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 -
$2$
$3$
$4$
$6$
Let $f: R \rightarrow R$ be the function $f(x)=\left(x-a_1\right)\left(x-a_2\right)$ $+\left(x-a_2\right)\left(x-a_3\right)+\left(x-a_3\right)\left(x-a_1\right)$ with $a_1, a_2, a_3 \in R$.Then, $f(x) \geq 0$ if and only if
If $\alpha $ and $\beta $ are the roots of the quadratic equation, $x^2 + x\, sin\,\theta -2sin\,\theta = 0$, $\theta \in \left( {0,\frac{\pi }{2}} \right)$ then $\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}}$ is equal to
The number of solutions for the equation ${x^2} - 5|x| + \,6 = 0$ is
Consider the equation $(1+a+b)^2=3\left(1+a^2+b^{2})\right.$ where $a, b$ are real numbers. Then,
The number of integers $k$ for which the equation $x^3-27 x+k=0$ has at least two distinct integer roots is