FN_JEE MAIN_08-04-2019_S1 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2019

08-04-2019 S1

Question

If $f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|<1$, then $f\left(\frac{2 x}{1+x^2}\right)$ is equal to :

Select the correct option:

A

2f(x)

B

2f(x2)

C

–2f(x)

D

(f(x))2

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

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