Which of the following is/are TRUE ?

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The equation $\sin^{- 1} (| \cos x |) - \cos^{- 1} (| \sin x |) = k,$ has at least one solution then the number of integers in range of k is 1

The equation $\sin^{- 1} (| \cos x |) - \cos^{- 1} (| \sin x |) = k,$ has at least one solution iff $k \in (0, π)$

If $\sin^{- 1} (x - 1) + \cos^{- 1} (x - 3) + \tan^{- 1} (\frac{x}{2 - x^{2}}) = \cos^{- 1} k + π,$ then the value of $k = \frac{1}{\sqrt{2}}$

$α_{1}$ and $α_{2}$ satisfies $\sin^{- 1} (\frac{2 x}{1 + x^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ and $| α_{1} - α_{2} | < k,$ for all $α_{1}$ and $α_{2} (k \in I)$ , then minimum value of k is equal to 2

answer is A, C, D.

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Detailed Solution

$(a) \sin^{- 1} (| \cos x |) - \cos^{- 1} (| \sin x |) = a \frac{π}{2} - \cos^{- 1} (| \cos x |) - (\frac{π}{2} - \sin^{- 1} (| \sin x |)) = a \Rightarrow \sin^{- 1} (| \sin x |) - \cos^{- 1} (| \cos x |) = a \Rightarrow a = 0, x \in (0, \frac{π}{2}) 0, x \in (\frac{π}{2}, π) a = {0}$

$(b) \sin^{- 1} (x - 1) + \cos^{- 1} (x - 3) + \tan^{- 1} (\frac{x}{2 - x^{2}}) = \cos^{- 1} (K) + π - 1 \leq x - 1 \leq 1 \Rightarrow 0 \leq x \leq 2 - 1 \leq x - 3 \leq 1 \Rightarrow 2 \leq x \leq 4 x = 2 \sin^{- 1} (1) + \cos^{- 1} (- 1) + \tan^{- 1} (\frac{2}{- 2}) = \frac{π}{2} + π - \frac{π}{4} = \frac{π}{2} + \frac{3 π}{4} = π + \frac{π}{4} \Rightarrow k = \frac{1}{\sqrt{2}}$

$(d) \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = \tan^{- 1} (\frac{2 x}{1 + x^{2}}) x = \tan θ \Rightarrow \sin^{- 1} (\sin 2 θ) = \tan^{- 1} (\tan 20) \Rightarrow 2 θ = 2 θ i f (2 θ \in (- \frac{π}{2}, \frac{π}{2}) \Rightarrow x \in (- 1, 1)) | α_{1} - α_{2} | < 2$