Let t1=(sin−1x)sin−1x,t2=(sin−1x)cos−1x,t3=(cos−1x)sin−1x,t4=(cos−1x)cos−1x Match the entries of list –I with the entries of list –II List – I List – II(I)x ∈ (0, cos 1)(P)t1>t2>t4>t3(II)x∈(cos1,12)(Q)t4>t3>t1>t2(III)x∈(12,sin1)(R)t2>t1>t4>t3(IV)x∈(sin1,1)(S)t3>t4>t1>t2 (T)t1>t4>t3>t2Which of the following is the only correct combination.

I → Q ; I I → S ; I I I → T ; I V → R

I → Q ; I I → S ; I I I → T ; I V → R

I → Q ; I I → S ; I I I → R ; I V → P

I → Q ; I I → S ; I I I → T ; I V → R

I → P ; I I → Q ; I I I → R ; I V → S

I → T ; I I → S ; I I I → R ; I V → P

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Let $t_{1} = {(\sin^{- 1} x)}^{\sin^{- 1} x}, t_{2} = {(\sin^{- 1} x)}^{\cos^{- 1} x}, t_{3} = {(\cos^{- 1} x)}^{\sin^{- 1} x}, t_{4} = {(\cos^{- 1} x)}^{\cos^{- 1} x}$
Match the entries of list –I with the entries of list –II

List – I List – II
(I) x $\in$ (0, cos 1) (P) $t_{1} > t_{2} > t_{4} > t_{3}$
(II) $x \in (\cos 1, \frac{1}{\sqrt{2}})$ (Q) $t_{4} > t_{3} > t_{1} > t_{2}$
(III) $x \in (\frac{1}{\sqrt{2}}, \sin 1)$ (R) $t_{2} > t_{1} > t_{4} > t_{3}$
(IV) $x \in (\sin 1, 1)$ (S) $t_{3} > t_{4} > t_{1} > t_{2}$
(T) $t_{1} > t_{4} > t_{3} > t_{2}$
Which of the following is the only correct combination.

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$I \to Q; I I \to S; I I I \to R; I V \to P$

$I \to Q; I I \to S; I I I \to T; I V \to R$

$I \to P; I I \to Q; I I I \to R; I V \to S$

$I \to T; I I \to S; I I I \to R; I V \to P$

answer is B.

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