JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 7)
If $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$, $$\left| x \right| < 1$$ then $$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ is equal to
2f(x2)
2f(x)
(f(x))2
-2f(x)
Explanation
Given, $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
$$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ = $$\ln \left( {{{1 - {{2x} \over {1 + {x^2}}}} \over {1 + {{2x} \over {1 + {x^2}}}}}} \right)$$
= $$\ln \left( {{{{x^2} - 2x + 1} \over {{x^2} + 2x + 1}}} \right)$$
= $$\ln {\left( {{{1 - x} \over {1 + x}}} \right)^2}$$
= $$2\ln \left( {{{1 - x} \over {1 + x}}} \right)$$
= 2f(x)
$$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ = $$\ln \left( {{{1 - {{2x} \over {1 + {x^2}}}} \over {1 + {{2x} \over {1 + {x^2}}}}}} \right)$$
= $$\ln \left( {{{{x^2} - 2x + 1} \over {{x^2} + 2x + 1}}} \right)$$
= $$\ln {\left( {{{1 - x} \over {1 + x}}} \right)^2}$$
= $$2\ln \left( {{{1 - x} \over {1 + x}}} \right)$$
= 2f(x)
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