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If $x$ is real, the function $\frac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided
$a > b > c$
$a < b < c$
$a > c < b$
$a < c < b$
Solution
(d) Let $y = \frac{{(x – a)(x – b)}}{{(x – c)}}$
or $y(x – c) = {x^2} – (a + b)x + ab$
or ${x^2} – (a + b + y)x + ab + cy = 0$
$\Delta = {(a + b + y)^2} – 4(ab + cy)$
$ = {y^2} + 2y(a + b – 2c) + {(a – b)^2}$
Since $x$ is real and $y$ assumes all real values, we must have $\Delta \ge 0$ for all real values of $y$.The sign of a quadratic in $y$ is same as of first term provided its discriminant ${B^2} – 4AC < 0$
This will be so if $4{(a + b – 2c)^2} – 4{(a – b)^2} < 0$
or $4(a + b – 2c + a – b)(a + b – 2c – a + b) < 0$
or $16(a – c)(b – c) < 0$ or $16(c – a)(c – b) = – $ve
$\therefore$ $c$ lies between $a$ and $b$ i.e., $a < c < b$…..$(i)$
Where $a < b$, but if $b < a$then the above condition will be $b < c < a$or $a > c >b$…..$(ii)$
Hence from $(i)$ and $(ii)$ we observe that $(d)$ is correct answer.
