Continuity and Differentiability
Limits Involving e
IIT-JEE 2011 Paper 2
Grade 12
Question:
If $\lim_{x \to 0} [1 + x \ln(1 + b^2)]^{1/x} = 2b \sin^2 \theta$, $b > 0$, $\theta \in (-\pi, \pi]$, then $b$ and $\theta$ satisfy:
(1) $b = 1, \theta = \pm \frac{\pi}{2}$
(2) $b = \frac{1}{\sqrt{2}}, \theta = \frac{\pi}{4}$
(3) $b = 2, \theta = \pm \frac{\pi}{3}$
(4) $b = 1, \theta = \frac{\pi}{6}$
Step-by-Step Solution
Key Concept: LHS $= e^{\ln(1+b^2)} = 1 + b^2$. So $1 + b^2 = 2b \sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1+b^2}{2b} \geq 1$ by AM-GM. Equality iff $b = 1$, giving $\sin^2 \theta = 1, \theta = \pm \pi/2$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)