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Mathematics Question on limits and derivatives

If limx→0[1+x log (1+b2)]1x=2bsin2 θ,b>0 and θ∈(−π,π),lim_{ x \to 0 } [ 1 + x \, log \, (1 + b^2) ]^{\frac{1}{x}} = 2 b sin^2 \, \theta, b > 0 \, and \, \theta \in ( - \pi , \pi), then the value of θ\theta is

A

±π4 \pm \frac{\pi}{4}

B

±π3 \pm \frac{\pi}{3}

C

±π6 \pm \frac{\pi}{6}

D

±π2 \pm \frac{\pi}{2}

Answer

±π2 \pm \frac{\pi}{2}

Explanation

Solution

Here, limx→01+x log(1+b2)1/x lim_{ x \to 0 } \\{ 1 + x \, log (1 + b^2) \\}^{1/x} = limx→0x log(1+b2).1xlim_{ x \to 0 } \\{ x \, log (1 + b^2) \\} . \frac{1}{x} = elog(1+b2)=(1+b2)e^{log (1 + b^2)} = (1 + b^2) Given, limx→01+x log(1+b2)1/x=2b sin2θlim_{ x \to 0 } \\{1 + x \, log (1 + b^2 ) \\}^{1/x} = 2b \, sin^2 \theta ⇒(1+b2)=2b sin2θ\Rightarrow (1 + b^2) = 2b \, sin^2 \theta ∴\therefore \hspace16mm sin2θ=1+b22bsin^2 \theta = \frac{1 + b^2}{2b} By AM ≥\ge GM, b+1b2≥(b.1b)1/2 \frac{ b + \frac{1}{b}}{ 2} \ge \bigg(b. \frac{1}{b}\bigg)^{1/2} ⇒b2+12b≥1\Rightarrow \frac{b^2 + 1}{2b} \ge 1 From Eqs. (ii) and (iii), \hspace16mm sin2θ=1⇒=±π2, as θ ∈ (−π,π]sin^2 \theta = 1 \Rightarrow = \pm \frac{\pi}{2}, \, as \, \theta \, \in \, (- \pi, \pi]