For any positive integer n, define ` f_(n) : (0, infty ) to R ` as `f_(n)(x) = sum_(j=1)^(n) tan^(-1). (1/(1+(x+j)(x+j-1))) ` for all ` x in (0, infty)`.
(Here, the inverse trigonometric function ` tan^(-1) ` x assumes values in `(- pi/2, pi/2) )`. Then, which of the following statement (s) is (are) TRUE?

Options

A. `sum_(j=1)^(5) tan^(2) (f_(j)(0)) = 55`
B. `sum_(j=1)^(10) (1+f'_(j) (0)) sec^(2) (f_(j) (0)) = 10`
C. For any fixed positive integer `n underset(x to infty) lim tan (f_(n) (x)) = 1/n`
D. For any fixed positive interger ` n, underset( x to infty) lim sec^(2) (f_(n)(x)) = 1`

Answer

D

Explanation

We have,
` f_(n) (x) = underset(j=1) overset(n) sum tan^(-1) (1/(1+(x+j)(x+j-1)))"for all " x in(0,infty)`
`rArr f_(n) (x) =underset (j=1)overset(n) sum tan^(-1). (((x+j)-(x+j-1))/(1+(x+j)(x+j-1)))`
` rArr f_(n)(x) = underset(j=1) overset(n) sum [tan^(-1)(x+j)-tan^(-1)(x+j-1)]`
` rArr f_(n)(x) = (tan^(-1)(x+1) - tan^(-1) x) `
`+(tan^(-1)(x+2)-tan^(-1)(x+1))`
`+(tan^(-1)(x+3)-tan^(-1)(x+2))`
`+...+ (tan^(-1)(x+n) - tan^(-1)(x+n-1))`
` rArr f_(n) (x) = tan^(-1) (x+n) - tan^(-1) x`
This statement is false as ` x ne 0.i.e., x in (0, infty)`.
(b) This statement is also false as ` 0 !in (0, infty)`
(c ) `f_(n) (x) = tan^(-1)(x+n) - tan^(-1) x`
`underset(x to infty) lim tan (f_(n) (x)) = underset(x to infty) lim tan (tan^(-1) (x+n) - tan^(-1) x)`
` rArr underset( x to infty) lim tan(f_(n)(x)) = underset (x to infty) lim tan (tan^(-1) n/(1+nx+x^(2))`
` = underset( x to infty) lim n/(1+ nx + x^(2)) = 0`
` :. ` (c ) statement is false.
(d) `underset(x to infty) lim sec^(2) (f_(n)(x)) = underset(x to infty) lim (1 + tan^(2) f_(n) (x))`
` = 1+ underset(x to infty) lim tan^(2) (f_(n) (x)) = 1 + 0 = 1`
`:.` (d) statement is true.