Question

HELO ❤️
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If f(x) = log (1+x/1-x),-1<x<1,thenf(3x+x^2/1+3x^2) - f(2x/1+x^2) is

a) [f(x)] ^3 b) [f(x)]^2
c) - f(x) d) f(x)

Find the correct option nd explain it step by step :
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Answer 1

Step-by-step explanation:

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Answer 2

Question:

If  $f(x)=\log\left(\dfrac{1+x}{1-x}\right)$  where  $-1<x<1,$

then  $f\left(\dfrac{3x+x^3}{1+3x^2}\right)-f\left(\dfrac{2x}{1+x^2}\right)\ =\ ?$

Solution:

We have  $f(x)\ =\ \log\left(\dfrac{1+x}{1-x}\right).$

So,

$\begin{aligned}&f\left(\dfrac{3x+x^3}{1+3x^2}\right)\\ \\ \Longrightarrow\ \ &\log\left(\dfrac{1+\left(\dfrac{3x+x^3}{1+3x^2}\right)}{1-\left(\dfrac{3x+x^3}{1+3x^2}\right)}\right)\\ \\ \Longrightarrow\ \ &\log\left(\dfrac{\left(\dfrac{1+3x+3x^2+x^3}{1+3x^2}\right)}{\left(\dfrac{1-3x+3x^2-x^3}{1+3x^2}\right)}\right)\\ \\ \Longrightarrow\ \ &\log\left(\dfrac{(1+x)^3}{(1-x)^3}\right)\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\log\left(\dfrac{1+x}{1-x}\right)^3\\ \\ \Longrightarrow\ \ &3\log\left(\dfrac{1+x}{1-x}\right)\\ \\ \Longrightarrow\ \ &3f(x)\end{aligned}$

And,

$\begin{aligned}&f\left(\dfrac{2x}{1+x^2}\right)\\ \\ \Longrightarrow\ \ &\log\left(\dfrac{1+\left(\dfrac{2x}{1+x^2}\right)}{1-\left(\dfrac{2x}{1+x^2}\right)}\right)\\ \\ \Longrightarrow\ \ &\log\left(\dfrac{\left(\dfrac{1+2x+x^2}{1+x^2}\right)}{\left(\dfrac{1-2x+x^2}{1+x^2}\right)}\right)\\ \\ \Longrightarrow\ \ &\log\left(\dfrac{(1+x)^2}{(1-x)^2}\right)\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\log\left(\dfrac{1+x}{1-x}\right)^2\\ \\ \Longrightarrow\ \ &2\log\left(\dfrac{1+x}{1-x}\right)\\ \\ \Longrightarrow\ \ &2f(x)\end{aligned}$

Now,

$\begin{aligned}&f\left(\dfrac{3x+x^3}{1+3x^2}\right)-f\left(\dfrac{2x}{1+x^2}\right)\\ \\ \Longrightarrow\ \ &3f(x)-2f(x)\\ \\ \Longrightarrow\ \ &\mathbf{f(x)}\end{aligned}$

Hence the answer is (d) f(x).