JEE Main 2024 — 1 February, Shift 2 — Question 7

$f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1$

$f^{\prime}(x)-\alpha . f(x)=3$

I.F $=\mathrm{e}^{-\alpha x}$

$y\left(e^{-\alpha x}\right)=\int 3 . e^{-\alpha x} d x$

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

$\mathrm{x}=0 \Rightarrow 2=\frac{-3}{\alpha}+\mathrm{c} \Rightarrow \frac{3}{\alpha}=\mathrm{c}-2$

$f(x)=\frac{-3}{\alpha}+c . e^{\alpha x}$

Case-I $\alpha>0$

$\mathrm{x} \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+\mathrm{c}(0)$

$\alpha=-3 \quad$ (rejected)

Case-II $\alpha<0$ as

$\lim _{\mathrm{x} \rightarrow-\infty} \mathrm{f}(\mathrm{x})=1 \Rightarrow \mathrm{c}=0$

and $\frac{-3}{\alpha}=1 \Rightarrow \alpha=-3$

$\Rightarrow \mathrm{f}(\mathrm{x})=1 \quad$

$f\left(-\log _{\mathrm{e}} 2\right)=1$