Question

Mark the correct alternative in the following question:

$$\begin{gathered} {\left[{\left\{{\left(-\frac{1}{3}\right)}^2\right\}}^{-2}\right]}^{-1}= \\ \left(\mathrm{a}\right)\frac{1}{81}\left(\mathrm{b}\right)\frac{1}{9}\left(\mathrm{c}\right)-\frac{1}{81}\left(\mathrm{d}\right)-\frac{1}{9} \end{gathered}$$

Answer 1

$$\begin{gathered} {\left[{\left\{{\left(-\frac{1}{3}\right)}^2\right\}}^{-2}\right]}^{-1} \\ =\left[{\left\{{\left(\frac{-1}{3}\right)}^2\right\}}^{\left(-2\right)\times \left(-1\right)}\right]\left[\mathrm{As},{\left(x^m\right)}^n=x^{mn}\right] \\ ={\left(\frac{-1}{3}\right)}^{2\times 2}\left[\mathrm{As},{\left(x^m\right)}^n=x^{mn}\right] \\ ={\left(\frac{-1}{3}\right)}^4 \\ =\frac{{\left(-1\right)}^4}{3^4}\left[\mathrm{As},{\left(\frac{x}{y}\right)}^m=\frac{x^m}{y^m}\right] \\ =\frac{1}{81} \end{gathered}$$

Hence, the correct alternative is option (a).