If _ x arrow 0 [1+x (1+b^2)]^ 1 / x =2 b ^2 θ b>0 and θ ∈(-π, π] , then the value of θ is

Correct answer: D. ± (π)/(2) Here, _ x arrow 0 \ 1+x (1+b^2)\ ^ 1 / x aligned Given, [1^ ∞ from ] & e^ _ x arrow 0 \ x (1+b^2)\ · (1)/(x) & e^ (1+b^2) =(1+b)^2 & _ x arrow 0 \ 1+x (1+b^2)\ ^ 1 / x =2 b ^2 θ & (1+b^2)=2 b ^2 θ & ^2 θ=(1+b^2)/(2 b) (ii) aligned By A M ≥ G M , (b+(1)/(b))/(2) ≥(b · (1)/(b))^ 1 / 2 (b^2+1)/(2 b) ≥1 From Eqs. (ii) and (iii), we get ^2 θ= 1 θ=± (π)/(2) as θ ∈(-π, π]

JEE Advanced 2011Paper 2

Exam: JEE Advanced
Year: 2011
Paper: Paper 2
Subject: Mathematics
Chapter: Limits

JEE Advanced 2011 · Paper 2 · Mathematics · Limits

If _ x arrow 0 [1+x (1+b^2)]^ 1 / x =2 b ^2 θ b>0 and θ ∈(-π, π] ,...

\lim _{x \rightarrow 0}\left[1+x \log \left(1+b^2\right)\right]^{1 / x}=2 b \sin ^2 \theta b>0

and

\theta \in(-\pi, \pi]

, then the value of

\theta

Options

A $\pm \frac{\pi}{4}$
B $\pm \frac{\pi}{3}$
C $\pm \frac{\pi}{6}$
D $\pm \frac{\pi}{2}$

Correct answer

D. $\pm \frac{\pi}{2}$

Step-by-step solution

Here,

\lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}

\begin{aligned} \text{ Given, } \quad \left[1^{\infty} \text{\text{ from } }\right]\\ & \Rightarrow \quad e^{\lim _{x \rightarrow 0}\left\{x \log \left(1+b^2\right)\right\} \cdot \frac{1}{x}} \\ & \Rightarrow \quad e^{\log \left(1+b^2\right)}=(1+b)^2 \\ & \lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}=2 b \sin ^2 \theta\\ & \Rightarrow \quad\left(1+b^2\right)=2 b \sin ^2 \theta\\ & \therefore \quad \sin ^2 \theta=\frac{1+b^2}{2 b} \quad \ldots \text{ (ii) } \end{aligned}

By

A M \geq G M

, \frac{b+\frac{1}{b}}{2} \geq\left(b \cdot \frac{1}{b}\right)^{1 / 2} \Rightarrow \frac{b^2+1}{2 b} \geq1

From

Eqs. (ii) and (iii), we get \sin ^2 \theta=

\Rightarrow \quad \theta=\pm \frac{\pi}{2} \text{ as } \theta \in(-\pi, \pi]

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