$$
\begin{aligned}
& T_{r+1} \text { in the expansion }\left[a x^2+\frac{1}{b x}\right]^{11}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{b x}\right)^r \\
& ={ }^{11} C_r(a)^{11-r}(b)^{-r}(x)^{22-2 r-r} \\
& \Rightarrow 22-3 r=7 \Rightarrow r=5 \\
& \therefore \text { coefficient of } x^7={ }^{11} C_5(a)^6(b)^{-5} ....(i)
\end{aligned}
$$
Again $T_{r+1}$ in the expansion $\left[a x-\frac{1}{b x^2}\right]^{11}={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{b x^2}\right)^r$ $={ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{11-\mathrm{r}}(-1)^{\mathrm{r}} \times(\mathrm{b})^{-\mathrm{r}}(\mathrm{x})^{-2 \mathrm{r}}(\mathrm{x})^{11-\mathrm{r}}$
Now $11-3 r=-7 \Rightarrow 3 r=18 \Rightarrow r=6$
$$
\begin{aligned}
& \therefore \text { coefficient of } x^{-7}={ }^{11} C_6 a^5 \times 1 \times(b)-6 \\
& \Rightarrow{ }^{11} C_5(a)^6(b)^{-5}={ }^{11} C_6 a^5 \times(b)^{-6} \\
& \Rightarrow a b=1
\end{aligned}
$$