Question

If $f\left(x\right)\mathrm{and}g\left(x\right)$ are two functions of $x$ such that $f\left(x\right)+g\left(x\right)=e^x$ and $f\left(x\right)-g\left(x\right)=e^{-x}$, then:

• A. $f\left(x\right)$ is odd, $g\left(x\right)$ is odd
• B. $f\left(x\right)$ is even, $g\left(x\right)$ is even
• C. $f\left(x\right)$ is even, $g\left(x\right)$ is odd
• D. $f\left(x\right)$ is odd, $g\left(x\right)$ is even

Answer 1

The correct option is C

$f\left(x\right)$ is even, $g\left(x\right)$ is odd

Explanation for the correct option.

Step 1: Find the value of $f\left(x\right)$.

$f\left(x\right)+g\left(x\right)=e^x...\left(1\right)$

$f\left(x\right)-g\left(x\right)=e^{-x}....\left(2\right)$

By adding $\left(1\right)$ and $\left(2\right)$, we get

$\begin{array}{rcl}2f\left(x\right)& =& e^x+e^{-x}\\ & \Rightarrow & f\left(x\right)=\frac{e^x+e^{-x}}{2}\end{array}$

Step 2: Find the value of $g\left(x\right)$.

Substituting the value of $f\left(x\right)$ in $\left(2\right)$, we get

$\begin{array}{rcl}g\left(x\right)& =& \frac{e^x+e^{-x}}{2}-e^{-x}\\ & =& \frac{e^x+e^{-x}-2e^{-x}}{2}\\ & =& \frac{e^x-e^{-x}}{2}\end{array}$

Step 3: Find $f\left(x\right)$ and $g\left(x\right)$ are odd or even.

$\begin{array}{rcl}f\left(-x\right)& =& \frac{e^{-x}+e^x}{2}=f\left(x\right)\end{array}$

So, $f\left(x\right)$ is even function.

$\begin{array}{rcl}g\left(-x\right)& =& \frac{e^{-x}-e^x}{2}=-g\left(x\right)\end{array}$

So, $g\left(x\right)$ is an odd function.

Hence, option C is correct.