1. We need to solve the differential equation $f'(x)=\alpha f(x)+3,$ with $f(0)=2, \qquad \lim_{x\to -\infty} f(x)=1,$ where $\alpha\neq 0$ is real.

2. Rewrite the differential equation in standard linear form: $f'(x)-\alpha f(x)=3.$

3. First find the general solution.

   The homogeneous equation is $f_h'(x)-\alpha f_h(x)=0,$ whose solution is $f_h(x)=Ce^{\alpha x}.$

   For a particular solution, try a constant $f_p(x)=k$. Then $f_p'(x)=0$, so $0=\alpha k+3 \implies k=-\frac{3}{\alpha}.$

   Hence the general solution is $f(x)=Ce^{\alpha x}-\frac{3}{\alpha}.$

4. Use the condition $\lim_{x\to -\infty} f(x)=1.$

   Now, $f(x)=Ce^{\alpha x}-\frac{3}{\alpha}.$

   For the limit as $x\to -\infty$ to be finite, we must have $\alpha>0$; otherwise $e^{\alpha x}$ would blow up as $x\to -\infty$ unless $C=0$, which would conflict with the initial condition later.

   So for $\alpha>0$, $e^{\alpha x}\to 0 \quad \text{as } x\to -\infty.$ Therefore, $\lim_{x\to -\infty} f(x)=-\frac{3}{\alpha}=1.$

   Hence, $\alpha=-3.$

5. But this seems to contradict the requirement $\alpha>0$. So let us check more carefully by using the limit condition directly in the differential equation.

   If $f(x)\to 1$ as $x\to -\infty$, then intuitively $f'(x)\to 0$ there, and from $f'(x)=\alpha f(x)+3,$ we get $0=\alpha(1)+3,$ so $\alpha=-3.$

   Now solve with $\alpha=-3$: $f'(x)=-3f(x)+3.$ Rewrite as $f'(x)+3f(x)=3.$

6. Solve this equation.

   The homogeneous solution is $f_h(x)=Ce^{-3x}.$

   A constant particular solution $f_p=k$ gives $0+3k=3 \implies k=1.$

   Thus, $f(x)=Ce^{-3x}+1.$

7. Use $f(0)=2$: $2=C+1 \implies C=1.$ Therefore, $f(x)=e^{-3x}+1.$

8. Now compute $f(-\ln 2)$: $f(-\ln 2)=e^{-3(-\ln 2)}+1=e^{3\ln 2}+1=2^3+1=8+1=9.$

9. Check options:

   • A: $7$
   • B: $9$ ✅
   • C: $3$
   • D: $5$

So the correct answer is B.