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(C) Given $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$.Dividing by $(x^2 - y^2) = (x+y)(x-y)$,we get $\frac{f(x+y)}{x+y} - \frac{f(x-y)}{x-y

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$\frac{f(4)}{16} 	ext{ sq. units}$

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(C) Given $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$.Dividing by $(x^2 - y^2) = (x+y)(x-y)$,we get $\frac{f(x+y)}{x+y} - \frac{f(x-y)}{x-y} = 4xy$.Let $u = x+y$ and $v = x-y$. Then $x = \frac{u+v}{2}$ and $y = \frac{u-v}{2}$.So $4xy = 4(\frac{u+v}{2})(\frac{u-v}{2}) = u^2 - v^2$.Thus,$\frac{f(u)}{u} - \frac{f(v)}{v} = u^2 - v^2 \Rightarrow \frac{f(u)}{u} - u^2 = \frac{f(v)}{v} - v^2 = c$.Therefore,$f(u) = cu + u^3$. Given $f(1) = 2$,we have $c(1) + 1^3 = 2 \Rightarrow c = 1$.So $f(x) = x + x^3$.The inequality becomes $\frac{|x^3|^{1/3}}{17} + \frac{|y^3|^{1/3}}{2} \le \frac{1}{4} \Rightarrow \frac{|x|}{17} + \frac{|y|}{2} \le \frac{1}{4}$.This represents a rhombus with vertices $(\pm \frac{17}{4}, 0)$ and $(0, \pm \frac{1}{2})$.Area $= 4 	imes 	ext{Area of triangle in first quadrant} = 4 	imes (\frac{1}{2} 	imes \frac{17}{4} 	imes \frac{1}{2}) = \frac{17}{4}$.Since $f(4) = 4 + 4^3 = 68$,we have $\frac{f(4)}{16} = \frac{68}{16} = \frac{17}{4}$.Thus,the area is $\frac{f(4)}{16}$.

Let $f$ be a differentiable function such that $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$ and $f(1) = 2$. Then the area enclosed by $\frac{|f(x) - x|^{1/3}}{17} + \frac{|f(y) - y|^{1/3}}{2} \le \frac{1}{4}$ is

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