Question

Let $f(x)={{a}^{x}}(a>0)$ be written as $f(x)={{f}_{1}}(x)+{{f}_{2}}(x)$, where ${{f}_{1}}(x)$ is an even function and ${{f}_{1}}(x)$ is an odd function. Then ${{f}_{1}}(x+y)+{{f}_{1}}(x-y)$ equals
(a)$2{{f}_{1}}(x){{f}_{1}}(y)$
(b) $2{{f}_{1}}(x){{f}_{2}}(y)$
(c) $2{{f}_{1}}(x+y){{f}_{2}}(x-y)$
(d) $2{{f}_{1}}(x+y){{f}_{1}}(x-y)$

Answer 1

Hint: To solve this question, first we will represent the function $f(x)={{a}^{x}}$ as sum of add and even function. Then, we will find the function ${{f}_{1}}(x)$ and ${{f}_{2}}(x)$. After that, using some exponential properties we will solve the ${{f}_{1}}(x+y)+{{f}_{1}}(x-y)$by doing some simplification and re-arrangement of terms. And after that, we will select the right option to form the question.

Complete step by step answer:
Now, we know that we can represent any function as sum of even and odd function, which means let we have function f(x), then $f(x)=\left( \dfrac{f(x)+f(-x)}{2} \right)+\left( \dfrac{f(x)-f(-x)}{2} \right)$ , where $\dfrac{f(x)+f(-x)}{2}$ is even function and $\dfrac{f(x)-f(-x)}{2}$ is odd function.
Now, in question it is given that $f(x)={{a}^{x}}(a>0)$.
So, we can write${{a}^{x}}=\left( \dfrac{{{a}^{x}}+{{a}^{-x}}}{2} \right)+\left( \dfrac{{{a}^{x}}-{{a}^{-x}}}{2} \right)$, where $\left( \dfrac{{{a}^{x}}+{{a}^{-x}}}{2} \right)$ is even function and $\left( \dfrac{{{a}^{x}}-{{a}^{-x}}}{2} \right)$ is odd function.
Now, we are given in question that $f(x)={{f}_{1}}(x)+{{f}_{2}}(x)$, where ${{f}_{1}}(x)$ is an even function and ${{f}_{1}}(x)$ is an odd function.
So, we can say that
${{f}_{1}}(x)=\left( \dfrac{{{a}^{x}}+{{a}^{-x}}}{2} \right)$ and ${{f}_{1}}(x)=\left( \dfrac{{{a}^{x}}-{{a}^{-x}}}{2} \right)$
Now, we are asked to find ${{f}_{1}}(x+y)+{{f}_{1}}(x-y)$.
So, ${{f}_{1}}(x+y)=\left( \dfrac{{{a}^{x+y}}+{{a}^{-(x+y)}}}{2} \right)$ and ${{f}_{1}}(x-y)=\left( \dfrac{{{a}^{x-y}}+{{a}^{-(x-y)}}}{2} \right)$
So, \[{{f}_{1}}(x+y)+{{f}_{1}}(x-y)=\left( \dfrac{{{a}^{x+y}}+{{a}^{-(x+y)}}}{2} \right)+\left( \dfrac{{{a}^{x-y}}+{{a}^{-(x-y)}}}{2} \right)\]
We know that exponential function holds the property of ${{a}^{x+y}}={{a}^{x}}.{{a}^{y}}$
So, on simplification we get
\[{{f}_{1}}(x+y)+{{f}_{1}}(x-y)=\left( \dfrac{{{a}^{x}}{{a}^{y}}+{{a}^{-x}}{{a}^{-y}}}{2} \right)+\left( \dfrac{{{a}^{x}}{{a}^{-y}}+{{a}^{-x}}{{a}^{y}}}{2} \right)\]
On solving, we get
\[{{f}_{1}}(x+y)+{{f}_{1}}(x-y)=\left( \dfrac{{{a}^{x}}{{a}^{y}}+{{a}^{-x}}{{a}^{-y}}}{2}+\dfrac{{{a}^{x}}{{a}^{-y}}+{{a}^{-x}}{{a}^{y}}}{2} \right)\]
Taking $\dfrac{1}{2}$ common, we get
So, \[{{f}_{1}}(x+y)+{{f}_{1}}(x-y)=\dfrac{1}{2}\left( {{a}^{x}}{{a}^{y}}+{{a}^{-x}}{{a}^{-y}}+{{a}^{x}}{{a}^{-y}}+{{a}^{-x}}{{a}^{y}} \right)\]
We know that, ${{a}^{-1}}=a$
So, \[=\dfrac{1}{2}\left( {{a}^{x}}{{a}^{y}}+\dfrac{1}{{{a}^{x}}{{a}^{y}}}+\dfrac{{{a}^{x}}}{{{a}^{y}}}+\dfrac{{{a}^{y}}}{{{a}^{x}}} \right)\]
On simplifying, we get
\[=\dfrac{1}{2}\left( {{a}^{y}}\left( {{a}^{x}}+\dfrac{1}{{{a}^{x}}} \right)+\dfrac{1}{{{a}^{y}}}\left( {{a}^{x}}+\dfrac{1}{{{a}^{x}}} \right) \right)\]
\[=\dfrac{1}{2}\left( {{a}^{x}}+\dfrac{1}{{{a}^{x}}} \right)\left( {{a}^{y}}+\dfrac{1}{{{a}^{y}}} \right)\]
Multiplying numerator and denominator by 4, we get
\[=\dfrac{4}{2}\left( \dfrac{{{a}^{x}}+{{a}^{-x}}}{2} \right)\left( \dfrac{{{a}^{y}}+{{a}^{-y}}}{2} \right)\]
But we already know that ${{f}_{1}}(x)=\left( \dfrac{{{a}^{x}}+{{a}^{-x}}}{2} \right)$ and ${{f}_{2}}(y)=\left( \dfrac{{{a}^{y}}+{{a}^{-y}}}{2} \right)$
So, $2{{f}_{1}}(x){{f}_{2}}(y)$
Hence, option ( a ) is correct.

Note:
Firstly, always read question twice and dig out all important points and hints which will help in solving the problem. To solve this question, one must know that we can represent any function as sum of even and odd function that is $f(x)=\left( \dfrac{f(x)+f(-x)}{2} \right)+\left( \dfrac{f(x)-f(-x)}{2} \right)$ , where $\dfrac{f(x)+f(-x)}{2}$ is even function and $\dfrac{f(x)-f(-x)}{2}$ is odd function. Also, remember that exponential function holds the property of ${{a}^{x+y}}={{a}^{x}}.{{a}^{y}}$. Try not to make any calculation mistake.