For any positive integer n , let S n : ( 0 , ∞ ) → R be defined by S n x = ∑ k = 1 n cot - 1 1 + k ( k + 1 ) x 2 x where for any x ∈ R , cot - 1 ( x ) ∈ ( 0 , π ) and tan - 1 ( x ) ∈ - π 2 , π 2 . Then which of the following statements is (are) TRUE ?
B. lim n → ∞ cot S n ( x ) = x , for all x > 0
S n x = ∑ k = 1 n cot - 1 1 + k ( k + 1 ) x 2 x = ∑ k = 1 n cot - 1 1 + k x ( k + 1 ) x ( k + 1 ) x - k x = ∑ k = 1 n tan - 1 ( k + 1 ) x - k x 1 + k x ( k + 1 ) x = ∑ k = 1 n tan - 1 ( k + 1 ) x - tan - 1 k x = tan - 1 ( n + 1 ) x - tan - 1 n x + … + tan - 1 3 x - tan - 1 2 x + tan - 1 2 x - tan - 1 x = tan - 1 ( n + 1 ) x - tan - 1 x = tan - 1 ( n + 1 ) x - x 1 + ( n + 1 ) x 2 = tan - 1 n x 1 + ( n + 1 ) x 2 1. S 10 x = tan - 1 10 x 1 + 11 x 2 = π 2 - cot - 1 10 x 1 + 11 x 2 = π 2 - tan - 1 1 + 11 x 2 10 x 2. lim n → ∞ cot S n ( x ) = lim n → ∞ cot tan - 1 n x 1 + ( n + 1 ) x 2 = lim n → ∞ cot cot - 1 1 + ( n + 1 ) x 2 n x = lim n → ∞ 1 + ( n + 1 ) x 2 n x = x 2 x = x 3. S 3 x = tan - 1 3 x 1 + 4 x 2 = π 4 ⇒ 3 x 1 + 4 x 2 = tan π 4 ⇒ 3 x 1 + 4 x 2 = 1 ⇒ 1 + 4 x 2 = 3 x ⇒ 4 x 2 - 3 x + 1 = 0 D < 0 Hence, no real roots. 4. tan S n ( x ) = tan tan - 1 n x 1 + ( n + 1 ) x 2 = n x 1 + ( n + 1 ) x 2 n x 1 + ( n + 1 ) x 2 ≤ 1 2 ⇒ 2 n x ≤ 1 + ( n + 1 ) x 2 ⇒ 2 n x ≤ ( n + 1 ) x 2 + 1 ⇒ ( n + 1 ) x 2 - 2 n x + 1 ≥ 0   ∀   n ≥ 1 ;   x > 0 Let, y = ( n + 1 ) x 2 - 2 n x + 1 D = 4 n 2 - 4 ( n + 1 ) and n ∈ N D < 0 for n = 1 Hence, no solution if n = 1 .
Related: Mathematics — Inverse Trigonometric Functions · All PYQ Banks