Question

Mark the correct alternative in the following question:

$$\begin{gathered} {\left(6^{-1}-8^{-1}\right)}^{-1}= \\ \left(\mathrm{a}\right)\frac{1}{24}\left(\mathrm{b}\right)24\left(\mathrm{c}\right)-24\left(\mathrm{d}\right)-\frac{1}{24} \end{gathered}$$

Answer 1

$$\begin{gathered} {\left(6^{-1}-8^{-1}\right)}^{-1} \\ ={\left(\frac{1}{6}-\frac{1}{8}\right)}^{-1}\left(\mathrm{As},x^{-1}=\frac{1}{x}\right) \\ ={\left(\frac{4-3}{24}\right)}^{-1} \\ ={\left(\frac{1}{24}\right)}^{-1} \\ =\frac{24}{1}\left(\mathrm{As},x^{-1}=\frac{1}{x}\right) \\ =24 \end{gathered}$$

Hence, the correct alternative is option (b).