Question

$y={\tan }^{-1}\frac{1}{1+x+x^2}+{\tan }^{-1}\frac{1}{x^2+3x+3}+{\tan }^{-1}\frac{1}{x^2+5x+7}+...+$ upto $n$ terms,then $y^{\prime \prime }\left(0\right)$ is equal to

• A. $-1/{(n^2+1)}^2$
• B. $-n^2/{(n^2+1)}^2$
• C. $n^2/{(n^2+1)}^2$
• D. None of these

Answer 1

The correct option is D None of these

$y={\tan }^{-1}\frac{1}{1+x+x^2}+{\tan }^{-1}\frac{1}{x^2+3x+3}+...+n$ terms

$={\tan }^{-1}\frac{(x+1)-x}{1+x(x+1)}+{\tan }^{-1}\frac{(x+2)-(x+1)}{1+(x+1)(x+2)}+...n$ terms

$={\tan }^{-1}(x+1)-{\tan }^{-1}x+{\tan }^{-1}(x+2)-{\tan }^{-1}(x+1)+...+{\tan }^{-1}(x+n)-{\tan }^{-1}(x+(n-1))={\tan }^{-1}(x+n)-{\tan }^{-1}x$

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

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