If sin−1(sin(2x2+41+x2))<π−3 then x satisfies

Let t=2x2+41+x2 ⇒x2=4−tt−2⇒t∈(2,4]⇒sin−1sint 3 ⇒3<t≤4⇒|x|<1

If sin−1(sin(2x2+41+x2))<π−3 then x satisfies

Let t=2x2+41+x2 ⇒x2=4−tt−2⇒t∈(2,4]⇒sin−1sint 3 ⇒3<t≤4⇒|x|<1

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If $\sin^{- 1} (\sin (\frac{2 x^{2} + 4}{1 + x^{2}})) < π - 3$ then x satisfies

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$| x | \geq 1$

$| x | > 1$

$| x | \leq 1$

$| x | < 1$

answer is B.

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Let $t = \frac{2 x^{2} + 4}{1 + x^{2}} \Rightarrow x^{2} = \frac{4 - t}{t - 2} \Rightarrow t \in (2, 4]$

$\Rightarrow \sin^{- 1} \sin t < π - 3 \Rightarrow \sin^{- 1} \sin (π - t) < π - 3 \Rightarrow t > 3 \Rightarrow 3 < t \leq 4 \Rightarrow | x | < 1$

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If sin−1(sin(2x2+41+x2))<π−3 then x satisfies

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If $\sin^{- 1} (\sin (\frac{2 x^{2} + 4}{1 + x^{2}})) < π - 3$ then x satisfies