Theory of Equations 1 Question 33
34. For real $x$, the function $\frac{(x-a)(x-b)}{(x-c)}$ will assume all real values provided
(1984, 3M)
(a) $a>b>c$
(b) $a<b<c$
(c) $a>c<b$
(d) $a \leq c \leq b$
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Solution:
- Let $y=\frac{x^{2}-(a+b) x+a b}{x-c}$
$\Rightarrow \quad y x-c y=x^{2}-(a+b) x+a b$
$\Rightarrow x^{2}-(a+b+y) x+(a b+c y)=0$
For real roots, $D \geq 0$
$\begin{array}{lrl}\Rightarrow & (a+b+y)^{2}-4(a b+c y) & \geq 0 \ \Rightarrow & (a+b)^{2}+y^{2}+2(a+b) y-4 a b-4 c y \geq 0 \ \Rightarrow & y^{2}+2(a+b-2 c) y+(a-b)^{2} \geq 0\end{array}$
which is true for all real values of $y$.
$\therefore \quad D \leq 0$
$4(a+b-2 c)^{2}-4(a-b)^{2} \leq 0$
$\Rightarrow 4(a+b-2 c+a-b)(a+b-2 c-a+b) \leq 0$
$\Rightarrow \quad(2 a-2 c)(2 b-2 c) \leq 0$
$\Rightarrow \quad(a-c)(b-c) \leq 0$
$\Rightarrow \quad(c-a)(c-b) \leq 0$
$\Rightarrow c$ must lie between $a$ and $b$
i.e. $a \leq c \leq b$ or $b \leq c \leq a$
