If Ltx→0[1+xlog(1+b2)]1x=2bsin2θ, b>0 and θ∈(−π,π] then the values of θ is

G.P=eLtx→01x[xlog(1+b2)] (1∞) =1+b2 2bsin2θ=1+b2 sin2θ=1+b22b A.M≥G.M θ=±π2

If Ltx→0[1+xlog(1+b2)]1x=2bsin2θ, b>0 and θ∈(−π,π] then the values of θ is

G.P=eLtx→01x[xlog(1+b2)] (1∞) =1+b2 2bsin2θ=1+b2 sin2θ=1+b22b A.M≥G.M θ=±π2

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If $\underset{x \to 0}{L t} {[1 + x \log (1 + b^{2})]}^{\frac{1}{x}} = 2 b \sin^{2} θ,$ $b > 0$ and $θ \in (- π, π]$ then the values of $θ$ is

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$\pm \frac{π}{6}$

$\pm \frac{π}{5}$

$\pm \frac{π}{4}$

$\pm \frac{π}{2}$

answer is D.

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$G . P = e^{\underset{x \to 0}{L t} \frac{1}{x} [x \log (1 + b^{2})]} (1^{\infty}) = 1 + b^{2} 2 b \sin^{2} θ = 1 + b^{2} \sin^{2} θ = \frac{1 + b^{2}}{2 b} A . M \geq G . M θ = \pm \frac{π}{2}$

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If Ltx→0[1+xlog(1+b2)]1x=2bsin2θ, b>0 and θ∈(−π,π] then the values of θ is

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If $\underset{x \to 0}{L t} {[1 + x \log (1 + b^{2})]}^{\frac{1}{x}} = 2 b \sin^{2} θ,$ $b > 0$ and $θ \in (- π, π]$ then the values of $θ$ is