Correct Answer: $\left[\left(\frac{11}{13}\right),\left(\frac{47}{55}\right)\right]$

Solution : Given:
$\left[\left(\frac{7}{9}\right),\left(\frac{31}{39}\right)\right];\left[\left(\frac{3}{5}\right),\left(\frac{15}{23}\right)\right]$

In the given sets, multiply both the numerator and the denominator by 4, and then add 3.
$\left[\left(\frac{7}{9}\right),\left(\frac{31}{39}\right)\right]$→$\left[\left(\frac{(7 × 4) + 3}{(9 × 4) + 3}\right) = (\frac{28 + 3}{36 + 3}) =\left(\frac{31}{39}\right)\right]$
$\left[\left(\frac{3}{5}\right),\left(\frac{15}{23}\right)\right]$→$\left[\left(\frac{(3 × 4) + 3}{(5 × 4) + 3}\right) = (\frac{12 + 3}{20 + 3}) =\left(\frac{15}{23}\right)\right]$
Let's check the options –
First option: $\left[\left(\frac{11}{13}\right),\left(\frac{47}{55}\right)\right]$→$\left[\left(\frac{(11 × 4) + 3}{(13 × 4) + 3}\right) = (\frac{44 + 3}{52 + 3}) =\left(\frac{47}{55}\right)\right]$
Second option: $\left[\left(\frac{9}{13}\right),\left(\frac{37}{55}\right)\right]$→$\left[\left(\frac{(9 × 4) + 3}{(13 × 4) + 3}\right) = (\frac{36 + 3}{52 + 3}) =\left(\frac{39}{55})\neq(\frac{37}{55}\right)\right]$
Third option: $\left[\left(\frac{9}{11}\right),\left(\frac{32}{37}\right)\right]$ →$\left[\left(\frac{(9 × 4) + 3}{(11 × 4) + 3}\right) = (\frac{36 + 3}{44 + 3}) =\left(\frac{39}{47})\neq(\frac{32}{37}\right)\right]$
Fourth option: $\left[\left(\frac{17}{19}\right),\left(\frac{36}{77}\right)\right]$ →$\left[\left(\frac{(17 × 4) + 3}{(19 × 4) + 3}\right) = (\frac{68 + 3}{76 + 3}) =\left(\frac{71}{79})\neq(\frac{36}{77}\right)\right]$

So, only the first option follows the same pattern as followed by the given set of numbers. Hence, the first option is correct.