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JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy
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Application +4 | -1
If f(x+ay,x−ay)=axyf(x+a y, x-a y)=a x y, then f(x,y)f(x, y) is equal to
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Correct option: C
x2−y24\frac{x^{2}-y^{2}}{4}
Option C (x2−y24\frac{{x^{2}-y^{2}}}{{4}}) is correct because by substituting u=x+ayu = x+ay and v=x−ayv = x-ay and solving for x and y, then substituting into the original equation, we get f(x,y)=x2−y24f(x, y) = \frac{{x^{2}-y^{2}}}{{4}}.

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