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. [0,1]
• D. [1, ∞]

Answer 1

The correct option is D

[1, ∞]

Let y=( $x^2$ –x)/(1-ax), x∈R

=> $x^2$-x = y-ayx =>$x^2$+(ay-1)x-y=0

Since y attains all values in R for real x,

D=${(ay-1)}^2$ +4y ⩾ 0 ∀ y ∈ R

=> $a^2$$y^2$ +(4-2a)y+1 ⩾0 ∀ y ∈ R

=> 4${(2-a)}^2$ -4$a^2$ ⩽0

=> 4-4a+$a^2$ -$a^2$ ⩽0

=> a⩾1

Hence a∈[1,$\infty$)