Explanation

(c)
Given expansion is $\left( \frac{a}{x} + bx \right)^{12}$
$\therefore$ General term, $T_{r + 1} = \ ^{12}C_{r}\left( \frac{a}{x} \right)^{12 - r}{(bx)}^{r}$
$= \ ^{12}C_{r}{(a)}^{12 - r}b^{r}x^{- 12 + 2r}$
For coefficient of $x^{- 10}$, put
$- 12 + 2r = - 10$
$\Rightarrow \ r = 1$
Now, the coefficient of $x^{- 10}$ is
$\ ^{12}C_{1}{(a)}^{11}{(b)}^{1} = 12a^{11}b$