If $$f(x)=kx(1-x),0<x<1$$

=0        otherwise

is the probability density function of certain distribution, then value of k is

Correct option is B

​$$\textbf{Solution: }\\ f(x) = \begin{cases}kx(1 - x), & 0 < x < 1 \\0, & \text{otherwise}\end{cases} \\[8pt]\text{Since } f(x) \text{ is a pdf, } \int_{-\infty}^{\infty} f(x)\,dx = 1 \\[8pt]\Rightarrow \int_0^1 kx(1 - x)\,dx = 1 \\[8pt]= k \int_0^1 (x - x^2)\,dx = k \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 \\[8pt]= k \left( \frac{1}{2} - \frac{1}{3} \right) = k \cdot \frac{1}{6} \\[8pt]\Rightarrow \frac{k}{6} = 1 \Rightarrow k = 6 \\[8pt]\boxed{k = 6}$$​​