# Given $f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\ and\ g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}},fog\left( x \right)$equals

A. $-f\left( x \right)$

B. $3f\left( x \right)$

C. ${{\left[ f\left( x \right) \right]}^{3}}$

D. None of these

Answer

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Hint: First of all, to find $fg\left( x \right)$, substitute $x=g\left( x \right)$ in $f\left( x \right)$ that it $fg\left( x \right)=\log \dfrac{1+g\left( x \right)}{1-g\left( x \right)}$. Then here put the value of $g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}}$ and simplifying the equation. Finally, find $fg\left( x \right)$in terms of $f\left( x \right)$.

Complete step-by-step answer:

Here we are given that $f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\ and\ g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}}$. We have to find the value of $fog\left( x \right)$.

Let us first take our given functions that are,

$\begin{align}

& f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\ \\

& and\ g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}} \\

\end{align}$

Here we have to find the value of $fog\left( x \right)$or $f\left( g\left( x \right) \right)$.

To find the value of $f\left( g\left( x \right) \right)$, we will have to substitute $x=g\left( x \right)$in expression of $f\left( x \right)$.

So by substituting the value of $x=g\left( x \right)$in expression of $f\left( x \right)$, we get,

$fg\left( x \right)=\log \left[ \dfrac{1+g\left( x \right)}{1-g\left( x \right)} \right]$……………… (1)

As we are given that $g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}}$, by substituting the value of $g\left( x \right)$in equation (1), we get,

$fg\left( x \right)=\log \left[ \dfrac{1+\left( \dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}} \right)}{1-\left( \dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}} \right)} \right]$

By simplifying the above equation, we get,

$fg\left( x \right)=\log \left[ \dfrac{\dfrac{\left( 1+3{{x}^{2}} \right)+\left( 3x+{{x}^{3}} \right)}{\left( 1+3{{x}^{2}} \right)}}{\dfrac{\left( 1+3{{x}^{2}} \right)-\left( 3x+{{x}^{3}} \right)}{\left( 1+3{{x}^{2}} \right)}} \right]$

By cancelling the like terms, we get,

$fg\left( x \right)=\log \left[ \dfrac{1+3{{x}^{2}}+3x+{{x}^{3}}}{1+3{{x}^{2}}-3x+{{x}^{3}}} \right]$

We can also write the above equation as,

$fg\left( x \right)=\log \left[ \dfrac{{{\left( 1 \right)}^{3}}+{{\left( x \right)}^{3}}+3x\left( 1+x \right)}{{{\left( 1 \right)}^{3}}-{{\left( x \right)}^{3}}-3x\left( 1-x \right)} \right]$

We know that, ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)and\ {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$. Applying these in above equation by taking a = 1 and b = x, we get,

$fg\left( x \right)=\log \left[ \dfrac{{{\left( 1+x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}} \right]$

We know that $\dfrac{{{a}^{n}}}{{{b}^{n}}}={{\left( \dfrac{a}{b} \right)}^{n}}$. By applying this in above equation, we get,

$fg\left( x \right)=\log \left[ {{\left( \dfrac{1+x}{1-x} \right)}^{3}} \right]$

We know that $\log {{a}^{m}}=m\log a$. By applying this in above equation, we get,

$fg\left( x \right)=3\log \left( \dfrac{1+x}{1-x} \right)$

As we are given that $\log \left( \dfrac{1+x}{1-x} \right)=f\left( x \right)$, by substituting the value in above equation, we get,

$fg\left( x \right)=3f\left( x \right)$

Therefore we get $fg\left( x \right)\ or\ fog\left( x \right)\ as\ 3f\left( x \right)$.

Hence option (B) is correct.

Note: Students often confuse between $fog\left( x \right)\ and\ gof\left( x \right)$ and some students even consider them as the same functions. But actually they are different functions. $fog\left( x \right)\ or\ fg\left( x \right)$means, we have to put $x=g\left( x \right)$ in expression of $f\left( x \right)$ whereas $gof\left( x \right)\ and\ gf\left( x \right)$means, we have to put $x=f\left( x \right)$ in expression of $g\left( x \right)$. In some cases, by completely solving $fg\left( x \right)\ and\ gf\left( x \right)$, they may come out to be the same, but generally they are different functions. Also if $fg\left( x \right)\ =gf\left( x \right)$then we can conclude that $f\left( x \right)$and $g\left( x \right)$are inverse of each other.

Complete step-by-step answer:

Here we are given that $f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\ and\ g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}}$. We have to find the value of $fog\left( x \right)$.

Let us first take our given functions that are,

$\begin{align}

& f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\ \\

& and\ g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}} \\

\end{align}$

Here we have to find the value of $fog\left( x \right)$or $f\left( g\left( x \right) \right)$.

To find the value of $f\left( g\left( x \right) \right)$, we will have to substitute $x=g\left( x \right)$in expression of $f\left( x \right)$.

So by substituting the value of $x=g\left( x \right)$in expression of $f\left( x \right)$, we get,

$fg\left( x \right)=\log \left[ \dfrac{1+g\left( x \right)}{1-g\left( x \right)} \right]$……………… (1)

As we are given that $g\left( x \right)=\dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}}$, by substituting the value of $g\left( x \right)$in equation (1), we get,

$fg\left( x \right)=\log \left[ \dfrac{1+\left( \dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}} \right)}{1-\left( \dfrac{3x+{{x}^{3}}}{1+3{{x}^{2}}} \right)} \right]$

By simplifying the above equation, we get,

$fg\left( x \right)=\log \left[ \dfrac{\dfrac{\left( 1+3{{x}^{2}} \right)+\left( 3x+{{x}^{3}} \right)}{\left( 1+3{{x}^{2}} \right)}}{\dfrac{\left( 1+3{{x}^{2}} \right)-\left( 3x+{{x}^{3}} \right)}{\left( 1+3{{x}^{2}} \right)}} \right]$

By cancelling the like terms, we get,

$fg\left( x \right)=\log \left[ \dfrac{1+3{{x}^{2}}+3x+{{x}^{3}}}{1+3{{x}^{2}}-3x+{{x}^{3}}} \right]$

We can also write the above equation as,

$fg\left( x \right)=\log \left[ \dfrac{{{\left( 1 \right)}^{3}}+{{\left( x \right)}^{3}}+3x\left( 1+x \right)}{{{\left( 1 \right)}^{3}}-{{\left( x \right)}^{3}}-3x\left( 1-x \right)} \right]$

We know that, ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)and\ {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$. Applying these in above equation by taking a = 1 and b = x, we get,

$fg\left( x \right)=\log \left[ \dfrac{{{\left( 1+x \right)}^{3}}}{{{\left( 1-x \right)}^{3}}} \right]$

We know that $\dfrac{{{a}^{n}}}{{{b}^{n}}}={{\left( \dfrac{a}{b} \right)}^{n}}$. By applying this in above equation, we get,

$fg\left( x \right)=\log \left[ {{\left( \dfrac{1+x}{1-x} \right)}^{3}} \right]$

We know that $\log {{a}^{m}}=m\log a$. By applying this in above equation, we get,

$fg\left( x \right)=3\log \left( \dfrac{1+x}{1-x} \right)$

As we are given that $\log \left( \dfrac{1+x}{1-x} \right)=f\left( x \right)$, by substituting the value in above equation, we get,

$fg\left( x \right)=3f\left( x \right)$

Therefore we get $fg\left( x \right)\ or\ fog\left( x \right)\ as\ 3f\left( x \right)$.

Hence option (B) is correct.

Note: Students often confuse between $fog\left( x \right)\ and\ gof\left( x \right)$ and some students even consider them as the same functions. But actually they are different functions. $fog\left( x \right)\ or\ fg\left( x \right)$means, we have to put $x=g\left( x \right)$ in expression of $f\left( x \right)$ whereas $gof\left( x \right)\ and\ gf\left( x \right)$means, we have to put $x=f\left( x \right)$ in expression of $g\left( x \right)$. In some cases, by completely solving $fg\left( x \right)\ and\ gf\left( x \right)$, they may come out to be the same, but generally they are different functions. Also if $fg\left( x \right)\ =gf\left( x \right)$then we can conclude that $f\left( x \right)$and $g\left( x \right)$are inverse of each other.

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