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Let $t_{1} = {(\sin^{- 1} x)}^{{\sin^{- 1}}_{x}}, t_{2} = {(\sin^{- 1} x)}^{{\cos^{- 1}}_{x}} x, t_{3} = {(\cos^{- 1} x)}^{\sin^{- 1} x},$ $t_{4} = {(\cos^{- 1} x)}^{\cos^{- 1} x}$
Match the items of column I with the items column II
Column-I
Column-II
(A) $x \in (0, \cos 1)$ (p) $t_{1} > t_{2} > t_{4} > t_{3}$
(B) $x \in [\cos 1, \frac{1}{\sqrt{2}}]$ (q) $t_{4} > t_{3} > t_{1} > t_{2}$
(C) $x \in [\frac{1}{\sqrt{2}}, \sin 1]$
(r) $t_{2} > t_{1} > t_{4} > t_{3}$
(D) $x \in (\sin 1, 1)$ (s) $t_{3} > t_{4} > t_{1} > t_{2}$

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$A \to R; B \to Q; \to C \to S; D \to P$

$A \to Q; B \to S; \to C \to R; D \to P$

$A \to S; B \to Q; \to C \to P; D \to R$

$A \to R; B \to Q; \to C \to P; D \to S$

answer is B.

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Detailed Solution

(i) when $x \in (0, \cos 1),$ then $\sin^{- 1} x \in (0, π / 2 - 1)$ and $\cos^{- 1} x \in (1, π / 2)$
(ii) when $x \in (\cos 1, 1 / \sqrt{2})$ , then $\sin^{- 1} x \in (π / 2 - 1, π / 4)$ and $\cos^{- 1} x \in (π / 4, 1)$
(iii) when $x \in (1 / \sqrt{2}, \sin 1),$ then $\sin^{- 1} x \in (1 / \sqrt{2}, 1)$ and $\cos^{- 1} x \in (π / 2 - 1, π / 4)$
(iv) when $x \in (\sin 1, 1),$ then $\sin^{- 1} x \in (1, π / 2)$ and $\cos^{- 1} x \in (0, π / 2 - 1)$

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