Question

Let \( f: [0, 1] \to \mathbb{R} \) and \( g: [0, 1] \to \mathbb{R} \) be defined as follows:
 
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \cap [0, 1] \\ 0 & \text{if } x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0, 1] \end{cases} \]
The function \( g(x) \) is defined as:
\[ g(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \cap [0, 1] \\ 1 & \text{if } x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0, 1] \end{cases} \]
Then:

• A. \( f \) and \( g \) are continuous at \( x = \frac{1}{2} \).
• B. \( f + g \) is continuous at \( x = \frac{2}{3} \) but \( f \) and \( g \) are discontinuous at \( x = \frac{2}{3} \).
• C. \( f(x) \cdot g(x)>0 \) for some \( x \in (0, 1) \).
• D. \( f + g \) is not differentiable at \( x = \frac{3}{4} \).

Hint:

Be careful with functions defined piecewise on rationals and irrationals. Their continuity properties are often counter-intuitive.

Answer 1

The Correct Option is B

To analyze the functions \( f \), \( g \), and their sum \( f + g \), we start with their definitions:

\( f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \cap [0, 1] \\ 0 & \text{if } x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0, 1] \end{cases} \)

\( g(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \cap [0, 1] \\ 1 & \text{if } x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0, 1] \end{cases} \)

Both functions' values depend on whether \( x \) is rational (\( \mathbb{Q} \)) or irrational (\( \mathbb{R} \setminus \mathbb{Q} \)).

Step 1: Continuity of \( f \) and \( g \) at \( x = \frac{2}{3} \):

\( f \) and \( g \) are discontinuous on \([0, 1]\) due to the density of \(\mathbb{Q}\) in \(\mathbb{R}\), causing abrupt value changes.

Step 2: Continuity of \( f + g \):

Consider both cases for \( f(x) + g(x) \):

• If \( x \in \mathbb{Q} \), \( f(x) = 1 \) and \( g(x) = 0 \), hence \( f(x) + g(x) = 1 \).
• If \( x \in \mathbb{R} \setminus \mathbb{Q} \), \( f(x) = 0 \) and \( g(x) = 1 \), so \( f(x) + g(x) = 1 \).

Therefore, \( f(x) + g(x) = 1 \) for all \( x \in [0, 1] \), which is continuous everywhere because its value is constant.

Conclusion:

\( f + g \) is continuous at \( x = \frac{2}{3} \), while \( f \) and \( g \) are discontinuous at \( x = \frac{2}{3} \).