If the function $f(x)=\dfrac{{{a}^{x}}+{{a}^{-x}}}{2},\left( a>2 \right),$ then \[f\left( x+y \right)+f\left( x-y \right)\] is equal to
A.\[2f\left( x \right)f\left( y \right)\]
B. \[f\left( x \right)f\left( y \right)\]
C. $\dfrac{f\left( x \right)}{f\left( y \right)}$
D. None of these

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Hint: Find f(x + y) and f(x – y) individually by the given function i.e. $f(x)=\dfrac{{{a}^{x}}+{{a}^{-x}}}{2},$
Try to simplify the relation f(x + y) + f(x – y) and then compare the simplified equation with f(x) and f(y) to get the correct answer.

Here, function f(x) is provided as
$f(x)=\dfrac{{{a}^{x}}+{{a}^{-x}}}{2},\left( a>2 \right).............\left( 1 \right)$
We need to determine the value of function f(x + y) + f(x – y)=?.
As, equation (1) is defined with respect to ‘x’, we can get f(x + y) and f(x – y) by replacing ‘x’ by (x + y) and (x – y) respectively to the given function or equation (1).
Hence, f(x + y) is given as
$f\left( x+y \right)=\dfrac{{{a}^{\left( x+y \right)}}+{{a}^{-(x+y)}}}{2}...........\left( 2 \right)$
And similarly f(x –y) can be written as
$f\left( x-y \right)=\dfrac{{{a}^{\left( x-y \right)}}+{{a}^{-\left( x-y \right)}}}{2}................\left( 3 \right)$
So, f(x + y) + f(x – Y) from equation (2) and (3), we get;
$f\left( x+y \right)+\left( x-y \right)=\dfrac{{{a}^{\left( x+y \right)}}+{{a}^{-\left( x+y \right)}}}{2}+\dfrac{{{a}^{\left( x-y \right)}}+{{a}^{-\left( x-y \right)}}}{2}..............\left( 4 \right)$
Now, we can use following property of surds to simplify equation (4);
  & {{m}^{-a}}=\dfrac{1}{{{m}^{a}}} \\
 & {{m}^{a}}.{{m}^{b}}={{m}^{a+b}} \\
 & \dfrac{{{m}^{a}}}{{{m}^{b}}}={{m}^{a-b}} \\
Hence, we can write equation (4) as
$f\left( x+y \right)+f\left( x-y \right)=\dfrac{{{a}^{x+y}}+\dfrac{1}{{{a}^{x+y}}}}{2}+\dfrac{{{a}^{x-y}}+\dfrac{1}{{{a}^{x-y}}}}{2}$
Now, using the relations ${{m}^{a+b}}={{m}^{a}}{{m}^{b}}\And {{m}^{a-b}}=\dfrac{{{m}^{a}}}{{{m}^{b}}}$, we get the above equation as;
  & f\left( x+y \right)+f\left( x-y \right)=\dfrac{1}{2}\left( {{a}^{x}}.{{a}^{y}}+\dfrac{1}{{{a}^{x}}.{{a}^{y}}}+\dfrac{{{a}^{x}}}{{{a}^{y}}}+\dfrac{{{a}^{y}}}{{{a}^{x}}} \right) \\
 & Or \\
 & f\left( x+y \right)+f\left( x-y \right)=\dfrac{1}{2}\left( {{a}^{x}}{{a}^{y}}+\dfrac{{{a}^{x}}}{{{a}^{y}}}+\dfrac{{{a}^{y}}}{{{a}^{x}}}+\dfrac{1}{{{a}^{x}}{{a}^{y}}} \right) \\
Now, taking ${{a}^{\left( x \right)}}$ common from first two brackets and $\dfrac{1}{{{a}^{x}}}$ from last two brackets, we get;
\[\dfrac{1}{2}\left( {{a}^{x}}\left( {{a}^{y}}+\dfrac{1}{{{a}^{y}}} \right)+\dfrac{1}{{{a}^{x}}}\left( {{a}^{y}}+\dfrac{1}{{{a}^{y}}} \right) \right)\]
Hence, $f\left( x+y \right)+f\left( x-y \right)$ can be written as
$f\left( x+y \right)+f\left( x-y \right)=\left( {{a}^{x}}+\dfrac{1}{{{a}^{x}}} \right)\left( {{a}^{y}}+\dfrac{1}{{{a}^{y}}} \right)$
Now, using property \[\dfrac{1}{{{m}^{n}}}={{m}^{-n}}\] ,
We can simplify above equation as;
$f\left( x+y \right)+f\left( x-y \right)=\dfrac{\left( {{a}^{x}}+{{a}^{-x}} \right)\left( {{a}^{y}}+{{a}^{-y}} \right)}{2}$
Now, let us multiply by ‘2’ in numerator and denominator both ;
$f\left( x+y \right)+f\left( x-y \right)=2\left( \dfrac{{{a}^{x}}+{{a}^{-x}}}{2} \right)\left( \dfrac{{{a}^{y}}+{{a}^{-y}}}{2} \right)........\left( 5 \right)$
Now, we have $f(x)=\dfrac{{{a}^{x}}+{{a}^{-x}}}{2}$from equation (1), hence we can replace $\dfrac{{{a}^{x}}+{{a}^{-x}}}{2}$by $f(x)$and $\dfrac{{{a}^{y}}+{{a}^{-y}}}{2}$by $f(y)$in equation (5), hence we can write equation (5) as
$f\left( x+y \right)+f\left( x-y \right)=2f\left( x \right)f\left( y \right)$
Therefore option (A) is the correct answer.

Note: Another approach for above question would be that one can simplify
$\left( \dfrac{{{a}^{x+y}}+{{a}^{-\left( x+y \right)}}}{2} \right)\And \left( \dfrac{{{a}^{x-y}}+{{a}^{-\left( x-y \right)}}}{2} \right)$ individually first before putting in equation $f\left( x+y \right)+f\left( x-y \right)$.
One can waste his/her time if he/she goes to solve the problem by solving the given options and then compare it with $f\left( x+y \right)+f\left( x-y \right)$. Hence, this procedure will take more time than provided in the solution.
One can apply property of surds as ${{a}^{-n}}=\dfrac{1}{{{a}^{\dfrac{1}{n}}}}\text{ or }{{a}^{\dfrac{1}{n}}}$ which are wrong. Correct property is given as ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. Hence be careful while applying the property of surds in the kinds of problems.