If \[ \lim _{x \rightarrow 0}\left[1+x \ln \left(1+b^{2}\right)\right]^{\frac{1}{x}}=2 b \sin ^{2} \theta, b>0 \text { and } \theta \in(-\pi, \pi] \] then the value of $\theta$ is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Limits XII

x \to 0 lim [1 + x ln (1 + b^{2})]^{\frac{1}{x}} = 2 b sin^{2} θ, b > 0 and θ \in (- π, π]

then the value of

θ

Answer: (D)

The limit is of the form

1^{\infty}

. Using

lim_{x \to 0} (1 + x f (x))^{1/ x} = e^{f (0)}

x \to 0 lim [1 + x ln (1 + b^{2})]^{1/ x} = e^{l n (1 + b^{2})} = 1 + b^{2}

1 + b^{2} = 2 b sin^{2} θ

.

Rearranging:

sin^{2} θ = \frac{1 + b ^{2}}{2 b}

.

By AM-GM inequality,

\frac{1 + b ^{2}}{2 b} \geq \frac{2 b}{2 b} = 1

, with equality when

b = 1

.

Since

sin^{2} θ \leq 1

, we must have

sin^{2} θ = 1

, giving

θ = \pm \frac{π}{2}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option