Question

If \[{e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,10} \right)\] and \[f\left( x \right) = k \cdot f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)\] then \[k = \]
A.\[0.5\]
B.\[0.6\]
C.\[0.7\]
D.\[0.8\]

Answer 1

Hint: Here, we will find the value of the variable. We will use the exponential formula and logarithmic formula to find the equation of the function and then by equating the equations, we will find the variable which is also a constant of the given function.

Formula Used:
We will use the following formula:
1.The square of the sum of two numbers is given by the formula: \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
2.The square of the difference of two numbers is given by the formula: \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
3.Exponential rule: \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\]
4.Logarithmic Rule: \[\log {a^b} = b\log a\]

Complete step-by-step answer:
We are given that \[{e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,10} \right)\]and \[f\left( x \right) = k \cdot f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)\]
We are given that \[f\left( x \right) = k \cdot f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)\]…………………………………………………..\[\left( 1 \right)\]
Now, we will consider the other equation
 \[ \Rightarrow {e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}}\]………………………………………………………………………………………….\[\left( 2 \right)\]
Now, taking the exponential function to the other side, the exponential function turns into logarithmic function, we get
\[ \Rightarrow f\left( x \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
Now, substituting \[x = \dfrac{{200x}}{{100 + {x^2}}}\]in the above equation, we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log \left( {\dfrac{{10 + \dfrac{{200x}}{{100 + {x^2}}}}}{{10 - \dfrac{{200x}}{{100 + {x^2}}}}}} \right)\]
By cross multiplying the terms in the numerators and in the denominators, we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log \left( {\dfrac{{\dfrac{{10\left( {100 + {x^2}} \right) + 200x}}{{100 + {x^2}}}}}{{\dfrac{{10\left( {100 + {x^2}} \right) - 200x}}{{100 + {x^2}}}}}} \right)\]
By cancelling the denominators in the fractions, we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log \left( {\dfrac{{10\left( {100 + {x^2}} \right) + 200x}}{{10\left( {100 + {x^2}} \right) - 200x}}} \right)\]
By dividing the numerator and the denominators by 10, we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log \left( {\dfrac{{100 + {x^2} + 20x}}{{100 + {x^2} - 20x}}} \right)\]
The square of the sum of two numbers is given by the formula: \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
The square of the difference of two numbers is given by the formula: \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
By using the above Algebraic Identities and simplifying the above equation, we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log \left( {\dfrac{{{{\left( {10 + x} \right)}^2}}}{{{{\left( {10 - x} \right)}^2}}}} \right)\]
By using the exponential rule \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log {\left( {\dfrac{{10 + x}}{{10 - x}}} \right)^2}\]
By using the logarithmic rule \[\log {a^b} = b\log a\], we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = 2\log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]

By rewriting the equation, we get
\[ \Rightarrow f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = 2f\left( x \right)\]
\[ \Rightarrow f\left( x \right) = \dfrac{1}{2}f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)\]…………………………………………………………………..\[\left( 3 \right)\]
By comparing equation \[\left( 1 \right)\] and equation \[\left( 3 \right)\], we get
\[ \Rightarrow k \cdot f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \dfrac{1}{2} \cdot f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)\]
By cancelling the term, we get
\[ \Rightarrow k = \dfrac{1}{2}\]
\[ \Rightarrow k = 0.5\]
Therefore, If \[{e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,10} \right)\] and \[f\left( x \right) = k \cdot f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)\] then \[k = 0.5\].
Thus Option (A) is the correct answer.

Note: We should remember that functions equated should always have the same variables substituted for the given function. We can equate two equations only when the functions on either of the sides are equal. A logarithmic is defined as the power to which number which must be raised to get some values. We know that the logarithmic and exponential are inverses to each other.