Functions Ques 28
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) $2 f _1(x) \cdot f _1(y)$
Show Answer
Answer:
Correct Answer: 28.(d)
Solution:
Formula:
Properties Of Even and Odd Function:
- 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, $f _1(x+y)+f _1(x-y)$
$=\frac{1}{2}\left[a^{x+y}+a^{-(x+y)}\right]+\frac{1}{2}\left[a^{x-y}+a^{-(x-y)}\right]$
$=\frac{1}{2} [a^{x} a^{y}+\frac{1}{a^{x} a^{y}}+\frac{a^{x}}{a^{y}}+\frac{a^{y}}{a^{x}}]$
$=\frac{1}{2} [a^{x} (a^{y}+\frac{1}{a^{y}})+\frac{1}{a^{x}} (\frac{1}{a^{y}}+a^{y})]$
$=\frac{1}{2} \quad (a^{x}+\frac{1}{a^{x}} )\quad (a^{y}+\frac{1}{a^{y}})$
$=2 (\frac{a^{x}+a^{-x}}{2}) \quad (\frac{a^{y}+a^{-y}}{2})=2 f _1(x) \cdot f _1(y)$
BETA
