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

Answer: (C)

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