Question

Which of the following function/functions have the graph symmetrical about the origin?

• A. $f\left(x\right)$ given by $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$
• B. $f\left(x\right)$ given by $f\left(x\right)+f\left(y\right)=f\left(x\sqrt{1-y^2}+y\sqrt{1-y^2}\right)$
• C. $f\left(x\right)$ given by $f\left(x+y\right)=f\left(x\right)+f\left(y\right)\forall x,y\in R$
• D. None of the above

Answer 1

The correct option is A $f\left(x\right)$ given by $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$

$\left(1\right)$ $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$

when $x$ and $y$ are interchanged.

$f\left(y\right)+f\left(x\right)=f\left(\frac{x+y}{1-xy}\right)$

Thus equation remains unchanged. Hence the graph is symmetrical about the origin.

$\left(2\right)$ $f\left(x\right)+f\left(y\right)=f(x\sqrt{1-y^2}+y\sqrt{1-y^2})$

when $x$ and $y$ are interchanged.

$f\left(y\right)+f\left(x\right)=f(y\sqrt{1-x^2}+x\sqrt{1-x^2})$

Since equation is altered. Hence the graph is not symmetrical about the origin.

$\left(3\right)$ $f(x+y)=f\left(x\right)+f\left(y\right)$

when $x$ and $y$ are interchanged.

$f(y+x)=f\left(y\right)+f\left(x\right)$

Since equation remains unchanged. So graph is symmetrical about origin.