Question

f(x) = $\frac{(x-b)(x-c)}{(x-a)}$, where a,b,c are distinct real numbers, will assume all real values provided :

• A. c lies between a and b
• B. a lies between b and c
• C. b lies between a and c
• D. None of these

Answer 1

The correct option is B

a lies between b and c

y = $\frac{(x^2-(b+c)x+bc)}{x-a}$ assume all real values.

$\Rightarrow$ $x^2-(b+c+y)x+bc+ay=0.xϵ$ R for $\forall$ y $ϵ$ R

$\Rightarrow$ D = $(b+c+y{)}^2-4bc-4ay\ge 0$ $\forall$ y $ϵ$ R

$\Rightarrow$ $y^2+(2b+2c-4a)y+(b-c{)}^2\ge 0$ $\forall$ y $ϵ$ R

$\Rightarrow$ 4 $(b+c-2a{)}^2-(b-c{)}^2\le 0$ $\forall$ y $ϵ$ R

$\Rightarrow$ 4 (b-a) ( c-a) $\le$ 0

$\Rightarrow$ ( a - b)(a - c) < 0 [ $\because$ a,b,c are distinct ]

$\Rightarrow$ b < a < c or c < a < b.

$\Rightarrow$ a lies between b and c .