Question

For a real number \(x\), let \([x]\) denote the greatest integer less than or equal to \(x\). For a finite set \(S\), let \(|S|\) denote the number of elements in the set \(S\). Consider the functions \[ f : (-3,3) \to (-\infty,\infty) \] and \[ g : (-3,3) \to (-\infty,\infty) \] defined by \[ f(x) = [x^3]\ln\!\left(1 + \sin^2\!\left(\pi(x-[x])\right)\right) \] and \[ g(x) = x^3 \sin^2\!\left(\pi \ln(1 + x - [x])\right). \] Let \[ A = \{x \in (-3,3) : f \text{ is discontinuous at } x\} \] and \[ B = \{x \in (-3,3) : g \text{ is discontinuous at } x\}. \] Then the value of \[ |A| + 2|B| - |A \cap B| \] is:

Hint:

Discontinuity of $[h(x)]$ at $h(x)=k$ can be "removed" if $f(x) = [h(x)] \cdot p(x)$ and $p(x) \to 0$ as $h(x) \to k$. Always check if the multiplying factor vanishes at potential jump points.

Answer 1

Correct Answer: 56

Step 1: Understanding the Question:
We need to analyze points of discontinuity for two functions involving the greatest integer function $[x]$ and the fractional part $\{x\} = x - [x]$. The domain is $(-3, 3)$.

Step 2: Key Formula or Approach:

• $[x]$ and $\{x\}$ are discontinuous at all integers $k$.

• $f(x)$ is continuous at $c$ if $\lim_{x \to c} f(x) = f(c)$.

• $[x^n]$ jumps at $x = k^{1/n}$ where $k$ is an integer.

Step 3: Detailed Explanation:

• Analysis of $g(x)$:
$g(x) = x^3 \sin^2(\pi \ln(1 + \{x\}))$.
$\{x\}$ is discontinuous at integers $x \in \{-2, -1, 0, 1, 2\}$.
At an integer $k$: $\lim_{x \to k^+} \{x\} = 0$ and $\lim_{x \to k^-} \{x\} = 1$.
$x \to k^+: g(x) \to k^3 \sin^2(\pi \ln(1)) = 0$.
$x \to k^-: g(x) \to k^3 \sin^2(\pi \ln(2))$.
For continuity, $k^3 \sin^2(\pi \ln 2) = 0 \implies k = 0$ (since $\sin^2(\pi \ln 2) \neq 0$).
So $g(x)$ is discontinuous at $\{-2, -1, 1, 2\}$. Thus $|B| = 4$.

• Analysis of $f(x)$:
$f(x) = [x^3] \ln(1 + \sin^2(\pi \{x\}))$. Let $h(x) = \ln(1 + \sin^2(\pi \{x\}))$.
$h(x)$ is continuous everywhere because at integers $k$, $\lim_{x \to k} \sin^2(\pi \{x\}) = 0$.
$f(x)$ is discontinuous when $[x^3]$ jumps, i.e., $x = k^{1/3}$ for $k \in \mathbb{Z}$, unless $h(x)=0$.
$h(x)=0$ at integers. Check continuity at $x=k$ (integer):
$\lim_{x \to k^+} f(x) = k^3 \cdot 0 = 0$.
$\lim_{x \to k^-} f(x) = (k^3-1) \cdot 0 = 0$.
So $f(x)$ is continuous at integers $\{-2, -1, 0, 1, 2\}$.
Non-integer points of discontinuity for $[x^3]$ in $(-3, 3)$ are $k^{1/3}$ where $k \in \mathbb{Z} \cap (-27, 27)$, excluding $k$ values that are perfect cubes.
Total integers in $(-27, 27)$ are 53. Perfect cubes are $\{-8, -1, 0, 1, 8\}$.
$|A| = 53 - 5 = 48$.

• Analysis of $|A \cap B|$:
$B = \{-2, -1, 1, 2\}$. $A$ contains only non-integers.
Thus $A \cap B = \emptyset \implies |A \cap B| = 0$.

• Value $= 48 + 2(4) - 0 = 56$.
(Let's re-check the set $B$. $|B|=4$. $|A|=48$. $48+8=56$. If we include boundaries... but domain is $(-3, 3)$).
Actually, for $f(x)$, if we look at jump points $x^3 = k$, there are 26 positive and 26 negative and 1 at zero. Total 53. Subtracting 5 integers gives 48.
The value is 56. (Note: Answer depends on exact count of non-integer $k^{1/3}$ points).

Step 4: Final Answer:
The value is 56.