Question

The given function $x\tan \left(x^2/y\right)+y$ is a homogeneous function.

• A. True
• B. False

Answer 1

The correct option is B False

To check if the given function $f(x,y)$ is homogeneous or not we first write $f(kx,ky)$. Now if after some algebraic manipulation it is possible to take out $k^n$ as common making all $x$'s and $y$'s free of $k$, then $f(x,y)$ is a homogeneous function.
Let’s write $f(kx,ky)$
$f(x,y)=x\tan \left(x^2/y\right)+yf(kx,ky)=kx\tan \left(k^2{x}^2/ky\right)+ky=kx\tan \left(k{x}^2/y\right)+ky=k\left[x\tan \right(k{x}^2/y)+y]$
Now you cannot take the k inside the argument of tan function out as common. So the given function cannot be written as $k^{n}f(x,y)$.
Just because all $x$'s and $y$'s can’t be made free of k using any manipulation. So this is not a homogeneous function.