If $\mathrm{\alpha },\mathrm{\beta }$ are roots of the equation $2{\mathrm{x}}^2+6\mathrm{x}+\mathrm{b}=0(\mathrm{b}<0)$ , then $\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}$ is less than

We have,

$\mathrm{\alpha }+\mathrm{\beta }=-3 \text{and }\mathrm{\alpha \beta }=\frac{\mathrm{b}}{2}$

Since b <0, therefore discriminant D = 36-4b> 0. 
So,$\mathrm{\alpha } \text{and }\mathrm{\beta }$ are real. 
Now, 

$\begin{array}{rlll} & & \frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }}=\frac{(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }}{\mathrm{\alpha \beta }}=\frac{(\mathrm{\alpha }+\mathrm{\beta }{)}^2}{\mathrm{\alpha \beta }}-2& =\frac{18}{\mathrm{b}}-2\\ \Rightarrow & & \frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}<-2& [\because \mathrm{b}<0]\end{array}$