Show that if x is real- the expression x2-bc2x-b-c has no value between b and c

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Standard XII

Mathematics

Selecting Consecutive Terms in GP

Show that if ...

Question

Show that if $x$ is real, the expression $\frac{x^{2} - b c}{2 x - b - c}$ has no value between $b$ and $c$

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Solution

Let
$\frac{x^{2} - b c}{2 x - b - c} = y$

$∴ x^{2} - b c = y (2 x - b - c)$
$x^{2} - b c - 2 x y + b y + c y = 0$
$x^{2} - (2 y) x + (b y + c y - b c) = 0$
As x is real, $D > 0$
$(- 2 y)^{2} - 4 (1) (b y + c y - b c) > 0$
$4 y^{2} - 4 b y - 4 c y + 4 b c > 0$
$y^{2} - b y - c y + b c > 0$
$(y - c) (y - b) > 0$
$(y - c) (y - b) = + v e$ only when $(y > c), (y > b)$
$(y - c) (y - b) = - v e$ only when $(y < c), (y < b)$

Therefore, the given expression has no real value between b and c

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Show that if x is real, the expression x2−bc2x−b−c has no value between b and c

Show that if $x$ is real, the expression $\frac{x^{2} - b c}{2 x - b - c}$ has no value between $b$ and $c$