Question

For real ${}^{\prime }{x}^{\prime }$,the expression $y=\frac{(x-a)(x-b)}{(x-c)}$ will assume all real values provided if

• A. $a\ge b\ge c$
• B. $a\le b\le c$
• C. $a\ge c\ge b$
• D. None of these

Answer 1

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

Given

$y=\frac{(x-a)(x-b)}{(x-c)},xϵR,yϵR$

$x^2-(a+b)x+ab=yx-yc$

$x^2-(a+b+y)x+(ab+yc)=0$

${(a+b+y)}^2-4(ab+yc)\ge 0$

${(a+b)}^2+2y(a+b)+y^2-4ab-4yc\ge 0$

$y^2+2y(a+b-2c)+{(a-b)}^2\ge 0$

$yϵR$

$\therefore 4{(a+b-2c)}^2-4{(a-b)}^2\le 0$

$\left(2(a-c)\left(2(b-c)\right)\right)\le 0$

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

$\therefore a\le c\le b$ (or) $a\ge c\ge b$