For any positive integer `n` , define `f_n :(0,oo)rarrR` as `f_n(x)=sum_(j=1)^ntan^(-1)(1/(1+(x+j)(x+j-1)))` for all `x in (0, oo)` . Here, the inverse trigonometric function `tan^(-1)x` assumes values in `(-pi/2,pi/2)dot` Then, which of the following statement(s) is (are) TRUE? `sum_(j=1)^5tan^2(f_j(0))=55` (b) `sum_(j=1)^(10)(1+fj '(0))sec^2(f_j(0))=10` (c) For any fixed positive integer `n` , `(lim)_(xrarroo)tan(f_n(x))=1/n` (d) For any fixed positive integer `n` , `(lim)_(xrarroo)sec^2(f_n(x))=1`

Options

A. `underset(j = 1)overset(5)sum tan^(2) (f_(j) (0)) = 55`
B. `underset(j =1)overset(10)sum (1 + f'_(j)(0)) sec^(2) (f_(j) (0)) = 10`
C. For any fixed positive inetger `n, underset(x rarr oo)("lim") tan (f_(n) (x)) = (1)/(n)`
D. For any fixed positive integer `n, underset(x rarr oo)("lim") sec^(2) (f_(n) (x)) = 1`

Answer

A, B, D

Explanation

`f_(n) (x) = underset(j =1)overset(n)sum tan^(-1) ((1)/(1 + (x + j) (x + j -1)))`
`= underset(j =1)overset(n)sum tan^(-1) (((x + j) - (x + j -1))/(1+(x +j) (x+j -1)))`
`= underset(j=1)overset(n)sum [tan^(-1) (x + j)-tan^(-1) (x + j -1)]`
`:. f_(n) (x) = tan^(-1) (x + n) - tan^(-1) (x)`
(1) `f_(n) (0) = tan^(-1) (n)`
`rArr tan^(2) (f_(n) (0)) = tan^(2) (tan^(-1) n) = n^(2)`
`underset(j =1)overset(5)sum tan^(2) (f_(j) (0)) = underset(j=1)overset(5)sum j^(2) = (5 xx 6 xx 11)/(6) = 55`
(2) `f'_(n) (x) = (1)/(1 + (x + n)^(2)) - (1)/(1 + x^(2))`
`rArr f'_(n) (0) = (1)/(1+ n^(2)) -1`
`rArr 1 + f'_(n) (0) = (1)/(1 + n^(2))`
Also, `sec^(2) (f_(n) (0)) = sec^(2) (tan^(-1) (n)) = 1 + n^(2)`
Hence, `(1 + f'_(n) (0)). sec^(2) (f_(n)(0)) = ((1)/(1 + n^(2))) (1 + n^(2)) =1`
So, `underset(j =1)overset(10)sum (1 + f'_(i)(0)) sec^(2) (f_(i) (0)) = underset(i=1)overset(10)sum 1 = 10`
(3) `underset(x rarr oo)("lim") tan (f_(n)(x)) = underset(x rarroo)("lim") (n)/(1 + x (n + x)) = 0`
(4) `underset(x rarroo)("lim") sec^(2) (f_(n) (x)) = underset(x rarr oo)("lim") (1 + tan^(2) (f_(n) (x))) = 1`