Question:
If \(\frac{\pi}{2} \le x \le \frac{3\pi}{2}\), then \(\sin^{-1}(\sin x)\) is equal to:
यदि \(\frac{\pi}{2} \le x \le \frac{3\pi}{2}\), तो \(\sin^{-1}(\sin x)\) किसके बराबर है?
Options:
- \(x\)
- \(-x\)
- \(\pi + x\)
- \(\pi – x\)
- \(x\)
- \(-x\)
- \(\pi + x\)
- \(\pi – x\)
Correct Answer: d)
Explanation:
The range of principal values of \(\sin^{-1}\theta\) is \([-\pi/2, \pi/2]\).
Given \(x \in [\pi/2, 3\pi/2]\), we can write:
\[\sin x = \sin(\pi – x)\]
Now check the interval for \(\pi – x\):
\[\frac{\pi}{2} \le x \le \frac{3\pi}{2} \implies -\frac{3\pi}{2} \le -x \le -\frac{\pi}{2} \implies -\frac{\pi}{2} \le \pi – x \le \frac{\pi}{2}\]
Since \(\pi – x\) lies in the principal value branch \([-\pi/2, \pi/2]\):
\[\sin^{-1}(\sin x) = \sin^{-1}(\sin(\pi – x)) = \pi – x\]
Hence, the correct option is (d).
\(\sin^{-1}\) की मुख्य शाखा \([-\pi/2, \pi/2]\) होती है।
\(\pi/2 \le x \le 3\pi/2\) होने पर, \(\pi – x \in [-\pi/2, \pi/2]\) होता है।
चूँकि \(\sin x = \sin(\pi – x)\) होता है:
\[\sin^{-1}(\sin x) = \sin^{-1}(\sin(\pi – x)) = \pi – x\]
अतः सही विकल्प (d) है।