A function is on action.
A continuous function f(x) on satisfies the relation where then which of the options are correct?
1. f(x) is an even function.
2. f(x) is an odd function.
3. f(x) is many-one .
4. f(x) is one-one.
5. f(x) is neither even nor odd.
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1 solution
The given condition is f ( x ) + f ( 2 x + y ) + 5 x y = f ( 3 x − y ) + 2 x 2 + 1 At first put x=y=0 in the given equation, then we get f ( 0 ) = 1 . If we put 2 x = − y we get f ( x ) + f ( 0 ) + 5 x × ( − 2 x ) = f ( 5 x ) + 2 x 2 + 1 f ( x ) − f ( 5 x ) = 1 2 x 2 ......... 1
If we put 3 x = y we get f ( x ) + f ( 5 x ) + 1 5 x 2 = f ( 0 ) + 2 x 2 + 1 f ( x ) + f ( 5 x ) = 2 − 1 3 x 2 .............. 2
If we add 1 & 2 we get f ( x ) = 1 − 2 x 2 The result suggests that the function is even as well as many-one.