Question

Determine the values of \(a\) and \(b\) for which the function \(f(x)\) defined as: \[ f(x) = \begin{cases} 1 + |\sin x|^{\left(\frac{a}{|\sin x|}\right)} & \text{if } \frac{-\pi}{6} < x < 0, \\ b & \text{if } x = 0, \\ e^{\left(\frac{\tan 2x}{\tan 3x}\right)} & \text{if } 0 < x < \frac{\pi}{6} \end{cases} \] is continuous at \(x = 0\).

• A. \( a = 1, b = \frac{2}{3} \)
• B. \( a = \frac{2}{3}, b = e^{\frac{2}{3}} \)
• C. \( a = \frac{2}{3}, b = \frac{3}{2} \)
• D. \( a = -1, b = e^{\frac{2}{3}} \)

Hint:

To ensure a piecewise function is continuous, match the limits from the left and right at each piece's boundary with the function's value at that point.

Answer 1

The Correct Option is B

Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0 from the left: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( 1 + \frac{1 + \sin(|x|/\pi)}{x} \right). \] Using the series expansion of \( \sin x \) and considering the behavior around \( x = 0 \), this limit evaluates to 1. 
Step 2: Find the limit of \( f(x) \) as \( x \) approaches 0 from the right: \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} a \tan\left( \frac{2\pi x}{3} \right). \] Since \( \tan\left( \frac{2\pi x}{3} \right) \) near 0 approximates as \( \frac{2\pi x}{3} \), the limit becomes \( \frac{2\pi a}{3} \cdot 0 = 0 \). Set \( a = \frac{2}{3} \) for continuity. 
Step 3: For continuity at \( x = 0 \), \( f(0) = b \) must equal the limits from both sides, which have been found to be 0. Thus, \( b = e^{\frac{2}{3}} \) ensures \( f(x) \) is continuous at \( x = 0 \).