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Let ƒ be a real valued function such that
`f(x)+2f(2002/x)`=3x for all x>0. the value of f(2)is
A. |
1000 |
B. |
2000 |
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C. |
3000 |
D. |
4000 |
Putting x = 2, we get
ƒ(2) + 2ƒ(1001) = 6 ..........(i)
Putting x = 1001, we get ƒ(1001) + 2ƒ(2) = 3003 .........(ii)
Solving for ƒ(2), we get ƒ(2) = 2000