Question

\[ \begin{array}{r}If\,\,\,\, y=\tan ^{-1}\left(\frac{1}{1+x+x^2}\right)+\tan ^{-1}\left(\frac{1}{x^2+3 x+3}\right) +\tan ^{-1}\left(\frac{1}{x^2+5 x+7}\right) \end{array} \]

then the value of \(y^{\prime}(0)\) is

• A. \(\frac{9}{10}\)
• B. \(\frac{1}{10}\)
• C. \(-\frac{9}{10}\)
• D. \(-\frac{1}{10}\)

Hint:

Each term is a difference of two inverse tangents, so the sum telescopes.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Use \(\tan^{-1}\dfrac{1}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n\) in the form \(\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}\).

Step 2: Rewrite each term
\(\tan^{-1}\dfrac{1}{1+x+x^{2}}=\tan^{-1}(x+1)-\tan^{-1}x\)
\(\tan^{-1}\dfrac{1}{x^{2}+3x+3}=\tan^{-1}(x+2)-\tan^{-1}(x+1)\)
\(\tan^{-1}\dfrac{1}{x^{2}+5x+7}=\tan^{-1}(x+3)-\tan^{-1}(x+2)\)

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

Step 4: Calculation
\(y^{\prime}=\dfrac{1}{1+(x+3)^{2}}-\dfrac{1}{1+x^{2}}\), so \(y^{\prime}(0)=\dfrac{1}{10}-1=-\dfrac{9}{10}\).

Final Answer:
The value of \(y^{\prime}(0)\) is \(-\frac{9}{10}\). \[ \boxed{\text{(C) }-\dfrac{9}{10}} \]