Let x*y = x2 + y3 and (x*1)*1 = x*(1*1). Then a value of 2sin-1x4 + x2 - 2x4 + x2 + 2 is

(x*1)*1 = x*(1*1) (x2 + 1)*1 = x*(2) (x2 + 1)2 + 1 = x2 + 8 x4 + x2 - 6 = 0⇒x2=2 ⇒2sin-1x4 + x2 - 2x4 + x2 + 2 = 2sin-148 =2sin-1 12=π3

Let x*y = x2 + y3 and (x*1)*1 = x*(1*1). Then a value of 2sin-1x4 + x2 - 2x4 + x2 + 2 is

(x*1)*1 = x*(1*1) (x2 + 1)*1 = x*(2) (x2 + 1)2 + 1 = x2 + 8 x4 + x2 - 6 = 0⇒x2=2 ⇒2sin-1x4 + x2 - 2x4 + x2 + 2 = 2sin-148 =2sin-1 12=π3

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$L e t x * y = x^{2} + y^{3} a n d (x * 1) * 1 = x * (1 * 1) . T h e n a v a l u e o f 2 \sin^{- 1} (\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}) i s$

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$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{4}$

answer is B.

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Detailed Solution

$(x * 1) * 1 = x * (1 * 1) (x^{2} + 1) * 1 = x * (2) {(x^{2} + 1)}^{2} + 1 = x^{2} + 8 x^{4} + x^{2} - 6 = 0 \Rightarrow x^{2} = 2 \Rightarrow 2 \sin^{- 1} (\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}) = 2 \sin^{- 1} |\frac{4}{8}| = 2 \sin^{- 1} (\frac{1}{2}) = \frac{π}{3}$