Question

If $y={\tan }^{-1}\frac{1}{1+x+x^2}+{\tan }^{-1}\frac{1}{x^2+3x+3}$
$+{\tan }^{-1}\frac{1}{x^2+5x+7}+\cdots +$ upto $n$ terms, then

• A. $y^{\prime }\left(0\right)=\frac{-n^2}{1+n^2}$
• B. $y^{\prime }\left(0\right)=\frac{n^2}{1+n^2}$
• C. $y^{\prime }(-n)=\frac{n^2}{1+n^2}$
• D. $y^{\prime }(-n)=\frac{-n^2}{1+n^2}$

Answer 1

Answer: (A), (C)

$y^{\prime }(-n)=\frac{n^2}{1+n^2}$

$y={\tan }^{-1}\frac{1}{1+x+x^2}+{\tan }^{-1}\frac{1}{x^2+3x+3}+\dots +$ upto $n$ terms
$={\tan }^{-1}\frac{(x+1)-x}{1+x(1+x)}+{\tan }^{-1}\frac{(x+2)-(x+1)}{1+(x+1)(x+2)}+\cdots +$ upto $n$ terms
$={\tan }^{-1}(x+1)-{\tan }^{-1}x+{\tan }^{-1}(x+2)-{\tan }^{-1}(x+1)$
$+\cdots +{\tan }^{-1}(x+n)-{\tan }^{-1}(x+(n-1\left)\right)$
$={\tan }^{-1}(x+n)-{\tan }^{-1}x$
$y^{\prime }\left(x\right)=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$
$\Rightarrow y^{\prime }\left(0\right)=\frac{1}{1+n^2}-1=\frac{-n^2}{1+n^2}$
$\Rightarrow y^{\prime }(-n)=1-\frac{1}{1+n^2}=\frac{n^2}{1+n^2}$