Inverse Trigonometric FunctionseasyFree

Inverse Trigonometric Functions: Terms Value

JEE Maths question with a full step-by-step solution.

Question
If f(x)=tan⁡−11x2+x+1+tan⁡−11x2+3x+3+tan⁡−11x2+5x+7+⋯f(x) = \tan^{-1}\dfrac{1}{x^2+x+1}+\tan^{-1}\dfrac{1}{x^2+3x+3}+\tan^{-1}\dfrac{1}{x^2+5x+7}+\cdots to ∞\infty terms, then the value of ∣f′(0)∣|f'(0)| is.
Solution
Answer: 1
Step 1: Identify the telescoping structure Each term satisfies:
tan⁡−11x2+(2k+1)x+(k2+k+1)=tan⁡−1(x+k+1)−tan⁡−1(x+k)\tan^{-1}\frac{1}{x^2+(2k+1)x+(k^2+k+1)} = \tan^{-1}(x+k+1)-\tan^{-1}(x+k)
 Step 2: Sum the telescoping series
f(x)=lim⁡n→∞[tan⁡−1(x+n)−tan⁡−1x]=π2−tan⁡−1xf(x) = \lim_{n\to\infty}\left[\tan^{-1}(x+n)-\tan^{-1}x\right] = \frac{\pi}{2}-\tan^{-1}x
 Step 3: Differentiate and evaluate
f′(x)=−11+x2  ⟹  f′(0)=−1  ⟹  ∣f′(0)∣=1f'(x) = -\frac{1}{1+x^2} \implies f'(0) = -1 \implies |f'(0)| = 1
 Answer: 1
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