Question

If \(\theta = \sin^{-1}x + \cos^{-1}x - \tan^{-1}x\), \(1 \le x < \infty\), then the smallest interval in which \(\theta\) lies is

• A. \(\frac{\pi}{2} \le \theta \le \frac{3\pi}{4}\)
• B. \(0 \le \theta \le \frac{\pi}{4}\)
• C. \(-\frac{\pi}{4} \le \theta \le 0\)
• D. \(\frac{\pi}{4} \le \theta \le \frac{\pi}{2}\)

Hint:

Remember the identity \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) for \(x \in [-1,1]\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use the identity \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) for \(x \in [-1,1]\). But here \(x \ge 1\), so careful with principal values.

Step 2: Detailed Explanation:
For \(x \ge 1\), \(\sin^{-1}x\) is not defined in reals. The domain is \(x \in [-1,1]\). However, the problem says \(1 \le x < \infty\). Possibly a misprint? For \(x=1\), \(\sin^{-1}1 = \pi/2\), \(\cos^{-1}1 = 0\), \(\tan^{-1}1 = \pi/4\). So \(\theta = \pi/2 + 0 - \pi/4 = \pi/4\). For \(x \to \infty\), \(\sin^{-1}x\) is undefined. So likely domain is \(x \ge 1\) but principal values are taken differently. For \(x > 1\), \(\sin^{-1}x\) and \(\cos^{-1}x\) are not defined. This suggests the intended domain is \(x \in [0,1]\) perhaps. If we take \(x \in [0,1]\), then \(\sin^{-1}x + \cos^{-1}x = \pi/2\), so \(\theta = \pi/2 - \tan^{-1}x\). As \(x\) goes from 0 to 1, \(\tan^{-1}x\) goes from 0 to \(\pi/4\), so \(\theta\) goes from \(\pi/2\) to \(\pi/4\). Thus \(\theta \in [\pi/4, \pi/2]\).

Step 3: Final Answer:
\(\frac{\pi}{4} \le \theta \le \frac{\pi}{2}\), which corresponds to option (D).