The number of solutions for the equation ${x^2} - 5|x| + \,6 = 0$ is

If $\alpha , \beta , \gamma$ are roots of equation $x^3 + qx -r = 0$ then the equation, whose roots are

$\left( {\beta \gamma  + \frac{1}{\alpha }} \right),\,\left( {\gamma \alpha  + \frac{1}{\beta }} \right),\,\left( {\alpha \beta  + \frac{1}{\gamma }} \right)$

Let $p$ and $q$ be two real numbers such that $p+q=$ 3 and $p^{4}+q^{4}=369$. Then $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$ is equal to

Let $x, y, z$ be positive reals. Which of the following implies $x=y=z$ ?

$I.$ $x^3+y^3+z^3=3 x y z$

$II.$ $x^3+y^2 z+y z^2=3 x y z$

$III.$ $x^3+y^2 z+z^2 x=3 x y z$

$IV.$ $(x+y+z)^3=27 x y z$

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 :

The number of integers $n$ for which $3 x^3-25 x+n=0$ has three real roots is