Question

Find the value of $f\left[ {\dfrac{{\left( {3x + {x^3}} \right)}}{{\left( {1 + 3{x^2}} \right)}}} \right] - f\left[ {\dfrac{{\left( {2x} \right)}}{{\left( {1 + {x^2}} \right)}}} \right]$ given that $f\left( x \right) = \log \left[ {\dfrac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}} \right]$where $ - 1 < x < 1$.
A. ${\left[ {f\left( x \right)} \right]^3}$
B. ${\left[ {f\left( x \right)} \right]^2}$
C. $ - f\left( x \right)$
D. $f\left( x \right)$

Answer 1

Hint: First we substitute $x$ by $\dfrac{{\left( {3x + {x^3}} \right)}}{{\left( {1 + 3{x^2}} \right)}}$ and $\dfrac{{\left( {2x} \right)}}{{\left( {1 + {x^2}} \right)}}$ in the given expression, then simplify the given function and after that we use logarithm properties and we get the required answer.

Formula Used:
Substitution process $x$ by $\dfrac{{\left( {3x + {x^3}} \right)}}{{\left( {1 + 3{x^2}} \right)}}$ and $\dfrac{{\left( {2x} \right)}}{{\left( {1 + {x^2}} \right)}}$
logarithm properties $\log {x^n} = n\log x$
Algebraic formulas ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ , ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and
${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ , ${\left( {a - b} \right)^3} = {a^3} + 3{a^2}b - 3a{b^2} + {b^3}$

Complete step by step solution:
Given we have: $f\left( x \right) = \log \left[ {\dfrac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}} \right]$
Now, for $f\left[ {\dfrac{{\left( {3x + {x^3}} \right)}}{{\left( {1 + 3{x^2}} \right)}}} \right] - f\left[ {\dfrac{{\left( {2x} \right)}}{{\left( {1 + {x^2}} \right)}}} \right]$, we have:
$ = \log \left[ {\dfrac{{1 + \left( {\dfrac{{3x + 3{x^3}}}{{1 + 3{x^2}}}} \right)}}{{1 - \left( {\dfrac{{3x + 3{x^3}}}{{1 + 3{x^2}}}} \right)}}} \right] - \log \left[ {\dfrac{{1 + \left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)}}{{1 - \left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)}}} \right]$
Simplifying and we get
$ = \log \left[ {\dfrac{{\dfrac{{1 + 3{x^2} + 3x + 3{x^3}}}{{1 + 3{x^2}}}}}{{\dfrac{{1 + 3{x^2} - 3x + 3{x^3}}}{{1 + 3{x^2}}}}}} \right] - \log \left[ {\dfrac{{\dfrac{{1 + {x^2} + 2x}}{{1 + {x^2}}}}}{{\dfrac{{1 + {x^2} - 2x}}{{1 + {x^2}}}}}} \right]$
Using the algebraic formulas ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ , ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and
${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ , ${\left( {a - b} \right)^3} = {a^3} + 3{a^2}b - 3a{b^2} + {b^3}$ and we get
$ = \log {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^3} - \log {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^2}$
using logarithm formula $\log {x^n} = n\log x$ and we get
$ = 3\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right) - 2\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
$ = \log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
$ = f\left( x \right)$

Option ‘D’ is correct

Note: If you're unsure what a function is, it is a special connection between every element in one set and just one element in another. Both sets, though, must not be empty. The conventional way to express function is $f\left( x \right)$.