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1)

For each positive integer n,  let yn=1n((n+1)(n+2)...(n+n))1n.. For x ε R, let [x] be the greatest integer less than or equal to x. If limnyn=L, then the value of [L] is ...........


A) 2

B) 5

C) 3

D) 1

Answer:

Option D

Explanation:

We have 

       yn=1n((n+1)(n+2)...(n+n))1n    and   

                                                                                                           limnyn=L

       L=  limn1n[(n+1)(n+2)(n+3)....(n+n)]1n

      L= limn1n[(1+1n)(1+2n)(1+3n)....(1+nn)]1n

    logL=limn1n[log(1+1n)+log(1+2n)+log(1+3n)....log(1+nn)]

     logL=limn1nnrlog(1+rn)

    logL=101×log(1+x)dx

   logL=(x.log(1+x)1010[ddx(log(1+x)dx]dx

                           [ by using integration by parts]

   logL=[xlog(1+x)1010x1+xdx

     logL=log210(x+1x+11x+1)dx

    logL=log2[x]10+[log(x+1)]10

     logL=log21+log20

   logL=log4loge=log4e

    L=4e

  [L]=[4e]=1