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COMEDK · Maths · 5. Sequences and Series

If \(y=\tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right)+\tan ^{-1}\left(\frac{1}{x^{2}+3 x+3}\right)\) \(+\tan ^{-1}\left(\frac{1}{x^{2}+5 x+7}\right)+\ldots+\) upto \(n\) terms then \(\frac{d y}{d x}\) at \(x=0\) and \(n=1\) is equal to

  1. A \(\frac{1}{2}\)
  2. B \(-\frac{1}{2}\)
  3. C 0
  4. D \(\frac{1}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

We have, \[ \begin{aligned} &y=\tan ^{-1}\left(\frac{1}{1+x(x+1)}\right)+ \\ &\tan ^{-1}\left(\frac{1}{1+(x+2)(x+3)}\right)+\ldots \\ &\tan ^{-1}\left(\frac{1}{1+(x+1)(x+2)}\right)+ \\ &+\tan ^{-1}\left(\frac{1}{1+(x+n-1)(x+n)}\right) \end{aligned} \] When \(n=1\), then…