A) when only \[p=q\]
B) when only \[p\ne q\]
C) when only \[p+q=0\]
D) for all \[p,\,\,q\,\,\in R\]
Correct Answer: D
Solution :
We have, \[f\left( \frac{x}{2008}+2009y,\,\frac{x}{2008}-2009y \right)=xy\] \[=1004\,\left( \frac{x}{2008}+2009y+\frac{x}{2008}-2009\,y \right)\] \[\times \frac{1}{2\times 2009}\,\left( \frac{x}{2008}+2009y-\frac{x}{2008}+2009y \right)\] \[\Rightarrow \] \[f\,(p,\,q)\,=\frac{502}{2009}\,(p+q)\,(p-q)\] \[=\frac{502}{2009}\,({{p}^{2}}-{{q}^{2}})\] \[\therefore \] \[f(p,\,\,q)+f(q,\,\,p)\] \[=\frac{502}{2009}\,({{p}^{2}}-{{q}^{2}}+{{q}^{2}}-{{p}^{2}})=0\] for all \[p,\,\,q\,\in \,R\].
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