Question

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

• A. $a\le c\le b$
• B. $b\ge a\ge c$
• C. $b\le c\le a$
• D. $a\ge b\ge c$

Answer 1

The correct option is B $b\ge a\ge c$

Let $y=\frac{(x-b)(x-c)}{(x-a)}$
$\therefore (x-a)y=x^2-(b+c)x+bc$
$\therefore x^2-(b+c+y)x+(bc+ay)=0$
For roots to be real,
$(b+c+y{)}^2-4(bc+ay)>0$
$\therefore (b+c{)}^2+y^2-(4bc+4ay)+2(b+c)y>0$
$\therefore y^2+2(b+c-2a)y+(b-c{)}^2>0$ ----------(1)
For any value of y $y^2\&(b-c{)}^2>0$
For (1) to be +ve for all y, 2(b+c-2a)=0
$\therefore b+c-2a=0$
$\therefore b-a+c-a=0$
$\therefore b-a=a-c$
$\therefore b,a,c$ are in AP.
$\therefore$ either b>a>c or b<a<c
Of the given options : $b\to a\to c$