Let α be a non-zero real number. Suppose f: R → R is a...

Given: $f\left(0\right) = 1, \lim_{x \to - \infty}f\left(x\right) = 1$ 

And ${f}^{'}\left(x\right) - \alpha \cdot f\left(x\right) = 3$

$⇒ \dfrac{dy}{dx} - \alpha \cdot y = 3$

Which is a linear differential equation, so

$⇒ I . F . = {e}^{- \alphax}$

Solution of the differential equation is given by,

$y\left({e}^{- \alphax}\right) = ∫ 3 \cdot {e}^{- \alphax}dx$

$⇒ f\left(x\right) ⋅ \left({e}^{- \alphax}\right) = \dfrac{3{e}^{- \alphax}}{- \alpha} + c$

$⇒ f\left(x\right) = \dfrac{- 3}{\alpha} + c \cdot {e}^{\alphax}$

Now, taking case I: when $\alpha > 0$ and $\lim_{x \to - \infty}f\left(x\right) = 1$ we get,

$⇒ 1 = \dfrac{- 3}{\alpha} + c\left(0\right)$

$\alpha = - 3$ (rejected)

Case-II: $\alpha < 0$

as $\lim_{x \to - \infty}f\left(x\right) = 1$

 $⇒ c = 0 \text{\text{ and } } \dfrac{- 3}{\alpha} = 1 \left\{\text{as}{e}^{- \left(- \infty\right)} \to \infty , \text{so}c = 0\right\}$

$⇒ \alpha = - 3$

$⇒ f\left(x\right) = 1$ and also $f\left(0\right) = 1$

Hence, $f\left(x\right) = 1$ is constant function, so $f\left(- \log_{e}2\right) = 1$