MCQ
$\bigg[\Big\{\big(-\frac{1}{3}\big)\Big\}^{-2}\bigg]^{-1}=$
- ✓$\frac{1}{81}$
- B$\frac{1}{9}$
- C$-\frac{1}{81}$
- D$-\frac{1}{9}$
$\bigg[\Big\{\big(-\frac{1}{3}\big)\Big\}^{-2}\bigg]^{-1}$
$\bigg[\Big\{\big(-\frac{1}{3}\big)^{2}\Big\}^{(-2)\times(-1)}\bigg]$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}\big]$
$=\Big(\frac{-1}{3}\Big)^{2\times2}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}\big]$
$=\Big(\frac{-1}{3}\Big)^4$
$=\frac{(-1)^4}{3^4}$ $\Big[\text{As, }\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{m}}=\frac{\text{x}^{\text{m}}}{\text{y}^{\text{m}}}\Big]$
$=\frac{1}{81}$
Hence, the correct alternative is option $(a)$.
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