A) \[a\]
B) \[{{a}^{2}}\]
C) \[1/a\]
D) \[{{a}^{2}}\]
Correct Answer: A
Solution :
Given, \[f(x)=\lambda {{e}^{-ax}},\] for \[0\le x<\infty \] and \[a>0\] \[\therefore \] \[\int_{0}^{\infty }{\lambda {{e}^{-ax}}=1}\] \[\Rightarrow \] \[\lambda \left[ \frac{{{e}^{-ax}}}{-a} \right]_{0}^{\infty }=1\] \[\Rightarrow \] \[\lambda \left[ 0+\frac{1}{a} \right]=1\] \[\Rightarrow \] \[\frac{\lambda }{a}=1\] \[\Rightarrow \] \[\lambda =a\]
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