For any positive integer n , define f n : 0 ,   ∞ → R as f n x = ∑ j = 1 n tan - 1 1…

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

fnx=∑j=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?

  1. ∑j=15tan2fj0=55
  2. ∑j=1101+fj'0sec2fj0=10
  3. For any fixed positive integer n,limx→∞tanfnx=1n
  4. 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

Asked in: JEE Advanced 2018 (Paper 2)

Practice more Trigonometric Functions questions on Aicharya