Question

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

• A. $a>b>c$
• B. $a<b<c$
• C. $a>c>b$
• D. $a<c<b$

Answer 1

Answer: (C), (D)

The correct options are
C $a>c>b$
D $a<c<b$

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

$y\left(x-c\right)=\left(x-a\right)\left(x-b\right)$

$x^2-\left(a+b+y\right)x+\left(ab+cy\right)=0$

$x$ is real if $Discriminant\ge 0$

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

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

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

In any quadratic $a{z}^2+bz+c$, if $a>0$, then $a{z}^2+bz+c\ge 0$ for every $z$ if $b^2-4ac<0$.

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

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

$c^2-\left(a+b\right)c+ab<0$

$\left(c-a\right)\left(c-b\right)<0$

This shows that $c$ lies between $a$ and $b$, therefore, $a<c<b$ or $a>c>b$.