Answer:
a) f(x) is a homogenous function of degree '1'
b) f(x) is a homogenous function of degree '2'
Step-by-step explanation:
Step(i):-
Homogenous function
If f(x) is a homogenous function of degree 'n' then
[tex]f(k x ,k y) = k^{n} f(x, y)[/tex]
a)
Given [tex]f(x, y) = \sqrt{x^{2} +3xy+2y^{2} }[/tex]
[tex]f(kx, ky) = \sqrt{(kx)^{2} +3kx(ky)+2(ky)^{2} }[/tex]
[tex]f(kx, ky) = \sqrt{(k)^{2} (x)^{2} +3k^{2} (xy)+2(k)^{2} y^{2} }[/tex]
[tex]f(kx, ky) =( k^{2})^{\frac{1}{2} } ( \sqrt{(x)^{2} +3 (xy)+2 y^{2} })[/tex]
f( k x , k y ) = k f( x, y)
∴ f(x) is a homogenous function of degree '1'
Step(ii):-
b)
[tex]f(x, y) = \sqrt{x^{4} +3x^{2} y+5y^{2} x^{2} -2y^{4} }[/tex]
[tex]f(kx, ky) = \sqrt{(k x)^{4} +3(k x)^{2} y+5(k y)^{2} (k x)^{2} -2(k y)^{4} }[/tex]
[tex]f(kx, ky) = (k^{4})^{\frac{1}{2} } \sqrt{( x)^{4} +3( x)^{3} y+5( y)^{2} ( x)^{2} -2( y)^{4} }[/tex]
f(k x , k y ) = k² f( x , y )
∴ f(x) is a homogenous function of degree '2'
