Question

The complete set of values of 'k' for which $\frac{x^2-x}{1-kx}$ attains all real values is

• A. $\left(0,\infty \right)$
• B. $[0,\infty )$
• C. $\left(1,\infty \right)$
• D. $[1,\infty )$

Answer 1

The correct option is D $[1,\infty )$

$\mathrm{Let}y=\frac{x^2-x}{1-\mathrm{kx}}\Rightarrow x^2-x=y-\mathrm{kxy}\Rightarrow x^2-x-y+\mathrm{kxy}=0\Rightarrow x^2+(\mathrm{ky}-1)x-y=0\because x\mathrm{is}\mathrm{real},\mathrm{we}\mathrm{get}D\ge 0\left(D\mathrm{is}\mathrm{discriminant}\right){\left(\mathrm{ky}-1\right)}^2+4y\ge 0\Rightarrow k^2{y}^2-2\mathrm{ky}+1+4y\ge 0\Rightarrow k^2{y}^2+2y\left(2-k\right)+1\ge 0\forall y\in R\mathrm{So},\mathrm{as}{k}^2>0,D\le 0\left(D\mathrm{is}\mathrm{discriminant}\right)4{\left(2-k\right)}^2-4k^2\le 0\Rightarrow 4-4k\le 0\Rightarrow k\in [1,\infty )$