Question

Mark the correct alternative in the following question:

$$\begin{gathered} \frac{{\left(144\right)}^{{\displaystyle \frac{1}{2}}}+{\left(256\right)}^{{\displaystyle \frac{1}{2}}}}{3^2-2}= \\ \left(\mathrm{a}\right)8\left(\mathrm{b}\right)4\left(\mathrm{c}\right)-4\left(\mathrm{d}\right)-8 \end{gathered}$$

Answer 1

Since,

$$\begin{gathered} \frac{{\left(144\right)}^{{\displaystyle \frac{1}{2}}}+{\left(256\right)}^{{\displaystyle \frac{1}{2}}}}{3^2-2} \\ =\frac{{\left({12}^2\right)}^{{\displaystyle \frac{1}{2}}}+{\left({16}^2\right)}^{{\displaystyle \frac{1}{2}}}}{9-2} \\ =\frac{{12}^{2\times {\displaystyle \frac{1}{2}}}+{16}^{{\displaystyle 2\times \frac{1}{2}}}}{7}\left[\mathrm{As},{\left(x^m\right)}^n=x^{mn}\right] \\ =\frac{12+16}{7} \\ =\frac{28}{7} \\ =4 \end{gathered}$$

Hence, the correct alternative is option (b).