Question

Complete set of values of a such that $\frac{x^2-x}{1-ax}$ attains all real values is

• A. [1, 4]
• B. [0, 4]
• C. $[1,\infty )$
• D. $[4,\infty )$

Answer 1

The correct option is C

$[1,\infty )$

Let $y=\frac{x^2-x}{1-ax}$
$\Rightarrow x^2-x=y-axy\Rightarrow x^2+x(ay-1)-y=0$
Since $xϵR\Rightarrow \Delta \ge 0$
$(ay-1{)}^2+4y\ge 0\Rightarrow a^2{y}^2+2y(2-a)+1\ge 0yϵR\Rightarrow a>0,4(2-a{)}^2-4a^2\le 0\Rightarrow 4+a^2-4a-a^2\le 0\Rightarrow 1\le a\therefore aϵ[1,\infty )$