Answer : A
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An Element $z \in N$ is called an Identity element for $f$ if $f(a,z) = f(z,a) = a, \forall a \in N$
$\text{I}$ : $f(x,y) = x + y -3$
We will say that $z$ is an identity element if $f(x,z) = x = f(z,x)$, Hence, $x + z -3 = x = z + x -3$
So, We have $z = 3$.
$\text{II}$ : $f(x,y) = \max(x,y)$
We will say that $z$ is an identity element if $f(x,z) = x = f(z,x)$, hence, $\max(x,z) = x = \max(z,x) , \forall x \in N$.
So, We have $z = 0$ (if $N$ consists of $0$)
Or We have $z = 1$ (if $N$ does not consist of $0$)
$\text{NOTE}$ : Whether $0$ belongs to Set of natural numbers or not, is an ambiguous thing. In Number theory, generally we don't consider $0$ as a Natural number. But in Computer Science, $0$ is considered as a Natural number by many authors.
In this question, It doesn't matter whether or not $0$ belong to $N$ or not. We can say that the Set $N$ will have an Identity element in both the cases. (Also See below NOTE 3)
$\text{III}$ : $f(x,y) = x^y$
We will say that $z$ is an identity element if $f(x,z) = x = f(z,x)$, hence, $x^z = x = z^x , \forall x \in N$.
But, we know that $x^y $ is Not necessarily equal to $y^x$. (See below NOTE 2)
Hence, there is No identity element for this Binary Operation.
$\textbf{Some Extra Points}$ :
$\text{NOTE 1: }$
For any Set $A$.
Saying that a Binary Operation $f : A \times A \rightarrow A$, is as good as saying that the Given Set $A$ is Closed under the Binary Operation $*$.
$\text{NOTE 2 : Solutions of the Equation } \bf{x^y = y^x}$
Obvious partial solution is $X = Y$, but there are other solutions like $2,4$ and $4,2$.
In fact, these solutions and $X = Y$ are the only solutions in positive integers. And the only integer solutions are $X = Y$ and $(2, 4), (4, 2), (-2,-4), (-4, -2)$
Refer Proof : http://mathforum.org/library/drmath/view/66166.html
$\text{NOTE 3 : }$
I had said in the Second Binary Operation that $0$ or $1$ could be the Identity element. But after seeing the Third Binary Operation, I can say that Whoever Professor set this Question, must have thought of $0$ as a Non-Natural Number because If $0 \in N$, then $f(0,0) $ will be Indeterminate and $f(0,0) = 0 \,\,or \,\,1$. Which will lead to $f$ Not being an Binary Operation.
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