Consider the function $f (x) = \sin^{- 1} x + \cos^{- 1} x + \sec^{- 1} 2 x$ then range of the function f(x) is

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$[\frac{π}{2}, \frac{7 π}{6}]$

$[\frac{π}{2}, \frac{7 π}{6}] \cup [\frac{4 π}{3}, \frac{3 π}{2}]$

$[\frac{π}{2}, \frac{5 π}{6}] \cup [\frac{7 π}{6}, \frac{3 π}{2}]$

$[\frac{π}{2}, \frac{3 π}{2}]$

answer is A.

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Detailed Solution

$domain - 1 \leq x \leq 1 \cap 2 x \leq - 1 \cup 2 x \geq 1 \Rightarrow x \in [1, - \frac{1}{2}] \cup [\frac{1}{2}, 1]$

$r (- 1) = \frac{π}{2} + \sec^{- 1} (- 2) = \frac{π}{2} + \frac{2 π}{3} = \frac{7 π}{6}$ $\begin{aligned} f (\frac{- 1}{2}) = \frac{π}{2} + \sec^{- 1} (- 1) = \frac{π}{2} + π = \frac{3 π}{2} \\ f (\frac{1}{2}) = \frac{π}{2} + \sec^{- 1} (1) = \frac{π}{2} + 0 = \frac{π}{2} \\ f (1) = \frac{π}{2} + \sec^{- 1} (2) = \frac{π}{2} + \frac{π}{3} = \frac{5 π}{6} \\ Range = [\frac{π}{2}, \frac{5 π}{6}] \cup [\frac{7 π}{6}, \frac{3 π}{2}] \end{aligned}$