If a random variable X has p.d.f. f (x)= cases a x^2 2 + b x & , if 1 x 3 0 & , otherwise cases and f(2)=2 , then the values of a and b are, respectively

Step-by-step solution

The probability density function is f(x) = cases a x^2 2 + b x & if 1 x 3 0 & otherwise cases with conditions f(2) = 2 and ₁^3 f(x) dx = 1 . Substituting x = 2 yields 4a 2 + 2b = 2 , simplifying to a + b = 1 . The integral becomes ₁^3 ( a x^2 2 + b x ) dx = [ a x^3 6 + b x^2 2 ]₁^3 = ( 27a 6 + 9b 2 ) - ( a 6 + b 2 ) = 13a 3 + 4b = 1 . Clearing fractions gives 13a + 12b = 3 . Solving the system a + b = 1 and 13a + 12b = 3 , substitute b = 1 - a to obtain 13a + 12(1 - a) = a + 12 = 3 , so a = -9 and b = 10 . These va