JEE Mains · Maths · STD 12 - 1. relation and function
If \(f\left( x \right) = {\log _e}\,\left( {\frac{{1 - x}}{{1 + x}}} \right)\), \(\left| x \right| < 1\), then \(f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\) is equal to
- A \(2f\left( x \right)\)
- B \({\left( {f\left( x \right)} \right)^2}\)
- C \(2f\left( x^2 \right)\)
- D \( - 2f\left( x \right)\)
Answer & Solution
Correct Answer
(A) \(2f\left( x \right)\)
Step-by-step Solution
Detailed explanation
\(f\left( x \right) = {\log _e}\left( {\frac{{1 - x}}{{1 + x}}} \right),\left| x \right| < 1\) \(f\left( {\frac{{2x}}{{1 + {x^2}}}} \right) = \ell n\left( {\frac{{1 - \frac{{2x}}{{1 + 2{x^2}}}}}{{1 + \frac{{2x}}{{1 + {x^2}}}}}} \right)\)…
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