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For each positive interger n, define `f_(n)(x)=` minimum `(x^(n)/(n!), ((1-x)^(n))/(n!))`, for `0 le x le 1`. Let `I_(n)= int_(0)^(1) f_(n) (x) dx, n ge 1`. Then `I_(n)=sum_(n=1)^(oo) I_(n)` is equal to -A. `2sqrt(e)-3`B. `2sqrt(e)-2`C. `2sqrt(e)-1`D. `2sqrt(e)` |
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Answer» Correct Answer - A `I_(n)=underset(0)overset(1//2)(int) x^(n)/(n!) dx +underset(1//2)overset(1)(int) ((1-x)^(n))/(n !) dx=1/((n+1)!) ((1/2)^(n+1)+(1/2)^(n+1))=((1/2)^(n))/((n+1)!)` `sum_(n=1)^(oo) I_(n)=((1//2)/(2!)+((1//2)^(2))/(3!)+....)=2 sqrt(e)-3` |
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