1.If limx→0[1+xlog(1+b2)]1/x=2bsin2θ\displaystyle\lim_{x\to0}[1+x\log(1+b^2)]^{1/x}=2b\sin^2\thetax→0lim[1+xlog(1+b2)]1/x=2bsin2θ, b>0b>0b>0 and θ∈(−π,π]\theta\in(-\pi,\pi]θ∈(−π,π], then the value of θ\thetaθ isa.±π/4\pm\pi/4±π/4b.±π/3\pm\pi/3±π/3c.±π/6\pm\pi/6±π/6d.±π/2\pm\pi/2±π/2Login to continueOnly logged in users canattempt or see the solution.
