Q.

For any positive integer n, let Sn:(0,∞)→ℝ be defined by Sn(x)=∑k=1ncot-1(1+k(k+1)x2x), where for any x∈ℝ,
cot-1x∈(0,π) and tan-1(x)∈(-π2,π2). Then which of the following statements is (are) TRUE                   [2021]

1 S10(x)=π2-tan-1(1+11x210x),  for all x>0  
2 limn→∞cot(Sn(x))=x,  for all x>0  
3 The equation S3(x)=π4 has a root in (0,∞)  
4 tan(Sn(x))≤12,  for all n≥1 and x>0  

Ans.

(1, 2)

Given that Sn(x)=∑k=1ncot-1(1+k(k+1)x2x)

         =∑k=1ntan-1(x1+kx(kx+x))

          =∑k=1ntan-1((kx+x)-(kx)1+(kx+x)(kx))

⇒ Sn(x)=tan-1(nx+x)-tan-1x=tan-1(nx1+(n+1)x2)

(1)    S10(x)=tan-1(10x1+11x2)=π2-tan-1(1+11x210x),    (x>0)

           Option (1) is correct

 (2)    limn→∞cot(Sn(x))=limn→∞cot[cot-1(1+(n+1)x2nx)]

                                         =limn→∞1n+(1+1n)x2x=x,  (x>0)

            Option (2) is correct

(3)    S3(x)=tan-1(3x1+4x2)=π4

           ⇒ 4x2-3x+1=0⇒x∉ℝ    [∵ D is negative]

            Option (3) is incorrect

(4)    For x=1,

           tan(Sn(x))=nn+2≥12,    for n≥3

            Option (4) is incorrect