Question

If $f\left(x\right)=\cosh x-\sinh x$, then $f\left(x_1+x_2+...+x_n\right)$ is equal to

• A. $f\left(x_1\right)·f\left(x_2\right)···f\left(x_n\right)$
• B. $f\left(x_1\right)+f\left(x_2\right)+...+f\left(x_n\right)$
• C. $0$
• D. $1$

Answer 1

The correct option is A

$f\left(x_1\right)·f\left(x_2\right)···f\left(x_n\right)$

Explanation for the correct option.

Step 1. Simplify the given function.

Writing $\cosh x$ and $\sinh x$ in terms of exponential function, the function $f\left(x\right)=\cosh x-\sinh x$ can be written as:

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

Step 2. Find the value of $f\left(x_1+x_2+...+x_n\right)$.

For the function $f\left(x\right)=e^{-x}$, the value of $f\left(x_1+x_2+...+x_n\right)$ is given as:

$\begin{array}{rcl}f\left(x_1+x_2+...+x_n\right)& =& e^{-\left(x_1+x_2+...+x_n\right)}\\ & =& e^{-x_1}{e}^{-x_2}...e^{-x_n}\\ & =& f\left(x_1\right)·f\left(x_2\right)···f\left(x_n\right)\end{array}$

Hence, the correct option is A.