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
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
\(f(x) = kx^2(1-x)\), for \(0 < x < 1 = 0\), otherwise, then the value of \(k\) is
Show Hint
A pdf must integrate to $1$.
Updated On: Oct 1, 2026
- \(-12\)
- \(\frac{1}{12}\)
- \(\frac{1}{6}\)
- \(12\)
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The Correct Option is D
Solution and Explanation
Step 1: Set the total to 1
\(\int_0^1kx^2(1-x)\,dx=1\).
Step 2: Integrate
\(k\left(\frac13-\frac14\right)=\frac{k}{12}=1\), so \(k=12\). Option (D).
Final Answer:
\(k=12\), option (D). \[ \boxed{\text{(D) }12} \]
\(\int_0^1kx^2(1-x)\,dx=1\).
Step 2: Integrate
\(k\left(\frac13-\frac14\right)=\frac{k}{12}=1\), so \(k=12\). Option (D).
Final Answer:
\(k=12\), option (D). \[ \boxed{\text{(D) }12} \]
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