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If `f(x)=(h_1(x)-h_1(-x))(h_2(x)-h_2(-x))(h_(2n+1)(-x)a n df(200)=0,` then prove that `f(x)` is many one function.

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If `f(x)=(h_1(x)-h_1(-x))(h_2(x)-h_2(-x))(h_(2n+1)(-x)a n df(200)=0,` then prove that `f(x)` is many one function.
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`f(-x)=(h_(1)(-x)-h_(1)(x))(h_(2)(-x)-h_(2)(x))…(h_(2n+1)(-x)-h_(2n+1)(x))`
` :. f(-x)=(-1)^(2n+1)f(x)=-f(x)`
or `f(x)+f(-x)=0`
So, f(x) is odd. Therefore,
`f(-200)=-f(200)=0`
So, f(x) is many-one.
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