For any positive integer n,let S_n: (0, infty | vedclass

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(C) We have $S_n(x) = \sum_{k=1}^n 	an^{-1}\left(\frac{x}{1+k(k+1)x^2}ight)$.Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A

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$A, B$

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(C) We have $S_n(x) = \sum_{k=1}^n 	an^{-1}\left(\frac{x}{1+k(k+1)x^2}ight)$.Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we can write the general term as:$	an^{-1}\left(\frac{(k+1)x - kx}{1 + ((k+1)x)(kx)}ight) = 	an^{-1}((k+1)x) - 	an^{-1}(kx)$.Summing from $k=1$ to $n$,we get a telescoping sum:$S_n(x) = (	an^{-1}(2x) - 	an^{-1}(x)) + (	an^{-1}(3x) - 	an^{-1}(2x)) + \dots + (	an^{-1}((n+1)x) - 	an^{-1}(nx))$$S_n(x) = 	an^{-1}((n+1)x) - 	an^{-1}(x) = 	an^{-1}\left(\frac{(n+1)x - x}{1 + (n+1)x^2}ight) = 	an^{-1}\left(\frac{nx}{1+(n+1)x^2}ight)$.$(A)$ For $n=10$,$S_{10}(x) = 	an^{-1}\left(\frac{10x}{1+11x^2}ight)$. Since $	an^{-1}(y) = \frac{\pi}{2} - \cot^{-1}(y) = \frac{\pi}{2} - 	an^{-1}(1/y)$ for $y > 0$,we have $S_{10}(x) = \frac{\pi}{2} - 	an^{-1}\left(\frac{1+11x^2}{10x}ight)$. Thus,$(A)$ is $TRUE$.$(B)$ $\cot(S_n(x)) = \frac{1}{	an(S_n(x))} = \frac{1+(n+1)x^2}{nx} = \frac{1}{nx} + \frac{n+1}{n}x$. As $n ightarrow \infty$,$\cot(S_n(x)) ightarrow 0 + 1 \cdot x = x$. Thus,$(B)$ is $TRUE$.$(C)$ $S_3(x) = 	an^{-1}\left(\frac{3x}{1+4x^2}ight) = \frac{\pi}{4} \implies \frac{3x}{1+4x^2} = 1 \implies 4x^2 - 3x + 1 = 0$. The discriminant $D = (-3)^2 - 4(4)(1) = 9 - 16 = -7 $(D)$ Let $f(x) = 	an(S_n(x)) = \frac{nx}{1+(n+1)x^2}$. To find the maximum,$f'(x) = \frac{n(1+(n+1)x^2) - nx(2(n+1)x)}{(1+(n+1)x^2)^2} = \frac{n - n(n+1)x^2}{(1+(n+1)x^2)^2}$. Setting $f'(x)=0$,$x^2 = \frac{1}{n+1}$,so $x = \frac{1}{\sqrt{n+1}}$. The maximum value is $f\left(\frac{1}{\sqrt{n+1}}ight) = \frac{n/\sqrt{n+1}}{1+(n+1)/(n+1)} = \frac{n}{2\sqrt{n+1}}$. For $n=3$,value is $3/(2\sqrt{4}) = 3/4 > 1/2$. Thus,$(D)$ is $FALSE$.

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