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

We are given the function \(f(x) = \left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\), where \([t]\) denotes the greatest integer less than or equal to \(t\). We need to find the sum of all points of discontinuity of this function in the interval \([0, 8]\).

Concept Used:

The greatest integer function, \([t]\), is discontinuous at every integer value of its argument \(t\). The function \(f(x)\) is a difference of two functions involving the greatest integer function. Therefore, \(f(x)\) will be discontinuous at a point \(a\) if either \(\left[\frac{x}{2} + 3\right]\) or \(\left[\sqrt{x}\right]\) is discontinuous at \(x=a\), unless the jump discontinuities of the two functions cancel each other out.

A function \(g(x)\) is discontinuous at a point \(x=a\) if the left-hand limit (LHL), right-hand limit (RHL), and the function's value at that point are not all equal. That is, if \( \lim_{x \to a^-} g(x) \neq \lim_{x \to a^+} g(x) \) or if these limits are not equal to \(g(a)\).

Step-by-Step Solution:

Step 1: Identify potential points of discontinuity from the first term, \( g(x) = \left[\frac{x}{2} + 3\right] \).

This term is discontinuous when its argument, \( \frac{x}{2} + 3 \), is an integer. Let \( \frac{x}{2} + 3 = k \), where \(k\) is an integer. This implies \( x = 2(k-3) \). We need to find the values of \(x\) in the interval \([0, 8]\).

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

The possible integer values for \(k\) are 3, 4, 5, 6, and 7. The corresponding values for \(x\) are:

• For \(k=4\): \(x = 2(4-3) = 2\)
• For \(k=5\): \(x = 2(5-3) = 4\)
• For \(k=6\): \(x = 2(6-3) = 6\)
• For \(k=7\): \(x = 2(7-3) = 8\)

The set of potential points of discontinuity from the first term in \((0, 8]\) is \( \{2, 4, 6, 8\} \).

Step 2: Identify potential points of discontinuity from the second term, \( h(x) = \left[\sqrt{x}\right] \).

This term is discontinuous when its argument, \( \sqrt{x} \), is an integer. Let \( \sqrt{x} = m \), where \(m\) is an integer. This implies \( x = m^2 \). We need to find the values of \(x\) in the interval \([0, 8]\).

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

The possible integer values for \(m\) are 1 and 2. (We check endpoints separately, \(m=0 \implies x=0\)). The corresponding values for \(x\) are:

• For \(m=1\): \(x = 1^2 = 1\)
• For \(m=2\): \(x = 2^2 = 4\)

The set of potential points of discontinuity from the second term in \((0, 8)\) is \( \{1, 4\} \).

Step 3: Combine the potential points and check for continuity at each one.

The combined set of potential points of discontinuity in \((0, 8]\) is \( S_{potential} = \{1, 2, 4, 6, 8\} \). We must test each point.

At \(x=1\):

LHL: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[3.5^-\right] - \left[1^-\right] = 3 - 0 = 3 \). RHL: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[3.5^+\right] - \left[1^+\right] = 3 - 1 = 2 \). Since LHL \(\neq\) RHL, \(f(x)\) is discontinuous at \(x=1\).

At \(x=2\):

LHL: \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[4^-\right] - \left[\sqrt{2}^-\right] = 3 - 1 = 2 \). RHL: \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[4^+\right] - \left[\sqrt{2}^+\right] = 4 - 1 = 3 \). Since LHL \(\neq\) RHL, \(f(x)\) is discontinuous at \(x=2\).

At \(x=4\):

LHL: \( \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[5^-\right] - \left[2^-\right] = 4 - 1 = 3 \). RHL: \( \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[5^+\right] - \left[2^+\right] = 5 - 2 = 3 \). Also, \( f(4) = \left[\frac{4}{2} + 3\right] - \left[\sqrt{4}\right] = [5] - [2] = 5 - 2 = 3 \). Since LHL = RHL = \(f(4)\), \(f(x)\) is continuous at \(x=4\).

At \(x=6\):

LHL: \( \lim_{x \to 6^-} f(x) = \lim_{x \to 6^-} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[6^-\right] - \left[\sqrt{6}^-\right] = 5 - 2 = 3 \). RHL: \( \lim_{x \to 6^+} f(x) = \lim_{x \to 6^+} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[6^+\right] - \left[\sqrt{6}^+\right] = 6 - 2 = 4 \). Since LHL \(\neq\) RHL, \(f(x)\) is discontinuous at \(x=6\).

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

We check the left-hand limit against the function value.

LHL: \( \lim_{x \to 8^-} f(x) = \lim_{x \to 8^-} \left(\left[\frac{x}{2} + 3\right] - \left[\sqrt{x}\right]\right) = \left[7^-\right] - \left[\sqrt{8}^-\right] = 6 - 2 = 4 \). Value at \(x=8\): \( f(8) = \left[\frac{8}{2} + 3\right] - \left[\sqrt{8}\right] = [7] - [2.828...] = 7 - 2 = 5 \). Since \( \lim_{x \to 8^-} f(x) \neq f(8) \), \(f(x)\) is not continuous from the left at \(x=8\), so it is discontinuous at \(x=8\).

Step 4: Sum the points of discontinuity.

The set of all points of discontinuity in the interval \([0, 8]\) is \( S = \{1, 2, 6, 8\} \).

The sum of these points is:

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

The final sum is 17.

Answer 2

\[\left\lfloor \frac{x}{2} + 3 \right\rfloor is  discontinuous  at  x = 2, 4, 6, 8\]
\[\sqrt{x} \text{ is discontinuous at } x = 1, 4\]
\[F(x) \text{ is discontinuous at } x = 1, 2, 6, 8\]
Summing the values:
\[\sum a = 1 + 2 + 6 + 8 = 17\]