f: R--> R f(x) = sin^2 (x)  +  cos^2 (x) to show: f is not a surjection  



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embizze's profile pic

Posted on (Answer #1)

Show that `f: RR->RR,f(x)=sin^2(x)+cos^2(x)` is not surjective:

Let X be the domain and Y the codomain (both of which are the set of reals in this case.) A function mapping from the domain to the codomain is surjective if for every `y in Y` there exists an `x in X` such that f(x)=y.

We need only show a counterexample to show that f is not surjective. Let y=2. There is no `x in RR` such that f(x)=2. (f(x) is identically 1 -- `sin^2(x)+cos^2(x)=1` for all `x in RR` )

Therefore the function is not surjective.

** Change to `f:RR ->{1}` and the function is surjective (onto).**

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rocky.goyal's profile pic

Posted on (Reply #1)

thank you sir...

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