Question

For what value of the constant \(k\), the function \[ f(x)= \begin{cases} kxe^{-2x}, & x>0 \\ 0, & \text{otherwise} \end{cases} \] is a probability density function?

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For PDF questions: \[ \boxed{\int_0^\infty x^n e^{-ax} dx = \frac{n!}{a^{n+1}}} \] Always ensure the total area under the curve equals 1.

Answer 1

The Correct Option is B

Concept: A function is a probability density function (PDF) if it satisfies: \[ \int_{-\infty}^{\infty} f(x)\,dx = 1 \quad \text{and} \quad f(x)\ge 0. \] Here, \[ f(x)=kxe^{-2x}, \quad x>0. \] So we only integrate from \(0\) to \(\infty\).

Step 1: Apply normalization condition.
\[ \int_0^\infty kxe^{-2x} dx = 1 \] Take \(k\) outside: \[ k \int_0^\infty xe^{-2x} dx = 1 \]

Step 2: Evaluate the integral.
Use standard result: \[ \int_0^\infty xe^{-ax} dx = \frac{1}{a^2} \] Here \(a=2\), so: \[ \int_0^\infty xe^{-2x} dx = \frac{1}{4} \]

Step 3: Find \(k\).
\[ k \cdot \frac{1}{4} = 1 \] \[ k = 4 \]

Step 4: Choose correct option.
\[ \boxed{k=4} \Rightarrow Option (D) \] Final Answer: \[ \boxed{k=4} \]