Question

If $y=\frac{(x-a)(x-c)}{x-b}$ assumes all real values for $x\in R-\left\{b\right\},$ then

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

Answer 1

The correct option is B $a<b<c$

Let $y=\frac{(x-a)(x-c)}{x-b}$
$\Rightarrow xy-by=x^2-(a+c)x+ac$
$\Rightarrow x^2-(a+c+y)x+(ac+by)=0$

$\because x\in R-\left\{b\right\}\Rightarrow D\ge 0$
$\Rightarrow (a+c+y{)}^2-4(ac+by)\ge 0\Rightarrow y^2+2(a+c-2b)y+(a-c{)}^2\ge 0$

$\because y\in R$ and Coefficient of $y^2>0,$
So, $D\le 0$
$\Rightarrow 4(a+c-2b{)}^2-4(a-c{)}^2\le 0\Rightarrow (a+c-2b{)}^2-(a-c{)}^2\le 0\Rightarrow (a+c-2b+a-c)(a+c-2b-a+c)\le 0\Rightarrow (2a-2b)(2c-2b)\le 0\Rightarrow (b-a)(b-c)\le 0\Rightarrow a\le b\le c\text{or}a\ge b\ge c$

When $a=b=c$, so
$y=\frac{(x-b)(x-b)}{(x-b)}=x-b$
As $x\ne b$, so $y\ne 0$
Therefore, $y$ can't take all $R$

Hence,
$a<b<c\text{or}a>b>c$

Alternate method:
For $y$ to take all values of $R.$
Root of the denominator must lie in between the roots of numerator.
Hence,
$a<b<c\text{or}a>b>c$