MHT CET · Maths · Inverse Trigonometric Functions
If \(y=\tan ^{-1}\left(\frac{1}{1+x+x^2}\right)+\tan ^{-1}\left(\frac{1}{x^2+3 x+3}\right)+\tan ^{-1}\left(\frac{1}{x^2+5 x+7}\right)\) then the value of \(\mathrm{y}^{\prime}(0)\) is
- A \(\frac{9}{10}\)
- B \(\frac{1}{10}\)
- C \(-\frac{9}{10}\)
- D \(-\frac{1}{10}\)
Answer & Solution
Correct Answer
(C) \(-\frac{9}{10}\)
Step-by-step Solution
Detailed explanation
\(y=\tan ^{-1}(x+1)-\tan ^{-1}(x)+\tan ^{-1}(x+2)-\tan ^{-1}(x+1)+\tan ^{-1}(x+3)-\tan ^{-1}(x+2)\) \(y=\tan ^{-1}(x+3)-\tan ^{-1}(x)\)
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