If y = tan ^{-1}left(frac{1}{1+x+x^{2}}right) | vedclass

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(D) The general term of the series is $T_r = 	an^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}ight)$ for $r=1, 2, \dots, n$. We can rewrite the arg

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$-\frac{n^{2}}{1+n^{2}}$

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(D) The general term of the series is $T_r = 	an^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}ight)$ for $r=1, 2, \dots, n$. We can rewrite the argument as: $\frac{1}{1 + (x+r)(x+r-1)} = \frac{(x+r) - (x+r-1)}{1 + (x+r)(x+r-1)}$. Thus,$T_r = 	an^{-1}(x+r) - 	an^{-1}(x+r-1)$. Summing these terms from $r=1$ to $n$: $y = \sum_{r=1}^{n} [	an^{-1}(x+r) - 	an^{-1}(x+r-1)]$. This is a telescoping series: $y = (	an^{-1}(x+1) - 	an^{-1}(x)) + (	an^{-1}(x+2) - 	an^{-1}(x+1)) + \dots + (	an^{-1}(x+n) - 	an^{-1}(x+n-1))$. $y = 	an^{-1}(x+n) - 	an^{-1}(x)$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{1+(x+n)^2} - \frac{1}{1+x^2}$. Evaluating at $x=0$: $y'(0) = \frac{1}{1+n^2} - \frac{1}{1+0^2} = \frac{1}{1+n^2} - 1 = \frac{1 - 1 - n^2}{1+n^2} = -\frac{n^2}{1+n^2}$.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$

If $y = \tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+2x+3}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+5x+7}\right) + \dots + n \text{ terms}$,then $y'(0)$ is

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