as f(x) is onto hence A is range of f(x) $\begin{aligned}
& \text { now } \begin{array}{l}
\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^2-30 \mathrm{x}+36 \
\quad=6(\mathrm{x}-2)(\mathrm{x}-3) \
\quad \mathrm{f}(2)=16-60+72+7=35 \
\mathrm{f}(3)=54-135+108+7=34 \
\mathrm{f}(0)=7
\end{array}
\end{aligned}hencerange\in[7,35]=\mathrm{A}alsoforrangeofg(x)\begin{aligned}
& g(x)=1-\frac{1}{x^{2025}+1} \in[0,1)=B \
& s={0,7,8, \ldots .35} \text { hence } n(s)=30
\end{aligned}$