Question

Which of the following is a function whose graph is symmetrical about the origin?

• A. $f\left(x\right)=\left(2^x+2^{-x}\right)$
• B. $f\left(x\right)={\left(\log \left(x+\sqrt{1+x^2}\right)\right)}^2$
• C. $f\left(x+y\right)=f\left(x\right)+f\left(y\right)\forall x,y\in R$
• D. None of these

Answer 1

Answer: (C)

A function whose graph is symmetrical about the origin must be odd.
$\left(2^x+2^{-x}\right)$ is an even function.
Since, $\log \left(x+\sqrt{1+x^2}\right)$ is an odd function,
$\therefore {\left(\log \left(x+\sqrt{1+x^2}\right)]\right)}^2$ is an even function.
If $f\left(x+y\right)=f\left(x\right)+f\left(y\right)\forall x,y\in R$ , then
Put $x=y=0\Rightarrow f\left(0\right)=0$
Now, put $y=-x\Rightarrow f\left(x\right)+f\left(-x\right)=0$
$\therefore f\left(x\right)$ is an odd function