Animated Solution for Mathematics - Functions: Consider two sets $A=\{x\in \mathbb{Z}:|(|x-3|-3)|\le 1\}$ and $B=\{x\in \mathbb{R}-\{1,2\}:\frac{(x-2)(x-4)}{x-1}{\log }_e(|x-2|)=0\}$. Then the number of onto functions $f:A\to B$ is equal to

Welcome, fellow explorer of the mathematical universe! Today, we are going to dissect a problem that might seem like a simple counting exercise at first glance, but it is actually a beautiful tapestry woven with inequalities, domain constraints, and the fundamental principles of combinatorics.

Let us embark on this journey together.

Imagine you are looking at the definition of Set $A$: $A=\{x\in \mathbb{Z}:|(|x-3|-3)|\le 1\}$. It looks like a Russian nesting doll, doesn't it? We have an absolute value inside another absolute value.

To solve this, we must peel the layers one by one. Recall the fundamental property of inequalities involving absolute values: if $|y|\le a$, then $-a\le y\le a$. Applying this to our expression, we can remove the outer modulus:

Now, we want to isolate the term $|x-3|$. By adding $3$ to all parts of the inequality, we get:

This is where the geometry becomes clear. We are looking for integers $x$ such that the distance between $x$ and $3$ is between $2$ and $4$ units. This splits into two distinct regions: $x-3$ must be in the interval $[-4,-2]$ or in the interval $[2,4]$.

Since we are restricted to integers, we can list them out: $x-3\in \{-4,-3,-2,2,3,4\}$. Adding $3$ back to each of these, we find our set $A=\{-1,0,1,5,6,7\}$.

Counting them up, we see that $|A|=6$. We have successfully populated our domain!

Now, let us turn our attention to Set $B$: $B=\{x\in \mathbb{R}-\{1,2\}:\frac{(x-2)(x-4)}{x-1}{\log }_e(|x-2|)=0\}$. This is where many students stumble.

The equation is a product of two parts: a rational expression and a logarithmic term. For the product to be zero, either the rational part must be zero, or the logarithmic part must be zero. However, we must be vigilant about the domain constraint: $x eq 1$ and $x eq 2$.

First, let us look at the rational part:

This gives us potential roots $x=2$ and $x=4$. But wait! Our constraint explicitly forbids $x=2$. So, we must reject $x=2$ and accept only $x=4$.

Next, let us examine the logarithmic part: ${\log }_e(|x-2|)=0$. This implies $|x-2|=1$, which means $x-2=1$ or $x-2=-1$.

This gives us $x=3$ or $x=1$. Again, we must check our constraints. We are told $x eq 1$. Therefore, we must reject $x=1$ and accept only $x=3$.

Thus, our set $B$ contains only two elements: $B=\{3,4\}$. So, $|B|=2$.

We have arrived at the final act. We need to find the number of onto functions $f:A\to B$. We have $|A|=6$ and $|B|=2$.

For any function from $A$ to $B$, each of the $6$ elements in $A$ has $2$ choices in $B$. Thus, the total number of functions is $2^6=64$.

But we are looking for onto functions. An onto function requires that every element in the codomain $B$ is mapped to by at least one element in the domain $A$.

The only way a function is NOT onto is if all elements of $A$ map to only one element of $B$. There are exactly two such cases: one where every element of $A$ maps to $3$, and one where every element of $A$ maps to $4$.

Subtracting these two non-onto cases from our total of $64$, we get $64-2=62$.

And there we have it! The elegance of the solution lies in the careful handling of the domain constraints and the logical subtraction of the non-onto cases. You have navigated the traps and arrived at the correct answer: 62.