Question

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

Answer 1

Hint: For solving this question we have to equate the question's equation equal to y and cross multiply. Then a quadratic equation will be formed. Use the properties of quadratic equations for further solution.

Complete step-by-step answer:

$ \Rightarrow y = \dfrac{{(x - a)(x - b)}}{{(x - c)}}$ eq1.
Now, on cross multiplication of above equation, we get
$
  \Rightarrow y(x - c) = (x - a)(x - b) \\
$
On rearranging above equation, we get
$
   \Rightarrow {x^2} - (a + b + y)x + (ab + cy) = 0 \\
$
Now for roots to be real, discriminant of above equation must be positive
$
  D > 0{\text{ }}\left[ {\because D = {b^2} - 4ac} \right] \\
   \Rightarrow {(a + b + y)^2} - 4(ab + cy) > 0 \\
   \Rightarrow {(a + b)^2} + {y^2} + 2y(a + b) - 4ab - 4cy > 0 \\
   \Rightarrow {y^2} + 2y(a + b - 2c) + {(a - b)^2} > 0{\text{ eq 2}}{\text{.}} \\
    \\
$
In eq 2. ${y^2}$ is always positive for any value of $y$ and similarly ${(a - b)^2}$ will always be positive . So, For eq 2. to be positive for all y , $(a + b - 2c)$ should be zero
$
   \Rightarrow a + b - 2c = 0 \\
   \Rightarrow a + b = 2c \\
    \\
 $
(we can see this is property of AP sum of first term and third term is equal to two times the second term)

Hence, $a,c,b$ are in A.P. (Arithmetic Progression)
Thus, we can say that
Either $a < c < b{\text{ or }}a > c > b$
Hence, option C is the correct answer.

Note: Whenever you get this type of question the key concept of solving is you have to resolve the given equation in quadratic equation by equating given equation to y and then Solve it and check various conditions of discriminant.