Question

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

• A. True
• B. False

Answer 1

The correct option is B False

What is a homogeneous function? OK. We will recall the concept. 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 xs and ys free of k, then f(x,y) is a homogeneous function.
Let’s write f(kx, ky)
$f(x,y)=xtan\left(x^2/y\right)+yf(kx,ky)=kxtan\left(k^2{x}^2/ky\right)+ky=kxtan\left(k{x}^2/y\right)+kyk\left[xtan\right(k{x}^2/y)+y]$
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 xs and ys can’t be made free of k using any manipulation. So this is not a homogeneous function.