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Question

For any positive integer n, define fn:(0,)R as
fn(x)=nj=1tan1(11+(x+j)(x+j1)) for all x(0,)
(Here, the inverse trigonometric function tan1x assumes values in (π2,π2) )
Then, which of the following statement(s) is (are) TRUE?

A
5j=1tan2(fj(0))=55
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B
10j=1(1+fj(0))sec2(fj(0))=10
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C
For any fixed positive integer n, limxtan(fn(x))=1n
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D
For any fixed positive integer n, limxsec2(fn(x))=1
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Solution

The correct option is D For any fixed positive integer n, limxsec2(fn(x))=1
fn(x)=nj=1tan1((x+j)(x+j1)1+(x+j)(x+j1))=nj=1[tan1(x+j)tan1(x+j1)]=tan1(x+n)tan1xfn(x)=tan1[n1+(x+n)x]tan[fn(x)]=[n1+(x+n)x]limxtan(fn(x))=0(C) is wrong.limxsec2(fn(x))=limx(1+tan2(fn(x))=1+limx[n1+(x+n)x]2=1(D) is correct.5j=1tan2(fj(0))=[51+0]2=25(A) is wrong.fj(x)=11+(n1+(x+n)x)2×(2x+n)fj(0)=n1+n210j=1(1+fj(0))sec2(fj(0))=10j=1(1+n1+n2)(1+[n1+0]2)=10j=1(1+n+n21+n2)(1+n2)=1+10+100=111(B) is wrong.

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