Question

Let $\alpha$ be a non-zero real number. Suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function such that $f(0) = 2$ and \[\lim_{x \to \infty} f(x) = 1.\]If $f'(x) = \alpha f(x) + 3$, for all $x \in \mathbb{R}$, then $f(-\log 2)$ is equal to ________ .

• A. 3
• B. 5
• C. 9
• D. 7

Answer 1

The Correct Option is C

To determine the value of \(f(-\log 2)\), the provided differential equation must be solved using the given boundary conditions.

The differential equation is:

\(f'(x) = \alpha f(x) + 3\)

This first-order linear differential equation can be solved using an integrating factor.

The standard form for a first-order linear differential equation is:

\(y' + P(x) y = Q(x)\)

By comparison, we identify:

\(P(x) = -\alpha, \quad Q(x) = 3\)

The integrating factor is calculated as:

\(\mu(x) = e^{\int P(x) \, dx} = e^{-\alpha x}\)

Multiplying the differential equation by the integrating factor yields:

\(e^{-\alpha x} f'(x) + \alpha e^{-\alpha x} f(x) = 3 e^{-\alpha x}\)

The left-hand side simplifies to the derivative of a product:

\(\frac{d}{dx} \left( e^{-\alpha x} f(x) \right) = 3 e^{-\alpha x}\)

Integrating both sides with respect to \(x\) results in:

\(e^{-\alpha x} f(x) = \int 3 e^{-\alpha x} \, dx = \frac{-3}{\alpha} e^{-\alpha x} + C\)

The general solution is therefore:

\(f(x) = \frac{-3}{\alpha} + Ce^{\alpha x}\)

Using the initial condition \(f(0) = 2\):

\(2 = \frac{-3}{\alpha} + C\)

Solving for \(C\) gives:

\(C = 2 + \frac{3}{\alpha}\)

Substituting \(C\) back into the general solution:

\(f(x) = \frac{-3}{\alpha} + \left(2 + \frac{3}{\alpha}\right) e^{\alpha x}\)

Given the condition \(\lim_{x \to \infty} f(x) = 1\):

\(\frac{-3}{\alpha} = 1\), which implies \(\alpha = -3\)

Substituting \(\alpha = -3\) into the solution:

\(f(x) = 1 + \left(2 - 1\right) e^{-3x} = 1 + e^{-3x}\)

Now, we compute \(f(-\log 2)\):

\(f(-\log 2) = 1 + e^{-3(-\log 2)} = 1 + e^{3\log 2}\)

Since \(e^{3\log 2} = 2^3\):

\(e^{3\log 2} = 8\)

Therefore, \(f(-\log 2) = 1 + 8 = 9\)

The value of \(f(-\log 2)\) is 9.