If _ x 0 [1 + x (1 + b^2) ]^ 1 x = 2b ^2 , b > 0 and (- , ], then the value of θ is

JEE Advanced 2011 Mathematics question

[1 + x(1 + = 2bθ,\ b > 0 and θ (-π, π],

then the value of θ is

Answer: (D)

Using with : the limit is . So \sin^2b^2\sin^2\dfrac${1 +

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Question ID VK-025988