If   y=tan−111+x+x2+tan−11×2+3x+3                 +tan−11×2+5x+7+tan−11×2+7x+13+…+ n terms then y’0  is equal to

# If    then ${y}^{\text{'}}\left(0\right)$  is equal to

1. A

$\frac{-1}{\left({n}^{2}+1\right)}$

2. B

$\frac{-{n}^{2}}{\left({n}^{2}+1\right)}$

3. C

$\frac{{n}^{2}}{\left({n}^{2}+1\right)}$

4. D

$\frac{1}{\left({n}^{2}+1\right)}$

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### Solution:

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

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

$={\mathrm{tan}}^{-1}\left(x+1\right)-{\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}\left(x+2\right)-{\mathrm{tan}}^{-1}\left(x+1\right)+...+$

${\mathrm{tan}}^{-1}\left(x+n\right)-{\mathrm{tan}}^{-1}\left(x+\left(n-1\right)\right)$

$={\mathrm{tan}}^{-1}\left(x+n\right)-{\mathrm{tan}}^{-1}x.$

${y}^{\text{'}}\left(x\right)=\frac{1}{1+{\left(x+n\right)}^{2}}\text{\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{1+{x}^{2}}$

$⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{y}^{\text{'}}\left(0\right)=\frac{1}{1+{n}^{2}}-1=-\frac{{n}^{2}}{1+{n}^{2}}$  +91

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