If the coefficient of $x^7$  in ${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$  equals the coefficient of $x^{-7}$  in ${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$ , then $a\text{and}b$  satisfy the relation

$={}^{11}{\text{C}}_r\frac{a^{11-r}}{b^r}.x^{22-2r-r}$

For $x^7$  coefficient $22-3r=7\Rightarrow r=5$

Coefficient of $x^7$  is $:{}^n{\text{C}}_5\frac{a^6}{b^5}$

$T_{r+1}={}^{11}{\text{C}}_r.{\left(ax\right)}^{11-r}.{\left(\frac{-1}{b{x}^2}\right)}^r={}^{11}{\text{C}}_r.\frac{a^{11-r}}{{\left(-b\right)}^r}.x^{11-3r}$

$x^{-7}$  coefficient is: $11-3r=-7\Rightarrow r=6$

Coefficient of $x^{-7}$  is: ${}^{11}{\text{C}}_6\frac{a^5}{b^6}$

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

${}^{11}{\text{C}}_5\frac{1}{b}={}^{11}{\text{C}}_5.a\left({}^{11}{\text{C}}_5={}^{11}{\text{C}}_6\right)$

$\frac{1}{b}=a$

$a.b=1$