The domain of the function \(f(x) = \sin^{-1}(3x + 1)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\left[-\displaystyle\frac{2}{3}, 0\right]\)
\(\sin^{-1} t\) is defined only for \(-1 \le t \le 1\). So the argument \(3x + 1\) must lie in \([-1, 1]\):
\(-1 \le 3x + 1 \le 1\)
Subtracting 1 throughout:
\(-2 \le 3x \le 0\)
Dividing by 3:
\( \displaystyle -\frac{2}{3} \le x \le 0 \)
Hence the domain is \(\left[-\frac{2}{3}, 0\right]\).