Question:

<p><strong>Analysis:</strong> The domain of <math>f(x)</math> is determined by the intersection of the domains of all three inverse trigonometric functions:</p><ul><li><math>\sin^{-1} x</math> has domain <math>[-1, 1]</math></li><li><math>\tan^{-1} x</math> has domain <math>\mathbb{R}</math></li><li><math>\sec^{-1} x</math> has domain <math>(-\infty, -1] \cup [1, \infty)</math></li></ul><p>The intersection is <math>\{-1, 1\}</math>.</p><p><strong>At x = 1:</strong></p><ul><li><math>\sin^{-1}(1) = \frac{\pi}{2}</math></li><li><math>\tan^{-1}(1) = \frac{\pi}{4}</math></li><li><math>\sec^{-1}(1) = 0</math></li><li><math>f(1) = \frac{\pi}{2} + \frac{\pi}{4} + 0 = \frac{3\pi}{4}</math></li></ul><p><strong>At x = -1:</strong></p><ul><li><math>\sin^{-1}(-1) = -\frac{\pi}{2}</math></li><li><math>\tan^{-1}(-1) = -\frac{\pi}{4}</math></li><li><math>\sec^{-1}(-1) = \pi</math></li><li><math>f(-1) = -\frac{\pi}{2} - \frac{\pi}{4} + \pi = \frac{\pi}{4}</math></li></ul><p>Therefore, the range is <math>\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]</math>.</p><p>∴ Answer is <strong>(b)</strong>.</p>