Question

If $y=\tan^{-1}(\frac{1}{1+x+x^{2}})+\tan^{-1}(\frac{1}{x^{2}+3x+3})+\tan^{-1}(\frac{1}{x^{2}+5x+7})$ then $y^{\prime}(0)$ is}

• A. $9/10$
• B. $1/10$
• C. $-9/10$
• D. $-1/10$

Hint:

In telescoping series for $\tan^{-1}$, only the last and first terms remain.

Answer 1

The Correct Option is C

Step 1: Concept
Use $\tan^{-1} \frac{x-y}{1+xy} = \tan^{-1} x - \tan^{-1} y$.

Step 2: Analysis
- Term 1: $\tan^{-1} \frac{(x+1)-x}{1+x(x+1)} = \tan^{-1}(x+1) - \tan^{-1} x$.
- Term 2: $\tan^{-1} \frac{(x+2)-(x+1)}{1+(x+1)(x+2)} = \tan^{-1}(x+2) - \tan^{-1}(x+1)$.
- Term 3: $\tan^{-1} \frac{(x+3)-(x+2)}{1+(x+2)(x+3)} = \tan^{-1}(x+3) - \tan^{-1}(x+2)$.

Step 3: Calculation
Summing gives $y = \tan^{-1}(x+3) - \tan^{-1} x$.
$y' = \frac{1}{1+(x+3)^2} - \frac{1}{1+x^2}$.
At $x=0$: $y'(0) = \frac{1}{1+9} - \frac{1}{1+0} = \frac{1}{10} - 1 = -9/10$.

Step 4: Conclusion
Hence, $y'(0) = -9/10$. Final Answer: (C)