(a + b - 2c)y$ has to be greater than or equal to 0.
$2(a + b - 2c)y \geqslant 0$
Since y is any real number, i.e., since it can be both positive and negative, $(a + b - 2c)$ must equal 0 for $2(a + b - 2c)y \geqslant 0$ to be true.
Therefore, $(a + b - 2c) = 0$
$a + b = 2c$. From this we can also say that c is the arithmetic mean of a and b. Since c is the arithmetic mean of a and b, either $a < c < b$ or $a > c > b$.
$a > c > b$ is not provided in the options.
Therefore, the correct option is D. $a < c < b$.

Note: The question asks us to assume that $\dfrac{{(x - a)(x - b)}}{{(x - c)}}$ is real. Therefore, y is real and subsequently ${y^2} \geqslant 0$. Similarly, a and b are real constants and therefore we can say that ${(a - b)^2} \geqslant 0$. Therefore, ${y^2} + {(a - b)^2} \geqslant 0$.