Medium MCQ +4 / -1 PYQ · JEE Mains 2024

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 :

  1. A 7
  2. B 9 Correct answer
  3. C 3
  4. D 5

Solution

$$ \begin{aligned} & f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 \\\\ & f^{\prime}(x)-x \cdot f(x)=3 \\\\ & \text { I.F }=e^{-\alpha x} \\\\ & y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x \\\\ & f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\\\ & x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\\\ & f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\ & x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\\\ & \alpha=-3 \therefore c=1 \\\\ & f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\ & =1+e^{3 \ln 2}=9 \end{aligned} $$

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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