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

The correct answer is Option (4) → \([0, \infty)\)

We need \(x \ne -1\) and \(-1 \le \frac{x - 1}{x + 1} \le 1\). The two sides are taken one at a time, without multiplying by \(x + 1\), whose sign is not known.

Right side:

    \( \displaystyle \frac{x - 1}{x + 1} - 1 \le 0 \)

    \( \displaystyle \frac{-2}{x + 1} \le 0 \)

This holds exactly when \(x + 1 \gt 0\), that is, \(x \gt -1\).

Left side:

    \( \displaystyle \frac{x - 1}{x + 1} + 1 \ge 0 \)

    \( \displaystyle \frac{2x}{x + 1} \ge 0 \)

With \(x \gt -1\) the denominator is positive, so this needs \(2x \ge 0\):

    \(x \ge 0\)

Both conditions together give \(x \ge 0\). Hence the domain is \([0, \infty)\).

Why the other options are incorrect

• Option (1) \((-1, \infty)\) — Incorrect. At \(x = -\frac{1}{2}\) the argument is \(-\frac{3}{2} \div \frac{1}{2} = -3\), which is less than \(-1\).

• Option (2) \([0, 1]\) — Incorrect. It misses values such as \(x = 3\), where the argument is \(\frac{2}{4} = \frac{1}{2}\) and \(\sin^{-1}\frac{1}{2} = \frac{\pi}{6}\) is defined.

• Option (3) \(\mathrm{R} - \{-1\}\) — Incorrect. At \(x = -3\) the argument is \(\frac{-4}{-2} = 2\), which is greater than 1.