Question

For real $x$, the expression $y=\frac{(x-a)(x-b)}{(x-c)}$ will assume all real values provided:(Given that a>b)

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

Answer 1

The correct option is C $a>c>b$

$y=\frac{(x-a)(x-b)}{(x-c)}$

$yx-yc=x^2-ax-bx+ab$

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

For $x$ to be real, $D\ge 0$,

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

$y^2+(2b+2a-4c)y+(a^2+b^2-2ab)\ge 0$

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

Here coefficient of $y>0$, and the expression is positve, hence $D<0$

$(2b+2a-4c{)}^2-4(a-b{)}^2<0$

$(b+a-2c{)}^2-(a-b{)}^2<0$

$(b+a-2c-a+b)(b+a-2c+a-b)<0$

$(2b-2c)(2a-2c)<0$

$(b-c)(a-c)<0$

EIther $b-c<0$ and $a-c>0$.

$\therefore b<c$ and $a>c$ $⟹a>c>b$

or $b-c>0$ and $a-c<0$

$\therefore b>c$ and $a<c$ $⟹a<c<b$