Question

Find the value of $\lambda$ such that the function f(x) is a valid probability density function.
$f\left(x\right)=\lambda (x-1)(2-x)\text{for}1\le x\le 2$
=0 otherwise

• A. 6

Answer 1

The correct option is A 6

1. probability of any function within; its whole domain is one. Thus, ${\int }_1^{2}f\left(x\right)dx=1$
   
   where,
   
   $f\left(x\right)=\lambda (x-1)(2-x)$
   
   $\Rightarrow \lambda {\int }_1^2(x-1)(2-x)dx=1$
   
   $\Rightarrow \lambda [{\int }_1^{2}2x-x^2-2+x)dx]=1$
   
   $\Rightarrow \lambda [-{\int }_1^2{x}^{2}dx+{\int }_1^{2}3xdx-{\int }_1^{2}2dx]=1$
   
   $\Rightarrow \lambda [\frac{(2^3-1)}{3}+\frac{3}{2}(2^2-1)-2(2-1\left)\right]=1$
   
   $\Rightarrow \lambda [-\frac{7}{3}+\frac{9}{2}-2]=1$
   
   $\Rightarrow$ $\lambda =6$