For \(\log_{[x-1]}\sin x\) to be defined:

• The base \([x-1]\) is an integer and must be positive and not 1, so \([x-1] \ge 2\).
• The argument must be positive: \(\sin x > 0\), so \(0 < \sin x \le 1\).

With base \(> 1\) and argument in \((0, 1]\), the log is \(\le 0\). It equals 0 when \(\sin x = 1\) and goes to \(-\infty\) as \(\sin x \to 0^+\).

Range = \((-\infty, 0]\)

Answer: (A)