The number of real solutions of the equation $$3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$$, is

The equation $${x^2} - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has :

If $$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$$, then the maximum value of $$\mathrm{a}$$ is :

$$ \text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $$

$$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. } $$

Then the number of elements in $$\mathrm{S} \cap \mathrm{T}$$ is :