Let $f(x)=2 x^{3}+15$ and $g(x)=9 x^{2}-12 x$ then $\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}$

$\therefore \mathrm{f}(\mathrm{x})$ is increasing function $\forall \mathrm{x} \in \mathrm{R}$

$$

\text { Also, } g^{\prime}(x)>0 \Rightarrow 18 x-12>0 \Rightarrow x>\frac{2}{3}

$$

Thus, $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ both increases for $\mathrm{x}>\frac{2}{3}$

$$

\text { Let } F(x)=f(x)-g(x), F^{\prime}(x)<0

$$

$(\because \mathrm{f}(\mathrm{x})$ increases less rapidly than the function $\mathrm{g}(\mathrm{x}))$

$$

\Rightarrow 6 \mathrm{x}^{2}-18 \mathrm{x}+12<0 \Rightarrow 1<\mathrm{x}<2

$$