# 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

Last updated date: 17th Mar 2023

•

Total views: 303.6k

•

Views today: 3.85k

Answer

Verified

303.6k+ views

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.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?