Question

If $f\left(x\right)=\frac{1}{2}(3^x+3^{-x}),g\left(x\right)=\frac{1}{2}(3^x-3^{-x})$ then $f\left(x\right)g\left(y\right)+f\left(y\right)g\left(x\right)$ is equal to

• A. $f(x+y)$
• B. $g(x+y)$
• C. $2f\left(x\right)$
• D. $2g\left(x\right)$

Answer 1

The correct option is B $g(x+y)$

$\Rightarrow$$f\left(x\right)g\left(y\right)=\frac{1}{4}(3^x+3^{-x})(3^y-3^{-y})$

$\Rightarrow$$\frac{1}{4}(3^{x+y}+3^{y-x}-3^{x-y}-3^{-x-y})$

$\Rightarrow$$f\left(y\right)g\left(x\right)=\frac{1}{4}(3^{x+y}-3^{y-x}+3^{x-y}-3^{-x-y})$

$\Rightarrow$$f\left(x\right)g\left(y\right)+f\left(y\right)g\left(x\right)=\frac{1}{4}\times 2(3^{x+y}-3^{-x-y})$

$\Rightarrow$$\frac{1}{2}(3^{x+y}-3^{-(x+y)})$

$\Rightarrow$$g(x+y)$