GATE (TF) Textile 2021 Question Paper Solution | GATE/2021/TF/02

If a continuous random variable $X$ has the following probability density function

$g(x)=\begin{cases} \frac{k}{4}x(2-x), \ \ 0<x<2 \\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise \end{cases}$

then the value of k is

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Given-

Probability density function=1

$\int_{-\infty}^{+\infty} \times \frac{K}{4} \times x(2-x)dx=1$

$\int_{-\infty}^{0} \frac{K}{4} \times x(2-x)dx+\int_{0}^{2} \frac{K}{4} \times x(2-x)dx+\int_{2}^{+\infty} \frac{K}{4} \times x(2-x)dx=1$

$0+\int_{0}^{2} \times \frac{K}{4} \times x(2-x)dx+0=1$

$\int_{0}^{2} \times \frac{K}{4} \times x(2-x)dx=1$

$\frac{K}{4}\times \int_{0}^{2} x(2-x)dx=1$

$\frac{K}{4}\times \int_{0}^{2} (2x-x^2)dx=1$

$\frac{K}{4}\times (\int_{0}^{2} 2xdx-\int_{0}^{2}x^2dx)=1$

$\frac{K}{4} \left ( \left [ x^2 \right ]_0^2 -\frac{1}{3}\left [ x^3 \right ]_0^2 \right )=1$

$\frac{K}{4} \left ( (4-0) - \frac{1}{3}(8-0)\right ) =1$

$\frac{K}{4}\left ( 4-\frac{8}{3} \right ) =1$

$\frac{K}{4}\left ( \frac{4}{3} \right ) =1$

$K = 3$