Question

If $f(x + y,x - y) = xy$, then the arithmetic mean of $f\left( {x,y} \right)$and $f\left( {y,x} \right)$is
$1)x$
$2)\,y$
$3)\,0$
4) None of these.

Answer 1

Hint: Arithmetic mean can be calculated by adding all the given terms and dividing it by the number of the total number of terms. In this question, the terms are in the form of functions. The given function should be expressed in terms of what we need to find. We will first find the expressions of the function in terms of the variables specified. So, we will make necessary substitutions and find out the answer.

Complete answer:
The given expression is,
\[f(x + y,x - y) = xy\]
So, the variables in the function are specified as $x + y$ and $x - y$ and the value of the function is $xy$. So, the expression of the function must involve product of the two variables in some form. Now, we also know that arithmetic mean of the two variables $x + y$ and $x - y$ is \[\left( {\dfrac{{x + y + x - y}}{2}} \right) = x\].
Similarly, we also get, \[\left( {\dfrac{{x + y - \left( {x - y} \right)}}{2}} \right) = \left( {\dfrac{{x + y - x + y}}{2}} \right) = y\].
So, we get the expression as,
\[f(x + y,x - y) = \left( {\dfrac{{\left( {x + y} \right) + \left( {x - y} \right)}}{2}} \right)\left( {\dfrac{{\left( {x + y} \right) - \left( {x - y} \right)}}{2}} \right)\].
Now, let us replace $x$ by $\dfrac{{x + y}}{2}$ and replace $y$by $\dfrac{{x - y}}{2}$
This will give us,
\[f\left( {x,y} \right) = \left( {\dfrac{{\left( {\dfrac{{\left( {x + y} \right)}}{2} + \dfrac{{\left( {x - y} \right)}}{2}} \right) + \left( {\dfrac{{\left( {x + y} \right)}}{2} - \dfrac{{\left( {x - y} \right)}}{2}} \right)}}{2}} \right)\left( {\dfrac{{\left( {\dfrac{{\left( {x + y} \right)}}{2} + \dfrac{{\left( {x - y} \right)}}{2}} \right) - \left( {\dfrac{{\left( {x + y} \right)}}{2} - \dfrac{{\left( {x - y} \right)}}{2}} \right)}}{2}} \right)\]
Now, we will simplify the expression,
\[ \Rightarrow f\left( {x,y} \right) = \left( {\dfrac{{\left( {\dfrac{{2x}}{2}} \right) + \left( {\dfrac{{2y}}{2}} \right)}}{2}} \right)\left( {\dfrac{{\left( {\dfrac{{2x}}{2}} \right) - \left( {\dfrac{{2y}}{2}} \right)}}{2}} \right)\]
\[ \Rightarrow f\left( {x,y} \right) = \left( {\dfrac{{x + y}}{2}} \right)\left( {\dfrac{{x - y}}{2}} \right)\]
Using the algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$, we get,
\[ \Rightarrow f\left( {x,y} \right) = \left( {\dfrac{{{x^2} - {y^2}}}{4}} \right)\]
So, $f\left( {x,y} \right) = \dfrac{{{x^2} - {y^2}}}{4}$
Now, let us replace $x$ by $\dfrac{{y - x}}{2}$ and replace $y$by $\dfrac{{x + y}}{2}$.
This will give us,
\[f\left( {y,x} \right) = \left( {\dfrac{{\left( {\dfrac{{\left( {y - x} \right)}}{2} + \dfrac{{\left( {x + y} \right)}}{2}} \right) + \left( {\dfrac{{\left( {y - x} \right)}}{2} - \dfrac{{\left( {x + y} \right)}}{2}} \right)}}{2}} \right)\left( {\dfrac{{\left( {\dfrac{{\left( {y - x} \right)}}{2} + \dfrac{{\left( {x + y} \right)}}{2}} \right) - \left( {\dfrac{{\left( {y - x} \right)}}{2} - \dfrac{{\left( {x + y} \right)}}{2}} \right)}}{2}} \right)\]
Simplifying the expression, we get,
\[ \Rightarrow f\left( {y,x} \right) = \left( {\dfrac{{\left( {\dfrac{{2y}}{2}} \right) + \left( {\dfrac{{ - 2x}}{2}} \right)}}{2}} \right)\left( {\dfrac{{\left( {\dfrac{{2y}}{2}} \right) - \left( {\dfrac{{ - 2x}}{2}} \right)}}{2}} \right)\]
\[ \Rightarrow f\left( {y,x} \right) = \left( {\dfrac{{y - x}}{2}} \right)\left( {\dfrac{{y + x}}{2}} \right)\]
\[ \Rightarrow f\left( {y,x} \right) = \left( {\dfrac{{{y^2} - {x^2}}}{4}} \right)\]
Now, we can find the arithmetic mean of these functions by adding them and dividing by two since there are only two terms involved.
Therefore, the arithmetic mean will be
$\dfrac{{f\left( {x,y} \right) + f\left( {y,x} \right)}}{2} = \dfrac{{\dfrac{{{x^2} - {y^2}}}{4} + \dfrac{{{y^2} - {x^2}}}{4}}}{2}$
All the terms cancel out each other.
Therefore, the final answer is
$\dfrac{{f\left( {x,y} \right) + f\left( {y,x} \right)}}{2} = 0$

Hence, option (3) is the right answer.

Additional Information:
The arithmetic mean is also called the average or mean. It can be calculated for two or more terms. It is the simplest calculation while finding out the average of any parameter. For example, average marks, average cost, etc.

Note:
To find the arithmetic mean in the given question, we first need to define the functions properly. So, find out the values of those functions first and then find the arithmetic mean. The values of the functions are a little confusing, so be careful while taking the LCM and solving to get to the final answer.