Answer:
Option A
Explanation:
The general term in the expansion of $\left( ax^{2}+\frac{1}{bx}\right)^{13}$
$T_{ (p+1)}=^{13}C_{p}(ax^{2})^{13-p}\frac{1}{(bx)^{p}}$
= $^{13}C_{p}\frac{a^{13-p}}{b^{p}}x^{26-3p}$
For, $x^{5}$, p=7
So, coefficient of $x^{5}$ in the expansion of
$\left(ax^{2}+\frac{1}{bx}\right)^{13} $ is $^{13}C_{7}\frac{a^{6}}{b^{7}}$ [$\because$ p=7]
Similarly, the general term in the expansion of $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$
$T_{q+1}=^{13}C_{q}(ax)^{13-q}\left(-\frac{1}{bx^{2}}\right)^{q}$
$=^{13}C_{q}(-1)^{q}\frac{a^{13-q}}{b^{q}}x^{13-q}$
for $x^{-5}, q=6$
So, coefficient of $x^{-5}$ in the expansion of $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$ is
$^{13}C_{6}\frac{a^{7}}{b^{6}}$
According to the question,
$^{13}C_{7}\frac{a^{6}}{b^{7}}=^{13}C_{6}\frac{a^{7}}{b^{6}}$
$\Rightarrow$ ab=1
