$\left(a x^2+\frac{1}{2 b x}\right)^{11}$

$T _{ r +1}={ }^{11} C _{ r }\left( ax ^2\right)^{11- r } \cdot\left(\frac{1}{2 bx }\right)^{ r }$

$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{1}{2 b }\right)^{ I } \cdot x ^{22-2 r – r }={ }^{11} C _{ r } a^{11- r } \cdot\left(\frac{1}{2 b }\right)^{ I } \cdot x ^{22-3 r }$

$\therefore 22-3 r=7$

$3 r =15$

$r=5$

Again $\left(a x-\frac{1}{3 b x^2}\right)^{11}$

$T _{ r +1}={ }^{11} C _{ r }( ax )^{11- r }\left(-\frac{1}{3 b x^2}\right)^{ r }$

$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{-1}{3 b }\right)^{ r } \cdot x ^{11- r -2 r }$ $\therefore 11-3 r=-7$

$3 r =18$

$r=6$

$\text { Now, } \frac{{ }^{11} C _5 a ^6}{32 b^5}=\frac{{ }^{11} C _6 \cdot a ^5}{3^6 \cdot b ^6}$

$729 ab =32$