Q19 / 54 · JEE Advanced 2018 (Paper 2) · Mathematics

For any positive integer n, define fn:0, ∞→R as

fnx=∑i=1ntan-111+x+jx+j-1 for all x∈0,∞ .

(Here, the inverse trigonometric function tan-1⁡x assumes values in -π2, π2)

Then, which of the following statement(s) is (are) TRUE?

  • A
    ∑j=15tan2fj0=55
  • B
    ∑j=1101+fj'0sec2fj0=10
  • C
    For any fixed positive integer n,limx→∞tanfnx=1n
  • Correct answer
    D
    For any fixed positive integer n, limx→∞sec2fnx=1

Solution

fnx=∑j=1ntan-1x+j-x+j-11+x+jx+j-1 fnx=∑j=1ntan-1x+j-tan-1x+j-1
fnx=tan-1x+n-tan-1x
∴ tanfnx=tantan-1x+n-tan-1x

tanfnx=tantan-1x+n-x1+xx+n
tanfnx=n1+x2+nx,

so that limx→∞tanfnx=0
∴ sec2fnx=1+tan2fnx
sec2fnx=1+n1+x2+nx2
limx→∞sec2fnx=limx→∞1+n1+x2+nx2=1

Note that since 0 is not in domain so first two options are wrong

Hence, the correct answer is option: 'D'.

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