Question:
If f(x) = log \(\left(\dfrac{1+x}{1-x}\right)\) and g(x) = \(\dfrac{3x+x^3}{1+3x^2}\) Then f(g(x)) is equal to
यदि \(f(x) = \log\left(\dfrac{1+x}{1-x}\right)\) और \(g(x) = \dfrac{3x+x^3}{1+3x^2}\) है, तो \(f(g(x))\) किसके बराबर है?
Options:
- -f(x)
- \([f(x)]^3\)
- 3 f(x)
- f(3x)
- -f(x)
- \([f(x)]^3\)
- 3 f(x)
- f(3x)
Correct Answer: c)
Explanation:
Given \(f(x) = \log\left(\dfrac{1+x}{1-x}\right)\) and \(g(x) = \dfrac{3x+x^3}{1+3x^2}\).
\(f(g(x)) = \log\left(\dfrac{1 + \frac{3x+x^3}{1+3x^2}}{1 – \frac{3x+x^3}{1+3x^2}}\right) = \log\left(\dfrac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3}\right) = \log\left(\dfrac{(1+x)^3}{(1-x)^3}\right) = 3\log\left(\dfrac{1+x}{1-x}\right) = 3f(x)\).
दिया गया है: \(f(x) = \log\left(\dfrac{1+x}{1-x}\right)\) और \(g(x) = \dfrac{3x+x^3}{1+3x^2}\)।
\(f(g(x)) = \log\left(\dfrac{1 + \frac{3x+x^3}{1+3x^2}}{1 – \frac{3x+x^3}{1+3x^2}}\right) = \log\left(\dfrac{(1+x)^3}{(1-x)^3}\right) = 3\log\left(\dfrac{1+x}{1-x}\right) = 3f(x)\)।