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 > b < c
• D. $a\le c\le b$

Answer 1

The correct option is D $a\le c\le b$

Let $y=\frac{x^2-(a+b)x+ab}{x-c}$
$\Rightarrow yx-cy=x^2-(a+b)x+ab\Rightarrow x^2-(a+b+y)x+(ab+cy)=0$
For real roots, $D\ge 0$
$\Rightarrow (a+b+y{)}^2-4(ab+cy)\ge 0\Rightarrow (a+b{)}^2+y^2+2(a+b)y-4ab-4cy\ge 0\Rightarrow y^2+2(a+b-2c)y+(a-b{)}^2\ge 0$
Which is true for all real values of y.
$\therefore D\le 04(a+b-2c{)}^2-4(a-b{)}^2\le 0\Rightarrow 4(a+b-2c+a-b)(a+b-2c-a+b)\le 0\Rightarrow (2a-2c)(2b-2c)\le 0$
$\Rightarrow (a-c)(b-c)\le 0\Rightarrow (c-a)(c-b)\le 0$
$\Rightarrow c$ must lie between a and b
i.e. $a\le c\le b$ or $b\le c\le a$