Key Concepts and Formulas

• Greatest Integer Function [t]: The greatest integer less than or equal to $t$. Key properties include:
  • $[n] = n$ for any integer $n$.
  • $[x]$ is constant over intervals of the form $[n, n+1)$.
  • $[x+1] = [x] + 1$.
• Absolute Value Function |a|: Defined as $a$ if $a \ge 0$ and $-a$ if $a < 0$.
• Continuity of a Function: A function $f(x)$ is continuous at a point $c$ if and only if:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists (i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$).
  3. $\lim_{x \to c} f(x) = f(c)$.
• Discontinuity: A function is discontinuous at a point if any of the conditions for continuity are not met. Points of discontinuity often occur where the function definition changes or where non-continuous functions (like the greatest integer function) are involved.

Step-by-Step Solution

The function is given by $f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$ for $x \in (-2, 2)$.

Step 1: Analyze the behavior of the greatest integer function [x] in the interval (-2, 2). The greatest integer function $[x]$ changes its value at integer points. In the interval $(-2, 2)$, the integer points are $-1$, $0$, and $1$. We need to consider the intervals defined by these points:

• For $-2 < x < -1$, $[x] = -2$.
• For $-1 \le x < 0$, $[x] = -1$.
• For $0 \le x < 1$, $[x] = 0$.
• For $1 \le x < 2$, $[x] = 1$.

Step 2: Simplify the function f(x) in each of these intervals.

• Interval 1: $-2 < x < -1$ Here, $[x] = -2$. Also, for this interval, $x^2 > 1$, so $|x^2 - 1| = x^2 - 1$. And $[x+1] = [-2+1] = [-1] = -1$. So, $f(x) = (-2)(x^2 - 1) + \sin\left(\frac{\pi}{-2 + 3}\right) - (-1)$ $f(x) = -2(x^2 - 1) + \sin\left(\frac{\pi}{1}\right) + 1$ $f(x) = -2x^2 + 2 + 0 + 1$ $f(x) = -2x^2 + 3$ However, the provided solution states $f(x) = -2|x^2-1| + 1$. Let's recheck the $|x^2-1|$ part. For $-2 < x < -1$, $x^2$ is between $1$ (exclusive, as $x \to -1$) and $4$ (exclusive, as $x \to -2$). So $x^2 - 1$ is positive. Thus $|x^2 - 1| = x^2 - 1$. The term $\sin(\frac{\pi}{[x]+3})$ becomes $\sin(\frac{\pi}{-2+3}) = \sin(\pi) = 0$. The term $[x+1]$ becomes $[-2+1] = [-1] = -1$. So, $f(x) = -2(x^2 - 1) + 0 - (-1) = -2x^2 + 2 + 1 = -2x^2 + 3$. Let's check the provided solution's expression for this interval: $-2|x^2-1| + 1$. If we assume $|x^2-1| = x^2-1$, this gives $-2(x^2-1) + 1 = -2x^2 + 2 + 1 = -2x^2 + 3$. This matches our calculation. So, for $-2 < x < -1$, $f(x) = -2(x^2 - 1) + 1$.

• Interval 2: $-1 \le x < 0$ Here, $[x] = -1$. For $-1 \le x < 0$, $x^2$ is between $0$ (inclusive, as $x \to 0$) and $1$ (inclusive, as $x = -1$). So $x^2 - 1$ is between $-1$ and $0$. Thus $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$. And $[x+1] = [-1+1] = [0] = 0$. So, $f(x) = (-1)(1 - x^2) + \sin\left(\frac{\pi}{-1 + 3}\right) - 0$ $f(x) = -(1 - x^2) + \sin\left(\frac{\pi}{2}\right)$ $f(x) = -1 + x^2 + 1$ $f(x) = x^2$ Let's check the provided solution's expression for this interval: $-|x^2-1| + 1$. For $-1 \le x < 0$, $|x^2 - 1| = 1 - x^2$. So, $-|x^2 - 1| + 1 = -(1 - x^2) + 1 = -1 + x^2 + 1 = x^2$. This matches our calculation. So, for $-1 \le x < 0$, $f(x) = -(x^2 - 1) + 1$.

• Interval 3: $0 \le x < 1$ Here, $[x] = 0$. For $0 \le x < 1$, $x^2$ is between $0$ and $1$. So $x^2 - 1$ is between $-1$ and $0$. Thus $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$. And $[x+1] = [0+1] = [1] = 1$. So, $f(x) = (0)|x^2 - 1| + \sin\left(\frac{\pi}{0 + 3}\right) - 1$ $f(x) = 0 + \sin\left(\frac{\pi}{3}\right) - 1$ $f(x) = \frac{\sqrt{3}}{2} - 1$ Let's check the provided solution's expression for this interval: $\sin(\pi/3) + 1$. There seems to be a discrepancy. Our calculation: $f(x) = \sin(\pi/3) - 1 = \frac{\sqrt{3}}{2} - 1$. The provided solution states $f(x) = \sin(\pi/3) + 1$. This implies a $+1$ instead of a $-1$ at the end. Let's re-examine $[x+1]$. For $0 \le x < 1$, $[x]=0$. So $[x+1] = [0+1] = [1] = 1$. The formula is $f(x) = [x]|x^2-1| + \sin(\frac{\pi}{[x]+3}) - [x+1]$. So, $f(x) = 0 \cdot |x^2-1| + \sin(\frac{\pi}{0+3}) - 1 = 0 + \sin(\frac{\pi}{3}) - 1 = \frac{\sqrt{3}}{2} - 1$. The provided solution's piecewise function has $\sin(\pi/3) + 1$ for $0 \le x < 1$. This suggests a potential error in the provided solution's piecewise definition or our interpretation. Let's proceed assuming our calculation is correct for now and verify continuity.

• Interval 4: $1 \le x < 2$ Here, $[x] = 1$. For $1 \le x < 2$, $x^2$ is between $1$ (inclusive, as $x=1$) and $4$ (exclusive, as $x \to 2$). So $x^2 - 1$ is between $0$ and $3$. Thus $|x^2 - 1| = x^2 - 1$. And $[x+1] = [1+1] = [2] = 2$. So, $f(x) = (1)(x^2 - 1) + \sin\left(\frac{\pi}{1 + 3}\right) - 2$ $f(x) = x^2 - 1 + \sin\left(\frac{\pi}{4}\right) - 2$ $f(x) = x^2 - 3 + \frac{1}{\sqrt{2}}$ Let's check the provided solution's expression for this interval: $|x^2 - 1| + \frac{1}{\sqrt{2}} - 2$. For $1 \le x < 2$, $|x^2 - 1| = x^2 - 1$. So, $|x^2 - 1| + \frac{1}{\sqrt{2}} - 2 = (x^2 - 1) + \frac{1}{\sqrt{2}} - 2 = x^2 - 3 + \frac{1}{\sqrt{2}}$. This matches our calculation. So, for $1 \le x < 2$, $f(x) = (x^2 - 1) + \frac{1}{\sqrt{2}} - 2$.

Let's re-evaluate the piecewise function based on our calculations: f(x) = \left\{ {\matrix{ {-2(x^2 - 1) + 1,} & { - 2 < x < - 1} \cr {-(x^2 - 1) + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} - 1,} & {0 \le x < 1} \cr {(x^2 - 1) + {\frac{1}{\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.

Now we check for continuity at the points where $[x]$ changes value: $x = -1$, $x = 0$, and $x = 1$.

Step 3: Check for continuity at x = -1. We need to compare the left-hand limit, the right-hand limit, and the function value at $x = -1$.

• Left-hand limit: As $x \to -1^-$, we are in the interval $-2 < x < -1$. So we use $f(x) = -2(x^2 - 1) + 1$. $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} (-2(x^2 - 1) + 1) = -2((-1)^2 - 1) + 1 = -2(1 - 1) + 1 = 0 + 1 = 1$

• Right-hand limit: As $x \to -1^+$, we are in the interval $-1 \le x < 0$. So we use $f(x) = -(x^2 - 1) + 1$. $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} (-(x^2 - 1) + 1) = -((-1)^2 - 1) + 1 = -(1 - 1) + 1 = 0 + 1 = 1$

• Function value at $x = -1$: Since $-1 \le x < 0$, we use $f(x) = -(x^2 - 1) + 1$. $f(-1) = -((-1)^2 - 1) + 1 = -(1 - 1) + 1 = 0 + 1 = 1$

Since $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = f(-1) = 1$, the function is continuous at $x = -1$.

Step 4: Check for continuity at x = 0. We need to compare the left-hand limit, the right-hand limit, and the function value at $x = 0$.

• Left-hand limit: As $x \to 0^-$, we are in the interval $-1 \le x < 0$. So we use $f(x) = -(x^2 - 1) + 1 = x^2$. $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} (x^2) = 0^2 = 0$

• Right-hand limit: As $x \to 0^+$, we are in the interval $0 \le x < 1$. So we use $f(x) = \sin(\frac{\pi}{3}) - 1 = \frac{\sqrt{3}}{2} - 1$. $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left(\frac{\sqrt{3}}{2} - 1\right) = \frac{\sqrt{3}}{2} - 1$

• Function value at $x = 0$: Since $0 \le x < 1$, we use $f(x) = \frac{\sqrt{3}}{2} - 1$. $f(0) = \frac{\sqrt{3}}{2} - 1$

Since $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 0$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \frac{\sqrt{3}}{2} - 1$, the left-hand limit and the right-hand limit are not equal. Therefore, the function is discontinuous at $x = 0$.

Let's re-examine the provided solution's calculation for $x=0$: It states: "at x = 0, $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$" This is significantly different from our calculation. Let's check their piecewise function for $0 \le x < 1$: $\sin(\pi/3) + 1$. If $f(x) = \sin(\pi/3) + 1$ for $0 \le x < 1$, then $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \sin(\pi/3) + 1 = \frac{\sqrt{3}}{2} + 1$. This matches the provided solution's right-hand limit. Now let's check their left-hand limit calculation at $x=0$: $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$. For $-1 \le x < 0$, their piecewise function is $-|x^2-1| + 1$. As $x \to 0^-$, $|x^2-1| \to |0-1| = |-1| = 1$. So, $-|x^2-1| + 1 \to -1 + 1 = 0$. There seems to be a consistent issue with the provided solution's piecewise definition or its limit calculations.

Let's assume for a moment that the provided piecewise function is correct and check the limits. Provided piecewise function: f(x) = \left\{ {\matrix{ { - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.

• Continuity at x = -1:

  • $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right)$: Use $-2|x^2-1|+1$. As $x \to -1^-$, $x^2-1 \to 0^+$. So $|x^2-1| \to 0$. Limit is $-2(0)+1 = 1$.
  • $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right)$: Use $-|x^2-1|+1$. As $x \to -1^+$, $x^2-1 \to 0^-$. So $|x^2-1| \to 0$. Limit is $-(0)+1 = 1$.
  • $f(-1)$: Use $-|x^2-1|+1$. $f(-1) = -|(-1)^2-1|+1 = -|1-1|+1 = 0+1 = 1$. Continuous at $x=-1$.

• Continuity at x = 0:

  • $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$: Use $-|x^2-1|+1$. As $x \to 0^-$, $|x^2-1| \to |-1| = 1$. Limit is $-1+1 = 0$.
  • $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$: Use $\sin(\pi/3)+1$. Limit is $\sin(\pi/3)+1 = \frac{\sqrt{3}}{2}+1$. Since $0 \ne \frac{\sqrt{3}}{2}+1$, discontinuous at $x=0$.

• Continuity at x = 1:

  • $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)$: Use $\sin(\pi/3)+1$. Limit is $\sin(\pi/3)+1 = \frac{\sqrt{3}}{2}+1$.
  • $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$: Use $|x^2-1|+\frac{1}{\sqrt{2}}-2$. As $x \to 1^+$, $x^2-1 \to 0^+$. So $|x^2-1| \to 0$. Limit is $0 + \frac{1}{\sqrt{2}} - 2 = \frac{1}{\sqrt{2}} - 2$. Since $\frac{\sqrt{3}}{2}+1 \ne \frac{1}{\sqrt{2}} - 2$, discontinuous at $x=1$.

This analysis, using the provided piecewise function, indicates discontinuities at $x=0$ and $x=1$. This matches the correct answer of 2 points.

Let's revisit the original function and derive the piecewise function carefully to understand the discrepancy in the provided solution's piecewise definition.

$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$

• Interval $-2 < x < -1$: $[x] = -2$. $x^2 - 1 > 0$, so $|x^2 - 1| = x^2 - 1$. $[x+1] = [-2+1] = -1$. $f(x) = -2(x^2 - 1) + \sin(\frac{\pi}{-2+3}) - (-1)$ $f(x) = -2(x^2 - 1) + \sin(\pi) + 1$ $f(x) = -2(x^2 - 1) + 0 + 1 = -2|x^2-1| + 1$. This matches the provided solution.

• Interval $-1 \le x < 0$: $[x] = -1$. $x^2 - 1 \le 0$, so $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$. $[x+1] = [-1+1] = 0$. $f(x) = -1(1 - x^2) + \sin(\frac{\pi}{-1+3}) - 0$ $f(x) = -(1 - x^2) + \sin(\frac{\pi}{2})$ $f(x) = -(1 - x^2) + 1 = -|x^2-1| + 1$. This matches the provided solution.

• Interval $0 \le x < 1$: $[x] = 0$. $x^2 - 1 < 0$, so $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$. $[x+1] = [0+1] = 1$. $f(x) = 0(1 - x^2) + \sin(\frac{\pi}{0+3}) - 1$ $f(x) = 0 + \sin(\frac{\pi}{3}) - 1 = \frac{\sqrt{3}}{2} - 1$. The provided solution has $\sin(\pi/3) + 1$. This is where the discrepancy lies. The correct term is $-[x+1]$, which is $-1$ in this interval. The provided solution seems to have added 1 instead of subtracting 1.

• Interval $1 \le x < 2$: $[x] = 1$. $x^2 - 1 \ge 0$, so $|x^2 - 1| = x^2 - 1$. $[x+1] = [1+1] = 2$. $f(x) = 1(x^2 - 1) + \sin(\frac{\pi}{1+3}) - 2$ $f(x) = (x^2 - 1) + \sin(\frac{\pi}{4}) - 2$ $f(x) = |x^2-1| + \frac{1}{\sqrt{2}} - 2$. This matches the provided solution.

So, the provided piecewise function is incorrect for the interval $0 \le x < 1$. However, the provided solution correctly identifies the points of discontinuity. Let's proceed with checking continuity using the correct piecewise function derived.

Corrected piecewise function: f(x) = \left\{ {\matrix{ {-2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} - 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {\frac{1}{\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.

Step 3 (Revised): Check for continuity at x = -1.

• $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} (-2|x^2-1| + 1) = -2(0) + 1 = 1$.
• $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} (-|x^2-1| + 1) = -(0) + 1 = 1$.
• $f(-1) = -|(-1)^2-1| + 1 = 0 + 1 = 1$. Continuous at $x = -1$.

Step 4 (Revised): Check for continuity at x = 0.

• $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} (-|x^2-1| + 1) = -|-1| + 1 = -1 + 1 = 0$.
• $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} (\sin(\pi/3) - 1) = \frac{\sqrt{3}}{2} - 1$.
• $f(0) = \sin(\pi/3) - 1 = \frac{\sqrt{3}}{2} - 1$. Since $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$, the function is discontinuous at $x = 0$.

Step 5: Check for continuity at x = 1.

• $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} (\sin(\pi/3) - 1) = \frac{\sqrt{3}}{2} - 1$.
• $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} (|x^2-1| + \frac{1}{\sqrt{2}} - 2) = (0) + \frac{1}{\sqrt{2}} - 2 = \frac{1}{\sqrt{2}} - 2$.
• $f(1) = |1^2-1| + \frac{1}{\sqrt{2}} - 2 = 0 + \frac{1}{\sqrt{2}} - 2 = \frac{1}{\sqrt{2}} - 2$. Since $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$, the function is discontinuous at $x = 1$.

The points of discontinuity are $x=0$ and $x=1$. There are 2 points of discontinuity.

Common Mistakes & Tips

• Careful Evaluation of Absolute Value: Pay close attention to the sign of the expression inside the absolute value function in each interval. For $|x^2-1|$, it is positive when $x^2 > 1$ (i.e., $x > 1$ or $x < -1$) and negative when $x^2 < 1$ (i.e., $-1 < x < 1$).
• Correctly Evaluating the Greatest Integer Function: Remember that $[x]$ is constant over intervals of the form $[n, n+1)$. Be precise with the endpoints. Also, remember the property $[x+1] = [x] + 1$.
• Checking All Three Conditions for Continuity: A function is continuous at a point only if the limit exists AND the limit equals the function value. Do not stop after checking only the limit existence.
• Discrepancies in Provided Solutions: If you encounter a discrepancy between your derivation and a provided solution's intermediate steps (like the piecewise function), trust your derivation based on the fundamental definitions. Then, use the provided solution's final answer to guide your re-check.

Summary

To find the points of discontinuity of the given function $f(x)$, we first analyzed the behavior of the greatest integer function $[x]$ in the specified interval $(-2, 2)$. This led to partitioning the domain into subintervals: $(-2, -1)$, $[-1, 0)$, $[0, 1)$, and $[1, 2)$. We then simplified the function $f(x)$ in each subinterval, paying careful attention to the absolute value term and the greatest integer function. The potential points of discontinuity are where the definition of the function changes, which are the integer points $x = -1, 0, 1$. We checked the continuity at each of these points by comparing the left-hand limit, the right-hand limit, and the function value. We found that the function is continuous at $x = -1$ but discontinuous at $x = 0$ and $x = 1$. Therefore, there are 2 points of discontinuity.

The final answer is $\boxed{2}$.