Question

Let \[f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}\] then find the value of \[f\left( x \right) + f\left( {1 - x} \right)\].
A. 0
B. 1
C. - 1
D. None of these

Answer 1

Hint: In this question, we need to find the value of \[f\left( x \right) + f\left( {1 - x} \right)\]. First, we have to find \[f\left( {1 - x} \right)\] by putting \[x = 1 - x\] in the function \[f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}\]. After that, we will add both the functions such as \[f\left( x \right)\] and newly obtained \[f\left( {1 - x} \right)\].

Complete step-by-step answer:
We know that \[f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}\]
Now, let us find the value of \[f\left( {1 - x} \right)\]
For this, put \[x = 1 - x\] in the function \[f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}\].
Thus, we get
\[f\left( {1 - x} \right) = \dfrac{{{4^{\left( {1 - x} \right)}}}}{{{4^{\left( {1 - x} \right)}} + 2}}\]
Now, we will apply the exponential identity which is shown below.
\[{a^{m - n}} = {a^m} \times {a^{ - n}} \Rightarrow \dfrac{{{a^m}}}{{{a^n}}}\]
Thus, we get
\[
   \Rightarrow f\left( {1 - x} \right) = \dfrac{{{4^1} \times {4^{ - x}}}}{{{4^1} \times {4^{ - x}} + 2}} \\
   \Rightarrow f\left( {1 - x} \right) = \dfrac{{\dfrac{4}{{{4^x}}}}}{{\left( {\dfrac{4}{{{4^x}}}} \right) + 2}} \\
   \Rightarrow f\left( {1 - x} \right) = \dfrac{{\dfrac{4}{{{4^x}}}}}{{\left( {\dfrac{{4 + 2 \times {4^x}}}{{{4^x}}}} \right)}} \\
 \]
By simplifying further, we get
\[
   \Rightarrow f\left( {1 - x} \right) = \dfrac{4}{{{4^x}}} \times \dfrac{{{4^x}}}{{4 + 2 \times {4^x}}} \\
   \Rightarrow f\left( {1 - x} \right) = \dfrac{4}{{4 + 2 \times {4^x}}} \\
   \Rightarrow f\left( {1 - x} \right) = \dfrac{4}{{2\left( {2 + {4^x}} \right)}} \\
   \Rightarrow f\left( {1 - x} \right) = \dfrac{2}{{\left( {2 + {4^x}} \right)}} \\
 \]
Now, we will find the value of \[f\left( x \right) + f\left( {1 - x} \right)\] by adding both the functions.
So, we get
\[
  f\left( x \right) + f\left( {1 - x} \right) = \dfrac{{{4^x}}}{{{4^x} + 2}} + \dfrac{2}{{\left( {2 + {4^x}} \right)}} \\
   = \dfrac{{{4^x}}}{{\left( {2 + {4^x}} \right)}} + \dfrac{2}{{\left( {2 + {4^x}} \right)}} \\
   = \dfrac{{2 + {4^x}}}{{\left( {2 + {4^x}} \right)}} \\
   = 1 \\
 \]
Hence, the value of \[f\left( x \right) + f\left( {1 - x} \right)\] is 1 if \[f\left( x \right) = \dfrac{{{4^x}}}{{{4^x} + 2}}\].
Therefore, the correct option is (B).

Additional Information: A function is defined as an expression, principle, or statement in mathematics that describes a connection among one variable (the independent variable) as well as another variable (the dependent variable). Functions are common in mathematics and are required for the formulation of physical connections in the fields of science. Every input in a function has just one outcome. A function's group of input values has one name, while its group of output values has another. The collection of input data is referred to as the function's domain. The collection of output values is referred to as the function's range.

Note: Many students make mistakes in the simplification part and applying properties of exponential terms. They may get wrong in simplifying the exponential functions So, so they will not get the desired result.