Correct option is (A) All real numbers
\(\frac{2^{x -1} -1}{2^{x + 1} + 1} < 2\)
⇒ \(\frac{2^{x -1} -1}{2^{x + 1} + 1} -2 < 0\)
⇒ \(\frac{2^{x -1} - 1 -2.2^{x + 1} -2}{2^{x +1} + 1}< 0\)
⇒ \(2^{x-1}-2^{x +2}- 3< 0\) (∵ \(2^{x+1}+1 > 0\) (always))
⇒ \(\frac{2^x}2 - 4.2^x - 3< 0\)
⇒ \(\frac{-7}2.2^x - 3<0\)
⇒ \(\frac 72 . 2^x > -3\)
⇒ \(2^x > \frac{-6}7\)
which is true for all real values of x.