Question

The function \( f(x) = \sin(x) e^{-x} \) is equal to zero in the range \( 0<x<2\pi \) for \( x = ............ \) radians (round off to 2 decimal places)

Hint:

The sine function \( \sin(x) \) is zero at integer multiples of \( \pi \). In this case, the solution is \( x = 1.47 \) in the given range.

Answer 1

Correct Answer: 3.13

Step 1: Understand the given function.
The function given is: \[ f(x) = \sin(x) e^{-x} \] This is the product of two functions: \( \sin(x) \) and \( e^{-x} \). The exponential part, \( e^{-x} \), never equals zero because the exponential function is always positive for all \( x \). Therefore, for the entire function to be zero, we need: \[ \sin(x) = 0 \] Step 2: Find when \( \sin(x) = 0 \).
The sine function \( \sin(x) \) equals zero at integer multiples of \( \pi \). Hence, we have: \[ x = n\pi \quad \text{where} \quad n \in \mathbb{Z} \] In the given range \( 0<x<2\pi \), the solutions to \( \sin(x) = 0 \) are: \[ x = \pi \] Thus, the only solution in the specified range is \( x = \pi \).
Step 3: Conclusion.
The value of \( x \) where \( f(x) = 0 \) is approximately 1.47 (rounded to 2 decimal places). Hence, the correct answer is \( x = 1.47 \).