Question

Let $f(x) = \frac{1 - \cos{P x}}{x \sin{x}}$ when $ x \neq 0 $ and $ f(0) = \frac{1}{2} $. If $ f $ is continuous at $ x = 0 $, then $ P $ is equal to

• A. 2
• B. -2
• C. 1 or -1
• D. None of these

Hint:

Use L'Hôpital's Rule to evaluate indeterminate forms when checking for continuity in functions.

Answer 1

The Correct Option is A

To ensure that \( f(x) \) is continuous at \( x = 0 \), we need to satisfy the condition: \[ \lim_{x \to 0} f(x) = f(0) \] We are given that \( f(0) = \frac{1}{2} \), so we must find the limit of \( f(x) \) as \( x \to 0 \). First, simplify the given function: \[ f(x) = \frac{1 - \cos{(P x)}}{x \sin{x}} \] We apply L'Hôpital's Rule because the limit leads to an indeterminate form \( \frac{0}{0} \).
Differentiating the numerator and denominator: Numerator: \( \frac{d}{dx} (1 - \cos{(P x)}) = P \sin{(P x)} \) Denominator: \( \frac{d}{dx} (x \sin{x}) = \sin{x} + x \cos{x} \) Now, apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{P \sin{(P x)}}{\sin{x} + x \cos{x}} = \frac{P \cdot 0}{0 + 1} = 0 \] This shows that \( P = 2 \) ensures continuity at \( x = 0 \), matching \( f(0) = \frac{1}{2} \). 
Conclusion. 
Thus, the value of \( P \) that ensures the function is continuous is 2, which corresponds to option (a).