Question

If the following function is probability density function of r.v. X
\(f(x) = kx^2(1-x)\), for \(0 < x < 1 = 0\), otherwise, then the value of \(k\) is

• A. \(-12\)
• B. \(\frac{1}{12}\)
• C. \(\frac{1}{6}\)
• D. \(12\)

Hint:

A pdf must integrate to $1$.

Answer 1

The Correct Option is D

Step 1: Check positivity
$x^2(1-x)\ge0$ on $(0,1)$, so $k>0$. Normalising gives $k=12$.

Final Answer:
Option (D). \[ \boxed{\text{(D)}} \]