Question

A function whose graph is symmetrical about the origin is given by

• A. $f\left(x\right)=e^x+e^{-x}$
• B. $f\left(x\right)={\log }_{e}x$
• C. $f(x+y)=f\left(x\right)+f\left(y\right)$
• D. None of these

Answer 1

The correct option is C $f(x+y)=f\left(x\right)+f\left(y\right)$

Consider $f\left(x\right)=e^x+e^{-x}$
Now $f(-x)=e^{-x}+e^x=f\left(x\right)$
Since
$f(-x)=f\left(x\right)$, hence it is an even function.
Therefore it is symmetric about the y axis.
However $f\left(x\right)$ intersects the y axis at $y=1$.
Hence the above f(x) is symmetric about the point (0,1).
Therefore it is not symmetric about the origin.
Now consider
$f\left(x\right)=ln\left(x\right)$
As $x\to 0$
$f\left(x\right)\to -\infty$
Furthermore the domain of f(x) is $x>0$.
The graph of $f\left(x\right)=ln\left(x\right)$ decreases steeply in the region (0,1) as compared to the interval x>0.
Hence the graph $f\left(x\right)=lnx$ is not symmetric about the origin.
Thirdly consider
$f(x+y)=f\left(x\right)+f\left(y\right)$
Then $f\left(x\right)$ is of the form $f\left(x\right)=\lambda x$ where $\lambda$ is a constant.
Now this represents a straight line passing through the origin, having a slope $\lambda$.
Hence it is symmetric about the origin.