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 $\mathrm{f}\left(\mathrm{x}_{1}\right)-\mathrm{f}\left(\mathrm{x}_{2}\right)=\mathrm{f}\left(\frac{\mathrm{x}_{1}-\mathrm{x}_{2}}{1-\mathrm{x}_{1} \mathrm{x}_{2}}\right)$ for $\mathrm{x}_{1}, \mathrm{x}_{2} \in(-1,1)$, then what
is $\mathrm{f}(\mathrm{x})$ equal to?
MathematicsFunctionsNDANDA 2016 (Phase 1)
Options:
  • A $\ln \left(\frac{1-x}{1+x}\right)$
  • B $\ln \left(\frac{2+x}{1-x}\right)$
  • C $\tan ^{-1}\left(\frac{1-x}{1+x}\right)$
  • D $\tan ^{-1}\left(\frac{1+x}{1-x}\right)$
Solution:
1790 Upvotes Verified Answer
The correct answer is: $\ln \left(\frac{1-x}{1+x}\right)$
$f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right)$
$x_{1}, x_{2} \in(-1,1)$
then $f(x)=\log \frac{(1-x)}{(1+x)}$
$\begin{aligned} f\left(x_{1}\right)=\log \frac{1-x_{1}}{1+x_{1}} & f\left(x_{2}\right)=\log \frac{1-x_{2}}{1+x_{2}} \\ f\left(x_{1}\right)-f\left(x_{2}\right) &=\log \frac{1-x_{1}}{1+x_{1}}-\log \frac{1-x_{2}}{1+x_{2}} \\=& \log \frac{\left(1-x_{1}\right)}{\left(1+x_{1}\right)} \times \frac{\left(1+x_{2}\right)}{\left(1-x_{2}\right)} \\=& \log \frac{\left(1-x_{1}+x_{2}-x_{1} x_{2}\right)}{\left(1+x_{1}-x_{2}-x_{1} x_{2}\right)} \\=& \log \frac{\left(1-x_{1} x_{2}\right)-\left(x_{1}-x_{2}\right)}{\left(1-x_{1} x_{2}\right)+\left(x_{1}-x_{2}\right)} \\ f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right) \end{aligned}$

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.