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 find the value of \(f(-\log 2)\), we need to solve the given differential equation and use the provided boundary conditions.

The differential equation given is:

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

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

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

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

Comparing, we get:

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

The integrating factor is given by:

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

Multiply the entire differential equation by the integrating factor:

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

The left-hand side can be rewritten as 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 \(x\) gives:

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

Therefore, the general solution is:

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

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

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

Thus, \(C = 2 + \frac{3}{\alpha}\)

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

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

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

\(\frac{-3}{\alpha} = 1\), leading to \(\alpha = -3\)

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

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

Now, calculate \(f(-\log 2)\):

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

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

Thus, \(f(-\log 2) = 1 + 8 = 9\)

Therefore, the value of \(f(-\log 2)\) is 9.

Answer 2

Given the differential equation:

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

This is a first-order linear differential equation. We can solve it using an integrating factor (IF).

The integrating factor is given by:

\[ IF = e^{\int \alpha dx} = e^{\alpha x} \]

Multiplying the differential equation by the integrating factor:

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

This simplifies to:

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

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

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

Thus, the general solution is:

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

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

\[ 2 = \frac{3}{\alpha} + C \] \[ C = 2 - \frac{3}{\alpha} \]

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

\[ \lim_{x \to \infty} \left(\frac{3}{\alpha} + \left(2 - \frac{3}{\alpha}\right)e^{-\alpha x}\right) = 1 \]

Since \( e^{-\alpha x} \to 0 \) as \( x \to \infty \), we have:

\[ \frac{3}{\alpha} = 1 \implies \alpha = 3 \]

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

\[ f(x) = 1 + (2 - 1)e^{-3x} = 1 + e^{-3x} \]

Now, we need to find \( f(-\log_2 2) \):

\[ f(-\log_2 2) = 1 + e^{-3(-\log_2 2)} = 1 + e^{3\log_2 2} = 1 + (2^3) = 1 + 8 = 9 \]

Conclusion: \( f(-\log_2 2) = 9 \).