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.
