Question

Let \( f(x) \) be a real valued function. If \( f'(x) \) is a constant for all \( x \in \mathbb{R}, f(0) = 1 \) and \( f'(0) = 2 \), then

• A. \( f(x) \text{ is not continuous on } \mathbb{R} \)
• B. \( f(x) \text{ is continuous at } x = 0, 1, 2 \text{ and 3 only} \)
• C. \( f(x) \text{ is continuous only on } [0, \infty) \)
• D. \( f(x) \text{ is continuous on } \mathbb{R} \)

Hint:

If \( f'(x) \) is constant, then \( f(x) \) is a linear function, which is continuous everywhere on \( \mathbb{R} \).

Answer 1

The Correct Option is D

We are given that \( f'(x) \) is a constant for all \( x \in \mathbb{R} \). This implies that \( f(x) \) is a linear function of the form: \[ f(x) = mx + c \] where \( m = f'(x) \) and \( c \) is a constant. We are also given that \( f(0) = 1 \), so: \[ f(0) = m(0) + c = 1 \implies c = 1 \] Thus, \( f(x) = mx + 1 \). Also, \( f'(0) = 2 \), so: \[ f'(x) = m \implies f'(0) = m = 2 \] Thus, the function becomes: \[ f(x) = 2x + 1 \] This function is continuous for all real values of \( x \), so the correct answer is that \( f(x) \) is continuous on \( \mathbb{R} \). Therefore, the correct answer is option (4).

Answer 2

Given that the derivative \( f'(x) \) is constant for all \( x \) in \(\mathbb{R}\) and it is provided \( f'(0) = 2 \), we can infer that \( f'(x) = 2 \) for all \( x \in \mathbb{R} \). This implies that the function \( f(x) \) is a linear function. The general form of a linear function is \( f(x) = mx + c \), where \( m \) is the slope (in this case, the derivative) and \( c \) is the y-intercept.

Since \( f'(x) = 2 \) for all \( x \), the slope \( m = 2 \). Therefore, the function can be written as \( f(x) = 2x + c \).

We also know from the problem statement that \( f(0) = 1 \). Substituting \( x = 0 \) into the equation \( f(x) = 2x + c \), we get \( f(0) = 2(0) + c = 1 \). Thus, \( c = 1 \).

Therefore, the function is \( f(x) = 2x + 1 \).

Because \( f(x) \) is a linear function, it is continuous over \(\mathbb{R}\) (the set of all real numbers). Linear functions are continuous at every point in their domain, and the domain here is \(\mathbb{R}\).

Therefore, \( f(x) \) is continuous on \(\mathbb{R}\).