LimitsHard
Question
If
[1 + ln(1 + b2)]1/x = 2b sin2 θ, b > 0 and θ ε (- π, π], then the value of θ is
[1 + ln(1 + b2)]1/x = 2b sin2 θ, b > 0 and θ ε (- π, π], then the value of θ is Options
A.

B.

C.

D.

Solution
eln(1 + b2) = 2b sin2θ
⇒ sin2 θ =
⇒ sin2 θ = 1 as
≥ 1

⇒ sin2 θ =
⇒ sin2 θ = 1 as
≥ 1
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