If in the expansion of ${\left[a{x}^2+\frac{1}{bx}\right]}^{11}$the coefficient of x⁷ is equal to the coefficient of x^-7 in the expansion of ${\left[ax-\frac{1}{b{x}^2}\right]}^{11}$, then ab=

In the expansion of ${\left[a{x}^2+\frac{1}{bx}\right]}^{11}$,

$T_{r+1}=$${11}_{C_r}$${\left[a{x}^2\right]}^{11-r}$${\left[\frac{1}{bx}\right]}^r$

$\therefore T_{r+1}={11}_{C_r}\frac{a^{11-r}}{b^r}$$x^{22-3r}$

$\therefore 22-3r=7\Rightarrow 3r=15\Rightarrow r=5$

$\therefore$The coefficient of x⁷ in the expansion of 

${\left[a{x}^2+\frac{1}{bx}\right]}^{11}$is = ${11}_{C_5}\frac{a^6}{b^5} \to \left(1\right)$

In the expansion of ${\left[a{x}^2-\frac{1}{bx}\right]}^{11}$,

$T_{k+1}={11}_{C_k}$${\left[ax\right]}^{11-k}$${\left[-\frac{1}{b{x}^2}\right]}^k$

$\therefore T_{k+1}={11}_{C_k}$$\frac{a^{11-k}}{b^k}{(-1)}^k{x}^{11-3k}$

$\therefore 11-3k=-7\Rightarrow 3k=18\Rightarrow k=6$

$\therefore$The coefficient of x^-7 int he expansion of 

x^-7 is = ${11}_{C_6}\frac{a^5}{b^6} \to \left(2\right)$

from (1) & (2) ${11}_{C_5}\frac{a^6}{b^5}={11}_{C_6}\frac{a^5}{b^6}$

$\Rightarrow ab=1 \left[\sin ce {11}_{C_5}={11}_{C_6}\right]$