The domain of the function f(x) = sin^-1≤ft(log_2 x) is

The correct answer is Option (1) → \(\left[\displaystyle\frac{1}{2}, 2\right]\)

Two conditions are needed: \(\log_{2} x\) must be defined, so \(x \gt 0\), and the argument of \(\sin^{-1}\) must lie in \([-1, 1]\):

    \(-1 \le \log_{2} x \le 1\)

Since \(\log_{2}\) is an increasing function, the inequality keeps its direction when we raise 2 to each part:

    \(2^{-1} \le x \le 2^{1}\)

    \( \displaystyle \frac{1}{2} \le x \le 2 \)

All these values are positive, so the first condition holds too. Hence the domain is \(\left[\frac{1}{2}, 2\right]\).