If the equation $\frac{1}{x} + \frac{1}{{x - 1}} + \frac{1}{{x - 2}} = 3{x^3}$ has $k$ real roots, then $k$ is equal to -

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

The sum of the roots of the equation, ${x^2}\, + \,\left| {2x - 3} \right|\, - \,4\, = \,0,$ is

If $|{x^2} - x - 6| = x + 2$, then the values of $x$ are

Let $\mathrm{S}=\left\{x \in R:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is :

If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be