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Question

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@basant kumar

is it always true f(x)+f(-x)=0 for many to one functions???

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no it is for only odd function .if your able to recognize odd function then we can directly calculate this by this approach.

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f(x)=x i.e odd function

f(x)+(-x)=0 but it is not many one.

???

Answer 1

h(x) =x³-x  i.e eg h(1)=h(-1)=0

g(x)=x²sinx i.e whenever sinx become 0

f(x)=lnx+x lets assume many to one

then lnx1+x1=lnx2+x2 =>lnx1-lnx2=x2-x1

=>ln(x1/x2)=x2-x1

=>x1/x2=e^{(x2-x1)     [Eq 1]}

As x1!=x2 so assume x1>x2 LHS of Eq1 become >1 but RHS <1

X2<X1 not possible LHS<1 but RHS>1,So f(x) is not many to one.

Comments

As x1!=x2

why? 

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@Gate fever

we assumed that its many to one i.e F[x1]=F[x2] and x1 x2 different if same it leads that its not many to one.