Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[x]$ denote the greatest integer function and $f(x)=\max \{1+x+[x],2+x,x+2[x]\}$, $0\le x\le 2$, where $m$ is the number of points in $[0,2]$ where $f$ is not continuous and $n$ be the number of points in $(0,2)$, where $f$ is not differentiable. Then $(m+n{)}^2+2$ is equal to

Unlocking the Greatest Integer Function

Welcome, fellow explorer of mathematics. Today, we are going to dissect a problem that, at first glance, might look like a chaotic mess of brackets and maximum functions. We have $f(x)=\max \{1+x+[x],2+x,x+2[x]\}$ defined on the interval $[0,2]$.

It is easy to feel intimidated by the Greatest Integer Function, $[x]$, but I want you to view it not as a barrier, but as a staircase. A staircase is predictable; it stays flat for a while, then jumps. Our goal is to map this staircase and find where the path breaks (discontinuity) and where it bends sharply (non-differentiability).

Deconstructing the Beast

The Greatest Integer Function, $[x]$, is the heartbeat of this problem. It changes its value only at integers. Therefore, our strategy is simple: we must respect the integers.

We will partition our domain $[0,2]$ into intervals where $[x]$ is constant. This is the secret to taming the beast. We will look at the interval $[0,1)$, the interval $[1,2)$, and finally, the specific point $x=2$.

The Journey Through $[0,1)$

Imagine you are standing in the interval $0\le x<1$. In this region, the greatest integer less than or equal to $x$ is always $0$. Let us substitute this into our function.

Our expressions become: 1. $1+x+0=x+1$ 2. $2+x$ 3. $x+2(0)=x$

Now, we ask ourselves: which of these is the maximum? We are comparing $x+1$, $x+2$, and $x$. For any $x$ in this range, $x+2$ is clearly the largest.

It is always greater than $x+1$ and $x$. Thus, in the interval $[0,1)$, our function simplifies beautifully to: It is just a straight, smooth line. No drama here.

The Journey Through $[1,2)$

Now, let us step onto the next segment: $1\le x<2$. Here, the value of $[x]$ is $1$. Let us substitute this into our original expressions again:

1. $1+x+1=x+2$ 2. $2+x$ 3. $x+2(1)=x+2$

Look at that! The expressions have converged. They are all identical. The maximum of $x+2$, $x+2$, and $x+2$ is, quite simply, $x+2$.

So, for the entire interval $[1,2)$, our function remains $f(x)=x+2$. The line continues seamlessly from the previous interval. We have been walking on the same line $f(x)=x+2$ from $x=0$ all the way to $x=2$.

The Boundary Trap at $x=2$

This is where the problem tests your vigilance. We have analyzed the intervals, but what happens at the very edge, at $x=2$? At this point, $[x]=2$.

Let us plug this into our original expressions: 1. $1+2+2=5$ 2. $2+2=4$ 3. $2+2(2)=6$

We are now comparing the values $5$, $4$, and $6$. The maximum is $6$. So, $f(2)=6$.

Now, let us look at the continuity. As we approached $x=2$ from the left, our function was $f(x)=x+2$. As $x\to 2^{-}$, the limit is $2+2=4$.

But the actual value of the function at $x=2$ is $6$. Since the limit ($4$) does not equal the function value ($6$), there is a jump discontinuity at $x=2$. This means $m=1$.

The Smoothness of the Path

Finally, we look for $n$, the number of points of non-differentiability in the open interval $(0,2)$. In this entire open interval, our function was consistently $f(x)=x+2$.

This is a linear polynomial. Linear functions are the definition of smoothness; they have a constant slope and no sharp corners. Therefore, there are no points of non-differentiability in $(0,2)$. Thus, $n=0$.

We have successfully navigated the terrain. We found $m=1$ and $n=0$. The problem asks us to calculate $(m+n{)}^2+2$.

Substituting our values, we get:

See how the complexity dissolved once we broke it down? Mathematics is not about memorizing formulas; it is about breaking down the problem into manageable, logical steps. You have mastered the staircase today. Keep that analytical mindset, and no function will ever be too intimidating again. The final answer is $3$.