# PLEASE EXPLAIN ME FASTLY33) Let f : R ------ R be a function defined by f(x) = (x-m)/(x-n) , where 'm' is not equal to nA) f is one - one onto B) f is one - one into C) f is many one onto D) f is many one into

A function $\mathrm{f}:\mathrm{A}\to \mathrm{B}$ is onto if for every element b in B, there is an element a in A such that f(a) = b. That is, for every element of the codomain, there is an element of the domain that maps to it under f. In other words, the image set (or range) of f is the entire codomain. A function is into simply if every element of A maps to something in B. In other words, if B is a codomain, then f is into. For a function to be into, it is not necessary for every element in B to have an element in A that maps to it, but if it does, f would be onto. So, onto functions are also into, but into functions aren't necessarily onto.

If for each x ε A there exist only one image y ε B and each y ε B has a unique pre-image x ε A (i.e. no two elements of A have the same image in B), then f is said to be one-one function. Otherwise f is many-to-one function.

So, from these, we can see that the given function $f\left(x\right)=\frac{x-m}{x-n}$ is one-one and into. It would have been onto if the domain of f(x) was defined as $R-\left\{n\right\}$, that is, if the function f was defined as $f:\left(R-\left\{n\right\}\right)\to R$, because if you plot the graph of the function you would ntice that f is undefined at x=n.

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