Question

If $y={\tan }^{-1}\left(\frac{1}{1+x+x^2}\right)+{\tan }^{-1}\left(\frac{1}{x^2+3x+2}\right)+{\tan }^{-1}\left(\frac{1}{x^2+5x+6}\right)+....+$ upto $n$ terms then $\frac{dy}{dx}$ at $x=0$ and $n=1$ is equal to

• A. $\frac{1}{2}d$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$-\frac{1}{2}$


$y={\tan }^{-1}\left(\frac{1}{1+x(x+1)}\right)+{\tan }^{-1}\left(\frac{1}{1+(x+1)(x+2)}\right)+{\tan }^{-1}\left(\frac{1}{1+(x+2)(x+3)}\right)+.....+{\tan }^{-1}\left(\frac{1}{1+(x+n-1)(x+n)}\right)$

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

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+(x+1{)}^2}-\frac{1}{1+x^2}\therefore \frac{dy}{dx}{]}_{x=0}=\frac{1}{1+1}-\frac{1}{1}=\frac{1}{2}-1=-\frac{1}{2}$