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Question

Find the value of λ such that the function f(x) is a valid probability density function.
f(x)=λ(x1)(2x) for 1x2
=0 otherwise


  1. 6

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Solution

The correct option is A 6
  1. probability of any function within; its whole domain is one. Thus, 21f(x)dx=1

    where,

    f(x)=λ(x1)(2x)

    λ21(x1)(2x)dx=1

    λ[212xx22+x)dx]=1

    λ[21x2dx+213xdx212dx]=1

    λ[(231)3+32(221)2(21)]=1

    λ[73+922]=1

    λ=6

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