QJEEM24Ma9e17: JEE Main 2024 Mathematics - Class 12 JEEM24Ma9e17 Solution: Let be a non-zero real number. Suppose is a differentiable…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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

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