First slide

Theory of expressions

Question

For real values of x, the expression $\frac{(x - b) (x - c)}{(x - a)}$ will assume all real values provided

Moderate

A

$a \leq c \leq b$
B

$b \geq a \geq c$
C

$b \leq c \leq a$
D

$a \geq b \geq c$

Solution

Let $m = \frac{(x - b)}{x - a} \frac{(x - c)}{}$
$\Rightarrow x^{2} - (b + c + m) x + (b c + a m) = 0$
Since $x$ is real, we must have
$(b + c + m)^{2} - 4 (b c + a m) \geq 0$
$\Rightarrow m^{2} + 2 (b + c - 2 a) m + (b - c)^{2} \geq 0$ for all $m$
$\begin{array}{l} \Rightarrow 4 (b + c - 2 a)^{2} - 4 (b - c)^{2} \leq 0 \\ \Rightarrow (b + c - 2 a)^{2} - (b - c)^{2} \leq 0 \\ \Rightarrow (b + c - 2 a + b - c) (b + c - 2 a - b + c) \leq 0 \\ \Rightarrow 2 (b - a) 2 (c - a) \leq 0 \\ \Rightarrow (a - b) (a - c) \leq 0 \\ \Rightarrow b \leq a \leq c or, c \leq a \leq b \end{array}$

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