A function \(F(x, y)\) is homogeneous of degree \(n\) if \(F(tx, ty) = t^n F(x, y)\) for some constant \(n\).
Let's check each option:
(A) \(F(x, y) = \cos x - \sin(\frac{y}{x})\)
\(F(tx, ty) = \cos(tx) - \sin(\frac{ty}{tx}) = \cos(tx) - \sin(\frac{y}{x})\). This is not of the form \(t^n F(x, y)\) because of the \(\cos(tx)\) term. Therefore, this is not a homogeneous function.
(B) \(F(x, y) = \frac{y}{x}\)
\(F(tx, ty) = \frac{ty}{tx} = \frac{y}{x} = t^0 \frac{y}{x}\). This is homogeneous of degree 0.
(C) \(F(x, y) = \frac{x^2 + y^2}{xy}\)
\(F(tx, ty) = \frac{(tx)^2 + (ty)^2}{(tx)(ty)} = \frac{t^2x^2 + t^2y^2}{t^2xy} = \frac{t^2(x^2 + y^2)}{t^2xy} = \frac{x^2 + y^2}{xy} = t^0 \frac{x^2 + y^2}{xy}\). This is homogeneous of degree 0.
(D) \(F(x, y) = \cos^2(\frac{x}{y})\)
\(F(tx, ty) = \cos^2(\frac{tx}{ty}) = \cos^2(\frac{x}{y}) = t^0 \cos^2(\frac{x}{y})\). This is homogeneous of degree 0.
Therefore, the function that is not homogeneous is option (A).
Correct Answer: cos~x-sin(y/x)
A function \(F(x, y)\) is homogeneous of degree \(n\) if \(F(tx, ty) = t^n F(x, y)\) for some constant \(n\).
Let's check each option:
(A) \(F(x, y) = \cos x - \sin(\frac{y}{x})\)
\(F(tx, ty) = \cos(tx) - \sin(\frac{ty}{tx}) = \cos(tx) - \sin(\frac{y}{x})\). This is not of the form \(t^n F(x, y)\) because of the \(\cos(tx)\) term. Therefore, this is not a homogeneous function.
(B) \(F(x, y) = \frac{y}{x}\)
\(F(tx, ty) = \frac{ty}{tx} = \frac{y}{x} = t^0 \frac{y}{x}\). This is homogeneous of degree 0.
(C) \(F(x, y) = \frac{x^2 + y^2}{xy}\)
\(F(tx, ty) = \frac{(tx)^2 + (ty)^2}{(tx)(ty)} = \frac{t^2x^2 + t^2y^2}{t^2xy} = \frac{t^2(x^2 + y^2)}{t^2xy} = \frac{x^2 + y^2}{xy} = t^0 \frac{x^2 + y^2}{xy}\). This is homogeneous of degree 0.
(D) \(F(x, y) = \cos^2(\frac{x}{y})\)
\(F(tx, ty) = \cos^2(\frac{tx}{ty}) = \cos^2(\frac{x}{y}) = t^0 \cos^2(\frac{x}{y})\). This is homogeneous of degree 0.
Therefore, the function that is not homogeneous is option (A).
Correct Answer: cos~x-sin(y/x)
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