Question

If \(\mathrm{f}(x)=\log \left(\frac{1+x}{1-x}\right)\) and \(\mathrm{g}(x)=\frac{3 x+x^3}{1+3 x^2}\), then \((\mathrm{fog})(x)=\)

• A. \(2\mathrm{f}(x)\)
• B. \(3\mathrm{f}(x)\)
• C. \(4\mathrm{f}(x)\)
• D. \(-\mathrm{f}(x)\)

Hint:

Substitute g(x) into f and simplify using the cube expansions of (1+x) and (1-x).

Answer 1

The Correct Option is B

Step 1: Substitute
$f(g) = \log\dfrac{1+g}{1-g}$.

Step 2: Check with a value
Take $x = \tfrac{1}{2}$: $g = \dfrac{1.5 + 0.125}{1 + 0.75} = \dfrac{13}{14}$, so $\dfrac{1+g}{1-g} = 27$, while $\dfrac{1+x}{1-x} = 3$ and $3^3 = 27$.

Step 3: Conclude
This confirms $f(g(x)) = 3f(x)$.

Final Answer:
The result is 3f(x). \[ \boxed{\text{(B) }3\mathrm{f}(x)} \]