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Question
If \[f\left( x \right) = \log \left( \frac{1 + x}{1 - x} \right) \text{ and} g\left( x \right) = \frac{3x + x^3}{1 + 3 x^2}\] , then f(g(x)) is equal to
Options
(a) f(3x)
(b) {f(x)}3
(c) 3f(x)
(d) −f(x)
Solution
(c) 3f(x)
\[\frac{1 + g(x)}{1 - g(x)} = \frac{1 + \frac{3x + x^3}{1 + 3 x^2}}{1 - \frac{3x + x^3}{1 + 3 x^2}}\]
\[ = \frac{1 + 3 x^2 + 3x + x^3}{1 + 3 x^2 - 3x - x^3}\]
\[ = \frac{(1 + x )^3}{(1 - x )^3}\]
\[\text{ Then } , f(g(x)) = \log \left( \frac{1 + g(x)}{1 - g(x)} \right)\]
\[ = \log \left( \frac{1 + x}{1 - x} \right)^3 \]
\[ = 3 \log \left( \frac{1 + x}{1 - x} \right)\]
\[ = 3f(x))\]
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