MCQ
For any positive integer $n$, let $S_n:(0, \infty) \rightarrow R$ be defined by

$S_n(x)=\sum_{k=1}^n \cot ^{-1}\left(\frac{1+k(k+1) x^2}{x}\right)$

where for any $x \in R , \cot ^{-1} x \in(0, \pi)$ and $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following

statements is (are) $TRUE$?

$(A)$ $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^2}{10 x }\right)$, for all $x >0$

$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=x$, for all $x>0$

$(C)$ The equation $S_3(x)=\frac{\pi}{4}$ has a root in $(0, \infty)$

$(D)$ $\tan \left( S _{ n }( x )\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x >0$

  • A
    $A,C$
  • B
    $A,D$
  • $A,B$
  • D
    $A,B,C$

Answer

Correct option: C.
$A,B$
c
$S_n(x)=\sum_{k=1}^n \tan ^{-1}\left(\frac{x}{1+k x(k x+x)}\right)$

$=\sum_{k=1}^n \tan ^{-1}\left(\frac{(k x+x)-(k x)}{1+(k x+x)(k x)}\right)$

$S_n(x)=\tan ^{-1}(n x+x)-\tan ^{-1} x=\tan ^{-1}\left(\frac{ nx }{1+(n+1) x^2}\right)$

$\text { (A) } S_{10}(x)=\tan ^{-1} \frac{10 x}{1+11 x^2}=\frac{\pi}{2}-\tan ^{-1}$$\left(\frac{1+11 x^2}{10 x}\right)(x>0)$

$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}+\left(1+\frac{1}{n}\right) x^2}{x}=x(x>0)$

$(C)$ $S_3(x)=\tan ^{-1} \frac{3 x}{1+4 x^2}=\frac{\pi}{4} \Rightarrow 4 x^2-3 x+1=0 \Rightarrow x \notin R$

$(D)$ $\tan \left( S _{ n }( x )\right)=\frac{ nx }{1+( n +1) x ^2} ; \forall n \geq 1 ; x >0$

We need to check the validity of $\frac{ nx }{1+( n +1) x ^2} \leq \frac{1}{2} \forall n \geq 1 ; x >0 ; n \in N$

$\Rightarrow 2 nx \leq( n +1) x ^2+1$

$\Rightarrow( n +1) x ^2-2 nx +1 \geq 0 \forall n \geq 1 ; x >0 ; n \in N$

Discriminant of $y=(n+1) x^2-2 n x+1$ is

$D=4 n^2-4(n+1) \text { and } n \in N$

$D >0$ for $n \geq 2 \Rightarrow \exists$ some $x >0$

for which $y <0$ as both roots of $y =0$ will be positive.

$y =( n +1) x ^2-2 nx +1, n \geq 2$

So, $y \geq 0 \forall n \geq 1 ; \forall x >0 ; n \in N$ is false.

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