Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), x \in(1, \infty)$, then $f^{\prime}(x)$
MathematicsDifferentiationMHT CETMHT CET 2022 (08 Aug Shift 2)
Options:
  • A $\frac{-4}{1+x^2}$
  • B 0
  • C $\frac{2 x}{1-x^2}$
  • D $\frac{4}{1+x^2}$
Solution:
1617 Upvotes Verified Answer
The correct answer is: 0
$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\ & \Rightarrow f(x)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\ & \Rightarrow f(x)=\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \theta) \\ & \Rightarrow f(x)=\pi-2 \theta+2 \theta=\pi \\ & \Rightarrow f^{\prime}(x)=0\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.