Question

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f: [0, \infty) \to \mathbb{R}$ be a function defined by \[ f(x) = \left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]. \] Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then \[ \sum_{a \in S} a \] is equal to ________.

Answer 1

Correct Answer: 17

Given the function \(f(x) = \left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\) on the interval \([0, 8]\), where \([t]\) is the greatest integer less than or equal to \(t\). We aim to find the sum of all points where \(f(x)\) is discontinuous within this interval.

Underlying Principle:

The greatest integer function \([t]\) is discontinuous at integer values of \(t\). The function \(f(x)\) is a difference of two such functions. Discontinuities in \(f(x)\) occur where either \(\left[\frac{x}{2} + 3\right]\) or \(\left[\sqrt{x}\right]\) is discontinuous, unless the jump discontinuities cancel each other out.

A function \(g(x)\) is discontinuous at \(x=a\) if the left-hand limit (\( \lim_{x \to a^-} g(x) \)), the right-hand limit (\( \lim_{x \to a^+} g(x) \)), and the function value (\(g(a)\)) are not all equal.

Solution Breakdown:

Step 1: Identify discontinuities from the term \( g(x) = \left[\frac{x}{2} + 3\right] \).

This term is discontinuous when \( \frac{x}{2} + 3 \) is an integer. Let \( \frac{x}{2} + 3 = k \), where \(k \in \mathbb{Z}\). Then \( x = 2(k-3) \). For \(x \in [0, 8]\):

\[ 0 \le 2(k-3) \le 8 \implies 0 \le k-3 \le 4 \implies 3 \le k \le 7 \]

Possible integer values for \(k\) are 3, 4, 5, 6, 7. The corresponding \(x\) values in \([0, 8]\) are:

• \(k=3 \implies x=0\)
• \(k=4 \implies x=2\)
• \(k=5 \implies x=4\)
• \(k=6 \implies x=6\)
• \(k=7 \implies x=8\)

Potential discontinuity points from this term in \((0, 8]\) are \( \{2, 4, 6, 8\} \).

Step 2: Identify discontinuities from the term \( h(x) = \left[\sqrt{x}\right] \).

This term is discontinuous when \( \sqrt{x} \) is an integer. Let \( \sqrt{x} = m \), where \(m \in \mathbb{Z}\). Then \( x = m^2 \). For \(x \in [0, 8]\):

\[ 0 \le m^2 \le 8 \implies 0 \le m \le \sqrt{8} \approx 2.828 \]

Possible integer values for \(m\) are 0, 1, 2. The corresponding \(x\) values in \([0, 8]\) are:

• \(m=0 \implies x=0\)
• \(m=1 \implies x=1\)
• \(m=2 \implies x=4\)

Potential discontinuity points from this term in \([0, 8)\) are \( \{0, 1, 4\} \).

Step 3: Consolidate and verify potential discontinuity points.

The union of potential points in \((0, 8]\) is \( S_{potential} = \{1, 2, 4, 6, 8\} \). We analyze each point:

At \(x=1\):

LHL: \( \lim_{x \to 1^-} f(x) = \left[\frac{1^-}{2} + 3\right] - \left[\sqrt{1^-}\right] = [3.5^-] - [1^-] = 3 - 0 = 3 \).

RHL: \( \lim_{x \to 1^+} f(x) = \left[\frac{1^+}{2} + 3\right] - \left[\sqrt{1^+}\right] = [3.5^+] - [1^+] = 3 - 1 = 2 \).

Since LHL \(eq\) RHL, \(f(x)\) is discontinuous at \(x=1\).

At \(x=2\):

LHL: \( \lim_{x \to 2^-} f(x) = \left[\frac{2^-}{2} + 3\right] - \left[\sqrt{2^-}\right] = [4^-] - [1.414...^-] = 3 - 1 = 2 \).

RHL: \( \lim_{x \to 2^+} f(x) = \left[\frac{2^+}{2} + 3\right] - \left[\sqrt{2^+}\right] = [4^+] - [1.414...^+] = 4 - 1 = 3 \).

Since LHL \(eq\) RHL, \(f(x)\) is discontinuous at \(x=2\).

At \(x=4\):

LHL: \( \lim_{x \to 4^-} f(x) = \left[\frac{4^-}{2} + 3\right] - \left[\sqrt{4^-}\right] = [5^-] - [2^-] = 4 - 1 = 3 \).

RHL: \( \lim_{x \to 4^+} f(x) = \left[\frac{4^+}{2} + 3\right] - \left[\sqrt{4^+}\right] = [5^+] - [2^+] = 5 - 2 = 3 \).

\( f(4) = \left[\frac{4}{2} + 3\right] - \left[\sqrt{4}\right] = [5] - [2] = 5 - 2 = 3 \).

Since LHL = RHL = \(f(4)\), \(f(x)\) is continuous at \(x=4\).

At \(x=6\):

LHL: \( \lim_{x \to 6^-} f(x) = \left[\frac{6^-}{2} + 3\right] - \left[\sqrt{6^-}\right] = [6^-] - [2.449...^-] = 5 - 2 = 3 \).

RHL: \( \lim_{x \to 6^+} f(x) = \left[\frac{6^+}{2} + 3\right] - \left[\sqrt{6^+}\right] = [6^+] - [2.449...^+] = 6 - 2 = 4 \).

Since LHL \(eq\) RHL, \(f(x)\) is discontinuous at \(x=6\).

At \(x=8\): (Endpoint)

LHL: \( \lim_{x \to 8^-} f(x) = \left[\frac{8^-}{2} + 3\right] - \left[\sqrt{8^-}\right] = [7^-] - [2.828...^-] = 6 - 2 = 4 \).

\( f(8) = \left[\frac{8}{2} + 3\right] - \left[\sqrt{8}\right] = [7] - [2.828...] = 7 - 2 = 5 \).

Since \( \lim_{x \to 8^-} f(x) eq f(8) \), \(f(x)\) is discontinuous at \(x=8\).

Step 4: Sum the points of discontinuity.

The set of discontinuities in \([0, 8]\) is \( S = \{1, 2, 6, 8\} \).

The sum is:

\[ \sum_{a \in S} a = 1 + 2 + 6 + 8 = 17 \]

The final sum is 17.