Question

If \( g(x) \) is an antiderivative of \( f(x) = 1 + 2 \log 2 \) and the graph of \( y = g(x) \) passes through the point \( \left( -1, -\frac{1}{2} \right) \), then the curve meets the Y-axis at

• A. \( (0, 1) \)
• B. \( (0, 2) \)
• C. \( (0, -2) \)
• D. \( (1, 1) \)

Hint:

When solving problems involving antiderivatives, use the given conditions to determine the constant of integration.

Answer 1

The Correct Option is B

Given that \( g(x) \) is the antiderivative of \( f(x) \), we can write: \[ g(x) = \int (1 + 2 \log 2) \, dx \] Since \( f(x) = 1 + 2 \log 2 \) is a constant function, we can integrate it as: \[ g(x) = (1 + 2 \log 2) x + C \] Using the condition that the graph passes through the point \( \left( -1, -\frac{1}{2} \right) \), we substitute \( x = -1 \) and \( g(x) = -\frac{1}{2} \) to find \( C \): \[ -\frac{1}{2} = (1 + 2 \log 2)(-1) + C \] Solving for \( C \), we get: \[ C = \frac{1}{2} + (1 + 2 \log 2) \] Now, to find where the curve meets the Y-axis, we set \( x = 0 \): \[ g(0) = (1 + 2 \log 2)(0) + C = C \] Substituting the value of \( C \), we get: \[ g(0) = \frac{1}{2} + (1 + 2 \log 2) \] Hence, the curve meets the Y-axis at \( (0, 2) \).