Discontinuity Points of Functions Involving Greatest Integer

To find the value of $|A| + 2|B| - |A \cap B|$, we need to determine the points of discontinuity for both functions $f(x)$ and $g(x)$ in the domain $(-3, 3)$.

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Step 1: Analyzing the continuity of $f(x)$

The function $f: (-3, 3) \to (-\infty, \infty)$ is defined as: $f(x) = [x^3] \log_e (1 + \sin^2(\pi \{x\}))$ where $\{x\} = x - [x]$ denotes the fractional part of $x$.

1. Continuity of $\log_e(1 + \sin^2(\pi \{x\}))$:

   • As $x \to k^+$ (where $k \in \mathbb{Z}$), $\{x\} \to 0^+$, so $\sin^2(\pi \{x\}) \to \sin^2(0) = 0$.
   • As $x \to k^-$, $\{x\} \to 1^-$, so $\sin^2(\pi \{x\}) \to \sin^2(\pi) = 0$.
   • At $x = k$, $\{k\} = 0$, so $\sin^2(\pi \{k\}) = 0$.

   Thus, $\sin^2(\pi \{x\})$ is continuous for all $x \in \mathbb{R}$, and $\log_e(1 + \sin^2(\pi \{x\})) = 0$ whenever $x \in \mathbb{Z}$.

2. Points of potential discontinuity of $[x^3]$:

   • For $x \in (-3, 3)$, we have $x^3 \in (-27, 27)$.
   • The greatest integer function $[x^3]$ is discontinuous at points where $x^3 = n \in \mathbb{Z}$, which corresponds to $x = n^{1/3}$ for $n \in \{-26, -25, \dots, 26\}$.
   • The total number of such integers $n$ is $26 - (-26) + 1 = 53$.

3. Checking continuity of $f(x)$ at $x = n^{1/3}$:

   • Case (i): When $x = n^{1/3}$ is an integer, i.e., $x \in \{-2, -1, 0, 1, 2\}$ (which corresponds to $n \in \{-8, -1, 0, 1, 8\}$): As $x \to k \in \{-2, -1, 0, 1, 2\}$, $[x^3]$ remains bounded, while $\log_e(1 + \sin^2(\pi \{x\})) \to 0$. Therefore, $\lim_{x \to k} f(x) = 0 = f(k)$ Thus, $f(x)$ is continuous at all 5 integer points.

   • Case (ii): When $x = n^{1/3}$ is NOT an integer, i.e., $n \in \{-26, \dots, 26\} \setminus \{-8, -1, 0, 1, 8\}$: Since $x$ is not an integer, $\{x\} \notin \mathbb{Z}$, which implies $\sin^2(\pi \{x\}) > 0$, so $\log_e(1 + \sin^2(\pi \{x\})) = C > 0$. However, $[x^3]$ has a jump discontinuity at $x = n^{1/3}$: $\lim_{x \to (n^{1/3})^+} [x^3] = n \quad \text{and} \quad \lim_{x \to (n^{1/3})^-} [x^3] = n - 1$ Consequently, $\lim_{x \to (n^{1/3})^+} f(x) = n C \quad \text{and} \quad \lim_{x \to (n^{1/3})^-} f(x) = (n - 1) C$ Since $C > 0$, these one-sided limits are unequal. Thus, $f(x)$ is discontinuous at these points.

Therefore, the set of points of discontinuity of $f(x)$ in $(-3, 3)$ is: $A = \{ n^{1/3} : n \in \{-26, \dots, 26\} \setminus \{-8, -1, 0, 1, 8\} \}$ $|A| = 53 - 5 = 48$

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Step 2: Analyzing the continuity of $g(x)$

The function $g: (-3, 3) \to (-\infty, \infty)$ is defined as: $g(x) = x^3 \sin^2(\pi \log_e(1 + \{x\}))$

1. Since $x^3$ and $\{x\}$ are continuous everywhere except possibly where $\{x\}$ is discontinuous (i.e., at integers), the only candidates for discontinuity of $g(x)$ are $x \in \{-2, -1, 0, 1, 2\}$.

2. Checking at $x = 0$: $\lim_{x \to 0} g(x) = 0 = g(0)$ So $g(x)$ is continuous at $x = 0$.

3. Checking at non-zero integers $k \in \{-2, -1, 1, 2\}$:

   • Right-Hand Limit (RHL): As $x \to k^+$, $\{x\} \to 0^+ \implies \log_e(1 + \{x\}) \to 0$. $\lim_{x \to k^+} g(x) = k^3 \sin^2(0) = 0$

   • Left-Hand Limit (LHL): As $x \to k^-$, $\{x\} \to 1^- \implies \log_e(1 + \{x\}) \to \log_e 2$. $\lim_{x \to k^-} g(x) = k^3 \sin^2(\pi \log_e 2)$

   Since $\log_e 2$ is not an integer, $\sin^2(\pi \log_e 2) \neq 0$. For $k \neq 0$, $k^3 \neq 0$, which means: $\text{LHL} \neq \text{RHL}$

   Therefore, $g(x)$ is discontinuous at $x \in \{-2, -1, 1, 2\}$.

Thus, the set of points of discontinuity of $g(x)$ is: $B = \{-2, -1, 1, 2\}$ $|B| = 4$

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Step 3: Calculating $|A \cap B|$ and the Final Value

• $A$ consists purely of non-integer points of the form $n^{1/3}$.
• $B$ consists purely of integer points $\{-2, -1, 1, 2\}$.

Therefore, $A \cap B = \emptyset$, so $|A \cap B| = 0$.

Now, substituting these values into the required expression: $|A| + 2|B| - |A \cap B| = 48 + 2(4) - 0 = 48 + 8 = 56$