f(x, y) = frac{1}{x + y} is a homogeneous fun | vedclass

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(B) function $f(x, y)$ is said to be homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$.Given $f(x, y) = \frac{1}{x + y}$.Substitute $x$ with $tx$

Vedclass · Accepted Answer

$-1$

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(B) function $f(x, y)$ is said to be homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$.Given $f(x, y) = \frac{1}{x + y}$.Substitute $x$ with $tx$ and $y$ with $ty$:$f(tx, ty) = \frac{1}{tx + ty} = \frac{1}{t(x + y)} = t^{-1} \cdot \frac{1}{x + y} = t^{-1} f(x, y)$.Comparing this with $f(tx, ty) = t^n f(x, y)$,we get $n = -1$.Therefore,the function is homogeneous of degree $-1$.

$f(x, y) = \frac{1}{x + y}$ is a homogeneous function of degree

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