Question

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

• A. $-\frac{1}{1+n^2}$
• B. $-\frac{n^2}{1+n^2}$
• C. $\frac{n}{1+n^2}$
• D. None of these

Answer 1

The correct option is B $-\frac{n^2}{1+n^2}$

$y=ta{n}^{-1}\frac{1}{1+x+x^2}+ta{n}^{-1}\frac{1}{x^2+3x+3}+\cdots +nterms$
$=ta{n}^{-1}\frac{(x+1)-x}{1+x(1+x)}+ta{n}^{-1}\frac{(x+2)-(x+1)}{1+(x+1)(x+2)}+\cdots +nterms$
$=ta{n}^{-1}(x+1)-ta{n}^{-1}x+ta{n}^{-1}(x+2)-ta{n}^{-1}(x+1)+\cdots +ta{n}^{-1}(x+n)-ta{n}^{-1}(x+(n-1\left)\right)$
$=ta{n}^{-1}(x+n)-ta{n}^{-1}x$
$y^{\prime }\left(x\right)=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$
$\Rightarrow y^{\prime }\left(0\right)=\frac{1}{1+n^2}-1=\frac{-n^2}{1+n^2}$