Key Concepts and Formulas

• Greatest Integer Function [t]: The greatest integer less than or equal to $t$. For example, $[3.7] = 3$, $[-2.1] = -3$, $[5] = 5$.
• Continuity: A function $f(x)$ is continuous at a point $x=a$ if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.
• Differentiability: A function $f(x)$ is differentiable at a point $x=a$ if the left-hand derivative (LHD) equals the right-hand derivative (RHD) at $x=a$.
  • LHD at $x=a$: $\lim_{h \to 0^-} \frac{f(a-h) - f(a)}{-h}$ or $\lim_{h \to 0^+} \frac{f(a-h) - f(a)}{-h}$
  • RHD at $x=a$: $\lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$
• Limit of $\frac{\sin \theta}{\theta}$: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Step-by-Step Solution

Step 1: Analyze the definition of the function $f(x)$ and simplify it over different intervals.

The function is defined piecewise. We need to determine the value of $[x]$ and $|x|$ in each specified interval.

• Interval 1: $x \in (-2, -1)$ For $x \in (-2, -1)$, we have $[x] = -2$. The first part of the function becomes $\frac{\sin(x - [x])}{x - [x]} = \frac{\sin(x - (-2))}{x - (-2)} = \frac{\sin(x+2)}{x+2}$.

• Interval 2: $|x| < 1$ This interval is equivalent to $x \in (-1, 1)$. This interval can be further split into $x \in (-1, 0]$ and $x \in (0, 1)$.

  • For $x \in (-1, 0]$: $|x| = -x$ (since $x \le 0$). So, $[|x|] = [-x]$. If $x=0$, $[|x|]=[0]=0$. If $x \in (-1, 0)$, then $-x \in (0, 1)$, so $[|x|] = [-x] = 0$. Thus, for $x \in (-1, 0]$, $[|x|] = 0$. The function is $\max\{2x, 3[|x|]\} = \max\{2x, 3(0)\} = \max\{2x, 0\}$. Since $x \in (-1, 0]$, $2x \in (-2, 0]$. Therefore, $\max\{2x, 0\} = 0$.
  • For $x \in (0, 1)$: $|x| = x$ (since $x > 0$). So, $[|x|] = [x]$. For $x \in (0, 1)$, $[x] = 0$. The function is $\max\{2x, 3[|x|]\} = \max\{2x, 3(0)\} = \max\{2x, 0\}$. Since $x \in (0, 1)$, $2x \in (0, 2)$. Therefore, $\max\{2x, 0\} = 2x$.

• Interval 3: Otherwise This means $x \notin (-2, -1)$ and $x \notin (-1, 1)$. So, $x \in (-\infty, -2] \cup [-1, -1] \cup [1, \infty)$. However, the second condition $|x|<1$ implies $x \in (-1, 1)$. The "otherwise" condition covers $x \le -2$ and $x \ge 1$. For these values, $f(x) = 1$.

Combining these, we can rewrite $f(x)$ as: f(x) = \left\{ {\matrix{ \frac{\sin(x+2)}{x+2} & , & {x \in ( - 2, - 1)} \cr 0 & , & {x \in ( - 1,0]} \cr 2x & , & {x \in (0,1)} \cr 1 & , & {x \le -2 \text{ or } x \ge 1} \cr } } \right.

Step 2: Identify potential points of discontinuity.

Points of discontinuity can occur where the definition of the function changes. These are the boundary points of the intervals: $x = -2$, $x = -1$, $x = 0$, and $x = 1$.

• Check continuity at $x = -2$: The function is defined as $f(x) = 1$ for $x \le -2$. So, $f(-2) = 1$. The limit from the right at $x=-2$ is from the interval $(-2, -1)$. $\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} \frac{\sin(x+2)}{x+2}$. Let $\theta = x+2$. As $x \to -2^+$, $\theta \to 0^+$. $\lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1$. Since $\lim_{x \to -2^-} f(x) = f(-2) = 1$ and $\lim_{x \to -2^+} f(x) = 1$, the function is continuous at $x=-2$.

• Check continuity at $x = -1$: We need to check the left-hand limit, right-hand limit, and the function value. The function is defined as $f(x) = 0$ for $x \in (-1, 0]$. So, $f(-1) = 0$. Left-hand limit: $\lim_{x \to -1^-} f(x)$. This is from the interval $(-2, -1)$, where $f(x) = \frac{\sin(x+2)}{x+2}$. $\lim_{x \to -1^-} \frac{\sin(x+2)}{x+2} = \frac{\sin(-1+2)}{-1+2} = \frac{\sin(1)}{1} = \sin(1)$. Right-hand limit: $\lim_{x \to -1^+} f(x)$. This is from the interval $(-1, 0]$, where $f(x) = 0$. $\lim_{x \to -1^+} f(x) = 0$. Since $\lim_{x \to -1^-} f(x) = \sin(1) \ne \lim_{x \to -1^+} f(x) = 0$, the function is discontinuous at $x = -1$.

• Check continuity at $x = 0$: We need to check the left-hand limit, right-hand limit, and the function value. The function is defined as $f(x) = 0$ for $x \in (-1, 0]$. So, $f(0) = 0$. Left-hand limit: $\lim_{x \to 0^-} f(x)$. This is from the interval $(-1, 0]$, where $f(x) = 0$. $\lim_{x \to 0^-} f(x) = 0$. Right-hand limit: $\lim_{x \to 0^+} f(x)$. This is from the interval $(0, 1)$, where $f(x) = 2x$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2x = 2(0) = 0$. Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$, the function is continuous at $x = 0$.

• Check continuity at $x = 1$: We need to check the left-hand limit, right-hand limit, and the function value. The function is defined as $f(x) = 2x$ for $x \in (0, 1)$. The limit from the left is $\lim_{x \to 1^-} 2x = 2(1) = 2$. The function is defined as $f(x) = 1$ for $x \ge 1$. So, $f(1) = 1$. The limit from the right is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 1 = 1$. Since $\lim_{x \to 1^-} f(x) = 2 \ne \lim_{x \to 1^+} f(x) = 1$, the function is discontinuous at $x = 1$.

Therefore, the points of discontinuity are $x = -1$ and $x = 1$. So, $m = 2$.

Step 3: Identify potential points of non-differentiability.

Points of non-differentiability can occur at points of discontinuity and at points where the definition of the function changes within a continuous interval, or where the derivative from the left does not equal the derivative from the right.

From Step 2, we know that $f(x)$ is discontinuous at $x = -1$ and $x = 1$. A function cannot be differentiable at a point where it is discontinuous. So, $f(x)$ is not differentiable at $x = -1$ and $x = 1$.

We need to check differentiability at $x = 0$, where the function is continuous but the definition changes from $0$ to $2x$.

• Check differentiability at $x = 0$: We need to calculate the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x = 0$. We know $f(0) = 0$.

  LHD at $x=0$: For $x < 0$ and close to 0 (i.e., $x \in (-1, 0)$), $f(x) = 0$. $\lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{f(-h) - 0}{-h}$. Since $-h \in (-1, 0)$ for small $h>0$, $f(-h) = 0$. LHD = $\lim_{h \to 0^+} \frac{0 - 0}{-h} = 0$.

  RHD at $x=0$: For $x > 0$ and close to 0 (i.e., $x \in (0, 1)$), $f(x) = 2x$. $\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - 0}{h}$. Since $h \in (0, 1)$ for small $h>0$, $f(h) = 2h$. RHD = $\lim_{h \to 0^+} \frac{2h - 0}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.

  Since LHD $= 0$ and RHD $= 2$, LHD $\ne$ RHD. Therefore, $f(x)$ is not differentiable at $x = 0$.

We also need to check differentiability at $x = -2$, where the function definition changes from $1$ to $\frac{\sin(x+2)}{x+2}$.

• Check differentiability at $x = -2$: We know $f(-2) = 1$. LHD at $x=-2$: For $x < -2$, $f(x) = 1$. $\lim_{h \to 0^+} \frac{f(-2-h) - f(-2)}{-h} = \lim_{h \to 0^+} \frac{1 - 1}{-h} = 0$.

  RHD at $x=-2$: For $x \in (-2, -1)$, $f(x) = \frac{\sin(x+2)}{x+2}$. $\lim_{h \to 0^+} \frac{f(-2+h) - f(-2)}{h} = \lim_{h \to 0^+} \frac{\frac{\sin(-2+h+2)}{-2+h+2} - 1}{h} = \lim_{h \to 0^+} \frac{\frac{\sin h}{h} - 1}{h}$. As $h \to 0^+$, $\frac{\sin h}{h} \to 1$. So the numerator approaches $1-1=0$. This limit is of the form $\frac{0}{0}$. We can use L'Hopital's rule or Taylor series. Using Taylor series for $\sin h \approx h - \frac{h^3}{6}$ for small $h$: $\frac{\sin h}{h} - 1 \approx \frac{h - \frac{h^3}{6}}{h} - 1 = 1 - \frac{h^2}{6} - 1 = -\frac{h^2}{6}$. So, $\lim_{h \to 0^+} \frac{-\frac{h^2}{6}}{h} = \lim_{h \to 0^+} -\frac{h}{6} = 0$. Alternatively, using L'Hopital's rule on $\lim_{h \to 0^+} \frac{\frac{\sin h}{h} - 1}{h}$: $\lim_{h \to 0^+} \frac{\frac{h \cos h - \sin h}{h^2}}{1} = \lim_{h \to 0^+} \frac{h \cos h - \sin h}{h^2}$. Applying L'Hopital's rule again: $\lim_{h \to 0^+} \frac{\cos h - h \sin h - \cos h}{2h} = \lim_{h \to 0^+} \frac{-h \sin h}{2h} = \lim_{h \to 0^+} -\frac{\sin h}{2} = 0$.

  Since LHD $= 0$ and RHD $= 0$, the function is differentiable at $x = -2$.

The points where $f(x)$ is not differentiable are $x = -1$, $x = 1$ (due to discontinuity), and $x = 0$ (due to differing LHD and RHD). So, $n = 3$.

Step 4: Determine the ordered pair (m, n).

From Step 2, we found $m = 2$ (discontinuities at $x=-1$ and $x=1$). From Step 3, we found $n = 3$ (non-differentiability at $x=-1$, $x=1$, and $x=0$).

The ordered pair is $(m, n) = (2, 3)$.

Common Mistakes & Tips

• Incorrectly evaluating [x] or |x|: Pay close attention to the interval for $x$ when determining the values of $[x]$ and $|x|$. For example, in $(-2, -1)$, $[x]$ is always $-2$. In $(-1, 0]$, $|x|$ is $-x$ and $[|x|]$ is $0$.
• Assuming differentiability at points of continuity: A function can be continuous at a point but not differentiable if the left and right derivatives do not match (e.g., sharp corners). Always check LHD and RHD at such points.
• Forgetting that discontinuity implies non-differentiability: If a function is not continuous at a point, it cannot be differentiable at that point. This is a quick way to identify some points of non-differentiability.
• Algebraic errors in limits or derivatives: Be careful with algebraic manipulations, especially when using L'Hopital's rule or Taylor series.

Summary

We first simplified the piecewise function $f(x)$ by carefully evaluating the greatest integer function and absolute value function over the given intervals. We then systematically checked for continuity at the boundary points of these intervals. We found discontinuities at $x = -1$ and $x = 1$, giving $m=2$. Next, we checked for differentiability. Points of discontinuity are automatically points of non-differentiability. We also checked differentiability at the point $x=0$ where the function is continuous but the definition changes. By comparing the left-hand and right-hand derivatives at $x=0$, we found that the function is not differentiable there. Thus, the total number of points of non-differentiability is $n=3$. The ordered pair $(m, n)$ is $(2, 3)$.

The final answer is \boxed{(2, 3)}.