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. $Always$
• D. $a<c<b$

Answer 1

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

Let $y=\frac{(x-a)(x-b)}{x-c}$
or $y(x-c)=x^2-(a+b)x+ab$
or $x^2-(a+b+y)x+ab+cy=0$
$\therefore D=(a+b+y{)}^2-4(ab+cy)$
$=y^2+2y(a+b-2c)+(a-b{)}^2$
Since $x$ is real and $y$ assumes all real values we must have $△\ge 0\forall$ real of $y$. The sign of a quadratic in $y$ is same as of first term provided its discriminant $D<0$
$\Rightarrow 4(a+b-2c{)}^2-4(a-b{)}^2<0$
$\Rightarrow 4(a+b-2c+a-b)(a+b-2c-a+b)<0$
or $16(a-c)(b-c)<0$ or $16(c-a)(c-b)=-ve$
$\therefore c$ lies between $a$ and $b$ i.e. $a<c<b...\left(i\right)$
where $a<b$ but if $b<a$ then above condition will be $b<c<a$ or $a>c>b...\left(ii\right)$
Hence from $\left(i\right)$ and $\left(ii\right)$ we observe $\left(d\right)$ is correct.