Let \alpha be a non-zero real number. Suppose f: \mathbf{R} \rightarrow \mathbf{R} is a differentiable function such that f(0)=2 and \lim\limits_{x \rightarrow-\infty} f(x)=1. If f^{\prime}(x)=\alpha f(x)+3, for all x \in \mathbf{R}, then f\left(-\log _{\mathrm{e}} 2\right) is equal to :

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Answer: B. 9 \begin{aligned} & f(0)=2, \lim _{x ightarrow-\infty} f(x)=1 \\ & f^{\prime}(x)-x \cdot f(x)=3 \\ & 	ext { I.F }=e^{-\alpha x} \\ & y\left(e^{-\alpha x}ight)=\int 3 \cdot e^{-\alpha x} d x \\ & f(x) \cdot\left(e^{-\alpha x}ight)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\ & x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\ & f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\ & x ightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\ & \alpha=-3 	herefore c=1 \\ & f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\ & =1+e^{3 \ln 2}=9 \end{aligned}

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