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Q.
For real $x$, the function $\frac{(x-a)(x-b)}{(x-c)}$ will assume all real values provided
IIT JEEIIT JEE 1984Complex Numbers and Quadratic Equations
Solution:
Let $y=\frac{x^2-(a+b)x+ab}{x-c}$
$\Rightarrow yx-cy=x^2-(a+b)x+ab$
$\Rightarrow x^2-(a+b+y)x+(ab+cy)=0$
For real roots,$D\ge0$
$\Rightarrow (a+b+y)^2-4(ab+cy)\ge0$
$\Rightarrow (a+b)^2+y^2+2(a+b)y-4ab-4cy\ge0$
$\Rightarrow y^2+2(a+b-2c)y+(a-b)^2\ge0$
which is true for all real values of $y$.
$\therefore D\le0$
$ 4(a+b-2c)^2-4(a-b)^2\le0$
$\Rightarrow 4(a+b-2c+a-b)(a+b-2c-a+b)\le0$
$\Rightarrow (2a-2c)(2b-2c)\le0$
$\Rightarrow (a-c)(b-c)\le0$
$\Rightarrow (c-a)(c-b)\le0$
$\Rightarrow c $ must lie between $a$ and $b$
i.e. $a\le c\le b\, $ or $ b\le c\le a $