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Let a≤tan⁡−1x+cot⁡−1x+sin⁡−1x≤ba \leq \tan^{-1} x + \cot^{-1} x + \sin^{-1} x \leq b. If α\alpha and β\beta denote the minimum and maximum possible values of a and b respectively, then :

Solution

✅ Correct Option: 1

tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} for all real xx. So expression =π2+sin⁡−1x= \frac{\pi}{2} + \sin^{-1}x. Since sin⁡−1x∈[−π2,π2]\sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], expression ranges from 00 to π\pi. Hence α=0\alpha = 0, β=π\beta = \pi.

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