JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 14)
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 :
7
9
3
5
Explanation
$$
\begin{aligned}
& f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 \\\\
& f^{\prime}(x)-x \cdot f(x)=3 \\\\
& \text { I.F }=e^{-\alpha x} \\\\
& y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x \\\\
& f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\\\
& x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\\\
& f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\
& x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\\\
& \alpha=-3 \therefore c=1 \\\\
& f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\
& =1+e^{3 \ln 2}=9
\end{aligned}
$$
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