Q95 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
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
- A$-12$
- B$\dfrac{1}{12}$
- C$\dfrac{1}{6}$
- Correct answerD$12$
Solution
For a function to be a probability density function, the total probability over its entire domain must be $1$.
$\int_{-\infty}^{\infty} f(x) dx = 1$
$\int_{0}^{1} kx^2(1-x) dx = 1$
$k \int_{0}^{1} (x^2 - x^3) dx = 1$
$k \left[ \dfrac{x^3}{3} - \dfrac{x^4}{4} \right]_{0}^{1} = 1$
$k \left( \dfrac{1}{3} - \dfrac{1}{4} \right) = 1$
$k \left( \dfrac{4 - 3}{12} \right) = 1$
$k \left( \dfrac{1}{12} \right) = 1$
$k = 12$
Answer: $12$
$\int_{-\infty}^{\infty} f(x) dx = 1$
$\int_{0}^{1} kx^2(1-x) dx = 1$
$k \int_{0}^{1} (x^2 - x^3) dx = 1$
$k \left[ \dfrac{x^3}{3} - \dfrac{x^4}{4} \right]_{0}^{1} = 1$
$k \left( \dfrac{1}{3} - \dfrac{1}{4} \right) = 1$
$k \left( \dfrac{4 - 3}{12} \right) = 1$
$k \left( \dfrac{1}{12} \right) = 1$
$k = 12$
Answer: $12$
Hence, the correct answer is option: 'D'.