Functions 1 Question 2
2. 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 $f _2(x)$ is an odd function. Then $f _1(x+y)+f _1(x-y)$ equals
(2019 Main, 8 April II)
(a) $2 f _1(x+y) \cdot f _2(x-y)$
(b) $2 f _1(x+y) \cdot f _1(x-y)$
(c) $2 f _1(x) \cdot f _2(y)$
(d) $2 f _1(x) \cdot f _1(y)$
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Answer:
Correct Answer: 2. (d)
Solution:
- Given, function $f(x)=a^{x}, a>0$ is written as sum of an even and odd functions $f _1(x)$ and $f _2(x)$ respectively.
Clearly, $f _1(x)=\frac{a^{x}+a^{-x}}{2}$ and $f _2(x)=\frac{a^{x}-a^{-x}}{2}$
So, $f _1(x+y)+f _1(x-y)$
$=\frac{1}{2}\left[a^{x+y}+a^{-(x+y)}\right]+\frac{1}{2}\left[a^{x-y}+a^{-(x-y)}\right]$
$=\frac{1}{2} a^{x} a^{y}+\frac{1}{a^{x} a^{y}}+\frac{a^{x}}{a^{y}}+\frac{a^{y}}{a^{x}}$
$=\frac{1}{2} a^{x} a^{y}+\frac{1}{a^{y}}+\frac{1}{a^{x}} \frac{1}{a^{y}}+a^{y}$
$$ \begin{gathered} =\frac{1}{2} \quad a^{x}+\frac{1}{a^{x}} \quad a^{y}+\frac{1}{a^{y}} \\ =2 \frac{a^{x}+a^{-x}}{2} \quad \frac{a^{y}+a^{-y}}{2}=2 f _1(x) \cdot f _1(y) \end{gathered} $$
