Question

If $f:R\to R$, then which of the following functions have the graph symmetrical about origin ?

• A. $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$
• B. $f\left(x\right)+f\left(y\right)=f(x\sqrt{1-y^2}+y\sqrt{1-x^2})$
• C. $f\left(x\right)+f\left(y\right)=f(x+y)$
• D. $f\left(x\right)+f\left(y\right)=f(x^3{e}^{x+y}+y^3{e}^{x+y})$

Answer 1

Answer: (A), (B), (C), (D)

$f\left(x\right)+f\left(y\right)=f(x^3{e}^{x+y}+y^3{e}^{x+y})$

We know that odd functions are symmetric about the origin, so we need to find which function is odd in the given ones,

$f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$
Replacing $y$ by $-x$, we get
$f\left(x\right)+f(-x)=f\left(0\right)$
Putting $x=y=0,$ we get
$f\left(0\right)+f\left(0\right)=f\left(0\right)$
$\Rightarrow f\left(0\right)=0$
$\therefore f\left(x\right)+f(-x)=0$
Hence, $f\left(x\right)$ is an odd function.

$f\left(x\right)+f\left(y\right)=f(x\sqrt{1-y^2}+y\sqrt{1-x^2})$
Replacing $y$ by $-x$, we get
$f\left(x\right)+f(-x)=f\left(0\right)$
Putting $x=y=0,$ we get
$f\left(0\right)+f\left(0\right)=f\left(0\right)$
$\Rightarrow f\left(0\right)=0$
$\therefore f\left(x\right)+f(-x)=0$
Hence, $f\left(x\right)$ is an odd function.

$f\left(x\right)+f\left(y\right)=f(x+y),\forall x,y\in R$
​​​​​​​Replacing $y$ by $-x$, we get
$f\left(x\right)+f(-x)=f\left(0\right)$
Putting $x=y=0,$ we get
$f\left(0\right)+f\left(0\right)=f\left(0\right)$
$\Rightarrow f\left(0\right)=0$
​​​​​​​$\therefore f\left(x\right)+f(-x)=0$
Hence, $f\left(x\right)$ is an odd function.

$f\left(x\right)+f\left(y\right)=f(x^3{e}^{x+y}+y^3{e}^{x+y})$
​​​​​​​​​​​​​​Replacing $y$ by $-x$, we get
$f\left(x\right)+f(-x)=f\left(0\right)$
Putting $x=y=0,$ we get
$f\left(0\right)+f\left(0\right)=f\left(0\right)$
$\Rightarrow f\left(0\right)=0$
​​​​​​​$\therefore f\left(x\right)+f(-x)=0$
Hence, $f\left(x\right)$ is an odd function.

So, all of the given functions are odd in nature.