A) \[\frac{{{x}^{2}}-{{y}^{2}}}{8}\]
B) \[\frac{{{x}^{2}}-{{y}^{2}}}{4}\]
C) \[\frac{{{x}^{2}}+{{y}^{2}}}{4}\]
D) \[\frac{{{x}^{2}}-{{y}^{2}}}{2}\]
Correct Answer: A
Solution :
We have, \[f(x+2y,x-2y)=xy\] ... (i) Let\[x+2y=u\]and \[x-2y=v\] Then,\[x=\frac{u+v}{2}\]and\[y=\frac{u-v}{4}\] Substituting the values of\[x\]and y in Eq. (i), we obtain \[f(u,v)=\frac{{{u}^{2}}-{{v}^{2}}}{8}\Rightarrow f(x,y)=\frac{{{x}^{2}}-{{y}^{2}}}{8}\]
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