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

$b \leq a \leq c$
C

$b \leq c \leq a$
D

$a \leq b \leq c$

Solution

The given expression is $\frac{(x - b) (x - c)}{(x - a)}$
Let $\frac{(x - b) (x - c)}{(x - a)} = m$
$\Rightarrow (x - b) (x - c) - m (x - a) = 0$
$\Rightarrow$ $x^{2} - (b + c + m) x + (b c + a m) = 0$ .
Since $x$ is real, $∴$ Discriminant $\geq 0$
$\Rightarrow$ ${(b + c + m)}^{2} - 4 (b c + a m) \geq 0$      $(∵ b^{2} - 4 a c \geq 0)$
$\Rightarrow$ $m^{2} + 2 (b + c - 2 a) m + {(b - c)}^{2} \geq 0$
$\Rightarrow$ ${[m + (b + c - 2 a)]}^{2} + {(b - c)}^{2} - {(b + c - 2 a)}^{2} > 0$
Since $m$ is real,
$∴$    ${(b - c)}^{2} - {(b + c - 2 a)}^{2} \geq 0$
$\Rightarrow$    $(b - a) (a - c) \geq 0$    $\Rightarrow$    $(a - b) (a - c) \leq 0$
$\Rightarrow$ $b ⩽ a ⩽ c$ .

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