If $|{x^2} - x - 6| = x + 2$, then the values of $x$ are
$-2, 2, -4$
$-2, 2, 4$
$3, 2, -2$
$4, 4, 3$
The solution of the equation $2{x^2} + 3x - 9 \le 0$ is given by
The maximum value $M$ of $3^x+5^x-9^x+15^x-25^x$, as $x$ varies over reals, satisfies
The equation${e^x} - x - 1 = 0$ has
The set of all real numbers $x$ for which ${x^2} - |x + 2| + x > 0,$ is
Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes