Question

If \(y = tan^{-1}(\frac{1}{x^2+x+1})+tan^{-1}(\frac{1}{x^2+3x+3})+tan^{-1}(\frac{1}{x^2+5x+7})+\ldots\) upto \(n\) terms, then \(y^'(0) =\)

• A. \(\frac{n^2}{n^2+1}\)
• B. \(-\frac{n^2}{n^2+1}\)
• C. \(0\)
• D. \(\frac{1}{n^2+1}\)

Hint:

Write each term as a difference of two arctangents so the series telescopes.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Each term has the form \(\tan^{-1}\frac{1}{1+k(k+1)}\) with \(k = x, x+1, x+2,\ldots\). Check: \(x^2+x+1 = 1 + x(x+1)\), \(x^2+3x+3 = 1 + (x+1)(x+2)\), \(x^2+5x+7 = 1+(x+2)(x+3)\).

Step 2: Use the difference formula:
\(\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A-B}{1+AB}\). With \(A = x+1\) and \(B = x\), we get \(\tan^{-1}\frac{1}{1+x(x+1)} = \tan^{-1}(x+1) - \tan^{-1}x\).
So the terms are \([\tan^{-1}(x+1)-\tan^{-1}x] + [\tan^{-1}(x+2)-\tan^{-1}(x+1)] + \ldots\)

Step 3: Telescope:
\[ y = \tan^{-1}(x+n) - \tan^{-1}x \]

Step 4: Differentiate:
\[ y' = \frac{1}{1+(x+n)^2} - \frac{1}{1+x^2} \]
At \(x=0\): \(y'(0) = \frac{1}{1+n^2} - 1 = -\frac{n^2}{1+n^2}\).

Final Answer:
\(y'(0) = -\dfrac{n^2}{n^2+1}\), option (B). \[ \boxed{-\frac{n^2}{n^2+1}} \]