$\{ x \in R:|x - 2|\,\, = {x^2}\} = $
$\{ -1, 2\}$
$\{1, 2\}$
$\{ -1, -2\}$
$\{1, -2\}$
The locus of the point $P=(a, b)$ where $a, b$ are real numbers such that the roots of $x^3+a x^2+b x+a=0$ are in arithmetic progression is
If ${x^2} + px + 1$ is a factor of the expression $a{x^3} + bx + c$, then
The sum of integral values of $a$ such that the equation $||x\ -2|\ -|3\ -x||\ =\ 2\ -a$ has a solution
The least integral value $\alpha $ of $x$ such that $\frac{{x - 5}}{{{x^2} + 5x - 14}} > 0$ , satisfies