Question

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

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

Answer 1

The correct option is D

$[1,\infty ]$

Let $y=\frac{x^2–x}{1-ax},x\in R$
$\Rightarrow x^2-x=y-ayx$
$\Rightarrow x^2+(ay-1)x-y=0$

Since $y$ attains all values in $R$ for real values of $x$,
$D={(ay-1)}^2+4y⩾0$ $\forall y\in R$
$\Rightarrow a^2$$y^2+(4-2a)y+1⩾0$
We must have $a^2\ge 0$ and discriminant $\le 0$
$\Rightarrow 4{(2-a)}^2$ -4$a^2$ ⩽0
$\Rightarrow 4-4a+a^2$ -$a^2$ ⩽0

$\Rightarrow a⩾1$

Hence $a\in [1,\infty )$